Algebraic Inequalities.pdf

Problem Books in Mathematics

Hayk Sedrakyan · Nairi Sedrakyan

Algebraic Inequalities

Problem Books in Mathematics Series Editor: Peter Winkler Department of Mathematics Dartmouth College Hanover, NH 03755 USA

More information about this series at http://www.springer.com/series/714

Hayk Sedrakyan Nairi Sedrakyan •

Algebraic Inequalities

123

Hayk Sedrakyan University Pierre and Marie Curie Paris, France

Nairi Sedrakyan Yerevan, Armenia

ISSN 0941-3502 ISSN 2197-8506 (electronic) Problem Books in Mathematics ISBN 978-3-319-77835-8 ISBN 978-3-319-77836-5 (eBook) https://doi.org/10.1007/978-3-319-77836-5 Library of Congress Control Number: 2018934928 Mathematics Subject Classification (2010): 97U40, 00A07, 26D05 © Springer International Publishing AG, part of Springer Nature 2018 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Printed on acid-free paper This Springer imprint is published by the registered company Springer International Publishing AG part of Springer Nature The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To Margarita, a wonderful wife and a loving mother To Ani, a wonderful daughter and a loving sister

Preface

In mathematics one often deals with inequalities. This book is designed to teach the reader new and classical techniques for proving algebraic inequalities. Moreover, each chapter of the book provides a technique for proving a certain type of inequality. The book includes techniques of using the relationship between the arithmetic, geometric, harmonic, and quadratic means, the principle of mathematical induction, the change of variable(s) method, techniques using the Cauchy–Bunyakovsky– Schwarz inequality, Jensen’s inequality, and Chebyshev’s properties of functions, among others. The main idea behind of the proof techniques discussed in this book is making the complicated simple, so that even a beginner can understand complicated inequalities, their proofs and applications. This approach makes it possible not only to prove a large variety of inequalities, but also to solve problems related to inequalities. To explain each technique of proof, we provide examples and problems with complete proofs or hints. At the end of each chapter there are problems for independent study. In Chapter 14 (Miscellaneous Inequalities) are included inequalities whose proofs employ various techniques not covered in the preceding chapters. In some cases, the proofs of Chapter 14 use several proof techniques from the preceding chapters simultaneously. One hundred selected inequalities and their hints are also provided in the end of Chapter 14, and interested readers are encouraged to choose and provide any methods of proofs they prefer. In each chapter we have tried to include inequalities belonging to the same topic and to present them in order of increasing difficulty, using principles similar to those in [11]. This allows the reader to try to prove these inequalities step by step and to refer to the provided proofs only when difficulties arise. We recommend to use the proofs provided in the book, paying more attention to the choice of the mathematical proof technique. Most of the inequalities in this book were created by the authors. Nevertheless, some of the inequalities were proposed in different mathematical olympiads in different countries or have been published elsewhere (including author-created inequalities). However, the provided solutions are different from the original ones. Most such inequalities are included in the books [2, 5, 6, 7, 8, 9], and since the vii

viii

Preface

name of the author of individual inequalities is unknown to us, we cite these books as the main references. However, for well-known inequalities we have tried to provide the name of the authors. This book was published in Seoul in Korean [13, 14] and is based on [16], which was later published in Moscow in Russian [15]. The historical origins provided at the beginning of some chapters are mostly based on [10] or our personal knowledge. It was considered appropriate to give the proofs of each chapter at the end of the same chapter. Paris, France Yerevan, Armenia

Hayk Sedrakyan Nairi Sedrakyan

Contents

1

Basic Inequalities and Their Applications . . . . . . . . . . . . . . . . . . . .

1

2

Sturm’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

3

The HM-GM-AM-QM Inequalities . . . . . . . . . . . . . . . . . . . . . . . . .

21

4

The Cauchy–Bunyakovsky–Schwarz Inequality . . . . . . . . . . . . . . .

45

5

Change of Variables Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

59

6

Using Symmetry and Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . .

71

7

The Principle of Mathematical Induction . . . . . . . . . . . . . . . . . . . .

81

8

A Useful Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

9

Using Derivatives and Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10 Using Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 11 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 12 Inequalities of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13 Algebraic Inequalities in Number Theory . . . . . . . . . . . . . . . . . . . . 187 14 Miscellaneous Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Appendix—Power Sums Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

ix

About the Authors

Hayk Sedrakyan is an IMO medal winner, a professor of mathematics in Paris, France, and a professional math olympiad coach in the greater Boston area, Massachusetts, USA. He received his doctorate in mathematics at the Université Pierre et Marie Curie, Paris, France. Hayk is a Doctor of Mathematical Sciences in USA, France, Armenia. He has been awarded master’s degrees in mathematics from Germany, Austria, and Armenia and completed part of his doctoral studies in Italy. Hayk has authored several books on the topic of problem-solving and olympiad-style mathematics published in USA and South Korea. Nairi Sedrakyan has long been involved in national and international mathematical olympiads, having served as an International Mathematical Olympiad (IMO) problem selection committee member and the president of Armenian Mathematics Olympiads. He is the author of one of the hardest problems ever proposed in the history of International Mathematical Olympiads (the fifth problem of the 37th IMO). He has been the leader of the Armenian IMO Team, jury member of the IMO, jury member and problem selection committee member of the Zhautykov International Mathematical Olympiad (ZIMO), jury member and problem selection committee member of the International Olympiad of Metropolises, and the president of the International Mathematical Olympiad Tournament of the Towns in Armenia. He is also the author of a large number of problems proposed in these olympiads and has authored several books on the topic of problem-solving and olympiad-style mathematics published in United States, Russia, Armenia, and South Korea. The students of Nairi Sedrakyan have obtained 20 medals (1 gold medal, 4 silver medals, 15 bronze medals) in IMO. For his outstanding teaching, Nairi Sedrakyan received the title of best teacher of the Republic of Armenia and was awarded special recognition by the prime minister.

xi

Overview

This book is designed to teach the reader new and classical mathematical proof techniques for proving inequalities, in particular, to prove algebraic inequalities. These proof techniques and methods are applied to prove inequalities of various types. The main idea behind this book and the proof techniques discussed is making the complicated simple, so that even a beginner can understand complicated inequalities, their proofs and applications. The book Algebraic Inequalities is also devoted to the topic of inequalities and can be considered a continuation of the book Geometric Inequalities: Methods of proving [12]. It can serve teachers, high-school students, and mathematical competitors.

xiii

Chapter 1

Basic Inequalities and Their Applications

Historical origins. According to [3], mathematical inequalities first were expressed verbally. Later, the following symbols were introduced: Less than and greater than: The mathematical symbols < and > first appear in the book Artis Analyticae Praxis ad Aequationes Algebraicas Resolvendas (The Analytical Arts Applied to Solving Algebraic Equations) posthumously published in 1631 and written by the English astronomer and mathematician Thomas Harriot (c. 1560–1621), who was born in Oxford, England, and died in London, England. Harriot initially used triangular symbols, but the editor modified them slightly, so that they resemble the modern less than and greater than symbols. In his book is stated the following: “The mark of the majority (signum majoritatis) as a > b, signifies a greater than b and the mark of the minority (signum minoritatis) to a < b signifies a lesser than b.” Less than or equal to and greater than or equal to: The double-bar style of the less than or equal to < and greater than or equal to > symbols was first employed in 1734 by the French mathematician, geophysicist, and astronomer Pierre Bouger (1698–1758) born in Le Croisic, France, and died in Paris, France. In 1670, a similar reduced single-bar notation was employed by the English mathematician John Wallis. Wallis used a single horizontal bar above instead of below the inequality symbols, leading to < and >. Wallis also introduced the symbol for infinity ∞. His academic advisor was the English mathematician William Oughtred (1574–1660), who introduced the multiplication symbol × and the abbreviations sin and cos for the sine and cosine. Oughtred was born in Eton, England, and died in Albury, England. Not equal to, not greater than, not less than: These symbols were employed by the Swiss mathematician, physicist, and astronomer Leonhard Euler (1707–1783), who was born in Basel, Switzerland, and died in Saint Petersburg, Russian Empire (now Russia). Euler introduced many modern mathematical notations, for example the notation for a mathematical function. He is considered one of the greatest mathematicians of all time. His doctoral advisor was the prominent Swiss mathematician Johann Bernoulli (1667–1748), who was born and died in Basel, Switzerland. Euler © Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_1

1

2

1 Basic Inequalities and Their Applications

was the doctoral advisor of six students, including the well-known Italian mathematician Joseph Louis Lagrange (1736–1813). He was born Giuseppe Lodovico Lagrangia in Turin, Kingdom of Sardinia (now Turin, Italy) and died in Paris, France, and his son Johann Euler (1734–1800) born and died in Saint Petersburg, Russia. In this chapter, basic but very useful inequalities are presented. We recommend that the reader pay special attention to these inequalities, which will lay a foundation for more challenging inequalities in the forthcoming sections. The main idea behind the proof techniques and methods presented here is making the complex simple, so that even a beginner can understand complex inequalities, their proofs and applications. A background in high-school-level algebra is sufficient for solving the inequalities presented below.

Problems Prove the following inequalities (1.1–1.26). 1.1. a 2 + b2 √ ≥ 2ab. 1.2. a +2 b ≥ ab, where a ≥ 0, b ≥ 0. √ ab ≥ 1 +2 1 , where a > 0, b > 0. 1.3. a b  a 2 + b2 a+b 1.4. ≥ . 2 2 1.5. 1.6. 1.7. 1.8. 1.9.

2 , where a > 0, b > 0. + b1   a 2 + b2 a+b 2 ≥ 2 . 2 a + b > 1 + ab, where b < 1 < a. a 2 + b2 > c2 + (a + b − c)2 , where b < c < a. (a) 2 ≤ ab + ab , where ab > 0, (b) ab + ab ≤ −2, where ab < 0. a+b 2

≥

1.10. x1 ≤ 1.11.

x1 y1

≤

1 a

x1 + ··· +xn n x1 + ··· +xn y1 + ··· +yn

≤ xn , where x1 ≤ · · · ≤ xn . ≤

xn , yn

where

x1 y1

≤ ··· ≤

xn yn

and yi > 0, i  1, . . . , n.

1 n

1.12. x1 ≤ (x1 · · · xn ) ≤ xn , where n ≥ 2, 0 ≤ x1 ≤ · · · ≤ xn . 1.13. |a1 | + · · · + |an | ≥ |a1 + a2 + · · · + an |. 1.14. a1 + ···n +an ≥ 1 + ···n + 1 , where n ≥ 2, ai > 0, i  1, . . . , n. an  a1 √ √ a+b 1.15. (a + b) 2 ≥ a b + b a, where a > 0, b > 0.  1.16. 21 (a + b) + 41 ≥ a +2 b , where a > 0, b > 0. 1.17. a(x + y − a) ≥ x y, where x ≤ a ≤ y. 1.18. x −1 1 + x +1 1 > 2x , where x > 1. 1.19. 1.20.

1 + 3k1+ 2 + 3k1+ 3 > 2k1+ 1 + 2k1+ 2 , where 3k + 1 a)(1 − b) ab ≤ ((1(1−−a)+(1 , where 0 < a ≤ 21 , − b))2 (a+b)2

k ∈ N. 0 < b ≤ 21 .

Problems

3

2k + 1 1.21. √3k1 + 1 · 2k < √3k1 + 4 , where k ∈ N. +2 1.22. 2n−1 ≥ n, where n ∈ N. 1.23. 13 + 23 · 15 + 23 · 45 · 17 + · · · + 23 · 45 · 67 · · · 100 · 1 < 1. 101 103 1−a 1−b a b 1.24. (a) 1 − b + 1 − a ≤ b + a , where 0 < a, b ≤ 21 , n n n n   1  1  (b) (1 − ai ) ≤ ai , where 0 < a1 , . . . , an ≤ 21 . 1 − ai ai i1

i1 i1 1 5 < , where n 3 n 4 a+b ab + , where 2 3

1.25. 1 + 213 + · · · + 1.26. 1 + a1 + b ≤ 1 −

i1

∈ N. 0 ≤ a ≤ 1, 0 ≤ b ≤ 1.

Proofs 1.1. We have a 2 + b2 − 2ab  (a − b)2 ≥ 0. Note that equality holds if and only if a  b. √ 1.2. The proof is similar to the proof of Problem 1.1. We have a + b − 2 ab  √ √ 2 a − b ≥ 0. Note that equality holds if and only if a  b. √

1.3. Multiplying both sides of the inequality of Problem 1.2 by 2a +ab , we obtain the b required inequality. Note that equality holds if and only if a  b. 1.4. We have 2ab ≤ a 2 + b2 (see Problem 1.1). We obtain the equivalent inequality 2 2 2 a 2 + b2 + 2ab ≤ 2a 2 + 2b2 . Thus, it follows that (a +4b) ≤ a +2 b . The last   a+b  ≤ a 2 + b2 . Therefore, a 2 + b2 ≥ a + b . inequality can be rewritten as  2

1.5. 1.6. 1.7. 1.8.

2

2

2

Note that equality holds if and only if a  b. This inequality follows from the inequalities of Problems 1.2 and 1.3. Note that equality holds if and only if a  b. See the proof of Problem 1.4. Note that equality holds if and only if a  b. Using that a + b − 1 − ab  (a − 1)(1 − b) and the condition b < 1 < a, it follows that (a − 1)(1 − b) > 0. Let us evaluate the difference between the left-hand side and the right-hand side of the given inequality:     a 2 + b2 − c2 − (a + b − c)2  a 2 − c2 − (a + b − c)2 − b2   (a − c)(a + c) − (a − c)(a + 2b − c)  2(a − c)(c − b) > 0,

since according to the assumption, we have b < c < a. 1.9. (a) Multiplying both sides of the inequality by ab, where ab > 0, we deduce that a 2 + b2 ≥ 2ab. (b) Dividing both sides of the inequality a 2 +b2 ≥ −2ab by ab, where ab < 0, we obtain the required inequality. 1.10. We have x1 ≤ x1 , . . . , xn ≤ xn , whence nx1 ≤ x1 + · · · + xn ≤ nxn . Thus, it follows that x1 ≤ x1 + ···n +xn ≤ xn .

4

1 Basic Inequalities and Their Applications

1.11. From the assumptions of the problem, it follows that yi xy11 ≤ xi ≤ yi xynn , i  1, . . . , n. Summing up these inequalities, we deduce that x1 xn (y1 + · · · + yn ) ≤ x1 + · · · + xn ≤ (y1 + · · · + yn ). y1 yn +xn Hence, it follows that xy11 ≤ xy11 ++ ··· ≤ xynn . ··· +yn 1.12. We have that x1 ≤ xi ≤ xn , i  1, . . . , n. Multiplying these inequalities, we 1 obtain x1n ≤ x1 · · · xn ≤ xnn . Thus, it follows that x1 ≤ (x1 · · · xn ) n ≤ xn . 1.13. If a1 + · · · + an ≥ 0, then |a1 + · · · + an |  a1 + · · · + an . Using a ≤ |a|, we deduce that

|a1 + · · · + an |  a1 + · · · + an ≤ |a 1 | + · · · + |an |. If a1 + · · · + an < 0, then |a1 + · · · + an |  −a1 − · · · − an . Using the inequality −a ≤ |a|, it follows that |a1 + · · · + an |  −a1 − · · · − an ≤ |a 1 | + · · · + |an |. 1.14. We have that ⎛ ⎞

⎜



1 an−1 1 an ⎟ ⎜ a1 a2 ⎟ ⎜ + ··· + + ··· + + + (a1 + · · · + an ) ⎟ + n, ⎝ a2 a1 a1 an an an−1 ⎠    n(n−1)/2

and using Problem 1.9 (a), it follows that

1 n(n − 1) 1 ≥n+2·  n2. + ··· + (a1 + · · · + an ) a1 an 2 

√ ≥ 2 ab · √ √ √ a+ b , which can be obtained by multiplying the inequalities a + b ≥ 2 ab 2  √ √ (Problem 1.2) and a +2 b ≥ a +2 b (see Problem 1.6). 1.16. We have 1.15. The given inequality is equivalent to the inequality (a + b)

1 1 (a + b) + − 2 4



a+b  2



a+b 1 − 2 2

a+b 2

2 ≥ 0.

 Therefore, 21 (a + b) + 41 ≥ a +2 b . 1.17. Since a(x + y − a) − x y  ax − x y + a(y − a)  (y − a)(a − x) and y ≥ a ≥ x, it follows that (y − a)(a − x) ≥ 0. Therefore, a(x + y − a) ≥ x y. 1.18. Using the inequality of Problem 1.5, it follows that

Proofs

5 1 x −1

+ 2

1 x +1

>

2 1 2 1 + > . , or x −1 x +1 x (x − 1) + (x + 1)

1.19. According to the inequality of Problem 1.18, we have that 1 + 3k1+ 3  (3k +12)−1 + (3k +12)+1 > 3k2+ 2 . Therefore, 3k1+ 1 + 3k + 1 3 . 3k + 2 Now let us prove that 3k3+ 2 > 2k1+ 1 + 2k1+ 2 . Indeed, 2k1+ 1 + 2k1+ 2 − 3k3+ 2  (2k + 1)(2k−k+ 2)(3k + 2) < 0.

1 3k + 2

+

1 3k + 3

>

1.20. The given inequality is equivalent to the following inequality:

2 −1 a +b



2 ≤



1 1 −1 −1 . a b

We have that

1 −1 a







2 1 2 1 1 4 1 4 −1 − −1  − − −  + b a +b ab a b (a + b)2 a + b

4 a +b 4 1 (a − b)2 (a − b)2 (1 − (a + b)) (a − b)2 − −   + − 2 2 ab a +b ab ab(a + b) ab(a + b) ab(a + b)2 (a + b)

b) (1−(a + b)) and 0 < a ≤ 21 , 0 < b ≤ 21 . Then (a − ab(a ≥ 0, and therefore, + b)2  2 2  1  1  −1 ≤ a −1 b −1 . a+b √ 1.21. The given 3k + 4 < √ inequality is equivalent to the inequality (2k + 1) (2k + 2) 3k + 1 or to the following inequality: (2k + 1)2 (3k + 4) < (2k + 2)2 (3k + 1). The last inequality holds because 2

(2k + 2)2 (3k + 1) − (2k + 1)2 (3k + 4)  k > 0. 1.22. Since 1 < 2 < 22 < · · · < 2n−1 and the number of positive integers 1, 2, 22 , . . . , 2n−1 is equal to n, it follows that 2n−1 ≥ n. 1.23. Consider a unit line segment and suppose on the first day, we paint 13 of thegiven 1 segment, the second day 15 of the rest of the segment, on the 51st day, 103 of the rest of the segment. Since every day there remains a part of the given segment, the sum of the painted parts must be less than 1. The first day we have painted 13 of the given segment, on the second day 23 · 15 , on the 51st day 23 · 45 · · · 100 · 1 . Hence, we deduce that 101 103 1 2 1 2 4 1 2 4 98 100 1 + · + · · + ··· + · ··· · · < 1. 3 3 5 3 5 7 3 5 99 101 103 1.24 (a) According to the inequality of Problem 1.20, we have that

6

1 Basic Inequalities and Their Applications

(a + b)2 1 − a 1 − b a b ((1 − a) + (1 − b))2 ≤ or + ≤ + . (1 − a)(1 − b) ab 1−b 1−a b a An alternative proof. Note that 1 − a ≥ a and 1 − b ≥ b, and therefore, 1−a 1−b (1 − a)2 + (1 − b)2 ((1 − a) − (1 − b))2 + 2(1 − a)(1 − b) +    1−b 1−a (1 − a)(1 − b) (1 − a)(1 − b) 

(a − b)2 (a − b)2 a b +2≤ +2 + . (1 − a)(1 − b) ab b a

a −b Hence, we deduce that 11 − + 11 − ≤ ab + ab . −b a n n  1  (b) Since (1 − ai )  1 − ai i1 i1



1 − an−1 1 − a1 1 − a2 1 − an + + +n, using the + ··· + 1 − a2 1 − a1 1 − an 1 − an−1    n(n−1)/2

inequality of Problem 1.24 (a), we obtain that



n n n   an−1 1 1  a1 a2 an + ··· + (1 − ai ) ≤ + + +n  ai . 1 − ai a2 a1 an an−1 a i1 i1 i1 i i1   

n 

n(n−1)/2

1.25. Note that if n ≥ 4, then 1+

1 2−1 n − (n − 1) 1 + ··· + 3  1 + + ··· + 3 3 23 n 2

n

2 n−1 1 5 1 1 5 − − ··· −  − − − < , 3 2 3 2 3 4 2 3 3 4 n 4

since (k +k 1)3 > (k +12)2 , where k ∈ N. 1.26. We have that (1 − a)(1 − b) ≥ 0, and therefore a + b − 1 ≤ ab. Then

a+b a+b 1 1 − 1−  (a + b − 1) ≤ ab. 1+a+b 2 2(1 + a + b) 3

Problems for Independent Study Prove the following inequalities (1–32). 1. |x − y| < |1 − x y|, where |x| < 1, |y| < 1. sin x−1 x 2. sin + 21 ≥ 2−sin . x−2 3−sin x a b c 2 3. bc + ca + ab ≥ a + b2 − 2c , where a > 0, b > 0, c > 0. 1 4. a1 + b1 − 1c < abc , where a 2 + b2 + c2  53 and a > 0, b > 0, c > 0.    2 5. 3 1 + a 2 + a 4 ≥ 1 + a + a 2 .

Problems for Independent Study

7

6. (ac + bd)2 + (ad − bc)2 ≥ 144, where a + b  4 , c + d  6 . √ 2 + na 2√≥ a 2(x1 + x2 + · · · + x2n ). 7. x12 + x22 + · · · + x2n 8.

1 + b +1 c + a 1+ c ≤ a +b   3 2 2 3 2

√ √ a+ b+ c √ , where a > 2  abc   2 3 2 2

0, b > 0, c > 0.

9. a b − c + b c − a + c a − b < 0, where 0 < a < b < c. 10. a 3 b + b3 c + c3 a ≥ a 2 b2 + b2 c2 + a 2 c2 , where a ≥ b ≥ c > 0. 2 11. xy + yz + x +y z ≤ (x x+zz) , where 0 < x ≤ y ≤ z.      √ √ √ 12. 1 + a + 1 + a + a 2 + · · · + 1 + a + · · · + a n < na, where n ≥ 2, a ≥ 2, n ∈ N. 1)2 , where tan α  n tan β , n > 0 , 13. (a) tan2 (α − β) ≤ (n − 4n (b) 1 + cos(α − β) ≥ cos α + cos β, where 0 ≤ α ≤ π2 , 0 ≤ β ≤ π2 . + [3x] + [4x] + [5x] , 14. [5x] ≥ [x] + [2x] 2 3 4 5 where [a] is the integer part of the real number a. 15. (n!)2 ≥ n n , where n ∈ N. 16. x 6 + x 5 + 4x 4 − 12x 3 + 4x 2 + x + 1 ≥ 0. 17. log2 α ≥ log β log γ , where α > 1, β > 1, γ > 1, α 2 ≥ βγ . 18. log4 5 + log5 6 + log6 7 + log7 8 > 4, 4. n < 21 , where n ∈ N. 19. 13 + · · · + 3·5···(2n + 1) 3

3

20. 223 −+ 11 · · · nn3 −+ 11 < 23 , where n ≥ 2, n ∈ N. 21. 1 · 1! + · · · + n ·n! < (n + 1)!  , where n ∈ N. 22. 1 + 212 1 + 312 . . . 1 + n12 < 2, where n ≥ 2, n ∈ N.      23. 1 − p12 1 − p12 . . . 1 − p12 > 21 , where 1 < p1 < p2 < · · · < pn , pi ∈ 1

n

2

N , i  1, . . . , n. 1 1 + 1000 < 25 . 24. 21 − 13 + 41 − 15 + · · · − 999 25. (sin x + 2 cos 2x)(2 sin 2x − cos x) < 4, 5. 26. (a) 1 +a a+ +b b ≤ 1 +a a + 1 +b b , where a ≥ 0, b ≥ 0.   (b) 2 +a a+ +b b ≥ 21 1 +a a + 1 +b b , where a ≥ 0, b ≥ 0. (c) n ≤

a1 + b1 a1 + b1 + 2

+ ... +

an + bn an + bn + 2

+

1 a1

1 a1

+

1 bi 1 1 bi + 2 1

+

+ ... +

1 1 an + bi n 1 1 + an bi n + 2

< 2n, where

i 1 , . . . , i n is some permutation of the numbers 1, . . . , n, ai , bi > 0, i  1, . . . , n. n n   a1 + 2a2 + ··· + iai ≤ 2 ai , where ai ≥ 0, i  1, . . . , n. 27. 2 i i1

i1

41 , where a1 + b1 + 1c < 1 , a, b, c ∈ N . 28. a1 + b1 + 1c ≤ 42 y 4x z 29. y + z + x + z + x + y > 2, where x, y, z > 0. 30. 1 < a + ab + d + a + bb + c + b + cc + d + a + dc + d < 2, where a > 0, b > 0, c > 0, d > 0. 31. a + b > c + d, where a, b, c, d ≥ 21 and a 2 + b > c2 + d, a + b2 > c + d 2 . Hint. If a+b ≤ c+d, then a ≤ c or b ≤ d. If a ≤ c, then b−d > (c−a)(c+a) ≥ c − a. √ 32. (b − a)(9 − a 2 ) + (c − a)(9 − b2 ) + (c − b)(9 − c2 ) ≤ 24 2, where 0 ≤ a ≤ b ≤ c ≤ 3.

8

1 Basic Inequalities and Their Applications

Hint. We have that (b − a)(9 − a 2 ) + (c − a)(9 − b2 ) + (c − b)(9 − c2 ) ≤ 9b + c(9 − b2 ) + (c − b)(9 − c2 )   18c − c3 + bc(c − b) ≤ 18c − c3 +

1 3 3 c  18c − c3 . 4 4

33. If 0 < a, b, c < 1, then one of the numbers (1 − a)b, (1 − b)c, (1 − c)a is not greater than 41 . 34. Let a > 0, b > 0, c > 0, and a + b + c  1. Prove that    (a) a + 41 (b − c)2 + b + 41 (c − a)2 + c + 41 (b − a)2 ≤ 2,  √ √ √ (b) a + 41 (b − c)2 + b + c ≤ 3.  Hint. (a) x + 41 (y − z)2 ≤ x + y+z if x, y, z > 0 and x + y + z  1. 2 35. Find the smallest possible value of the following expression: 4 2 2 a4 + ab4 − ab2 − ab2 + ab + ab , where a > 0, b > 0. b4

Chapter 2

Sturm’s Method

Historical origins. Sturm’s method was proposed by the Prussian mathematician Friedrich Otto Rudolf Sturm, born 6 January 1841 in Breslau, Prussia (now Wrocław, Poland), died 12 April 1919 in Breslau, Germany. Sturm obtained his doctorate in mathematics from the University of Breslau (now University of Wrocław) in 1863 under the supervision of the well-known Prussian mathematician Heinrich Eduard Schröter. Sturm was the advisor of 19 doctoral students, including the wellknown Prussian mathematician Otto Toeplitz, born 1 August 1881 in Breslau, Prussia, died 15 February 1940 in Jerusalem, Mandatory Palestine (now Israel). Toeplitz studied mathematics at the University of Breslau, where under Sturm’s supervision he obtained his doctorate in 1905. Besides its various applications, Sturm’s method provides an opportunity to prove a large number of different inequalities under certain conditions. Example 2.1 Prove that if the product of positive numbers x1 , . . . , xn (n ≥ 2) is equal to 1, then x1 + · · · + xn ≥ n. Proof If x1  · · ·  xn  1, then x1 + · · · + xn  n. Suppose that among the considered numbers there are at least two different numbers. Then among the numbers x1 , . . . , xn there are two numbers such that one of them is greater than 1 and the other one is less than 1 (Problem 1.12). Without loss of generality one can assume that those numbers are x1 and x2 , and that x1 < 1 < x2 . Note that x1 + x2 > 1 + x1 x2 (Problem 1.7). If one substitutes the given numbers by numbers 1, x1 x2 , x3 , . . . , xn , then their product is again equal to 1, and the sum satisfies 1+ x1 x2 + x3 +· · ·+ xn < x1 +· · ·+ xn . Doing the same with the numbers 1, x1 x2 , x3 , . . . , xn , in a similar way we obtain a new sequence such that two numbers in it are equal to 1. Doing the same at most n − 1 times, we obtain a sequence such that n − 1 numbers in it are equal to 1, and the nth number is equal to x1 · · · xn . On the other hand, x1 · · · xn  1.

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_2

9

10

2 Sturm’s Method

Hence, we obtain that n < x1 + x2 + · · · + xn . From the proof, it follows that equality holds if and only if x1  · · ·  xn  1. Example 2.2 Prove that if the sum of the numbers x1 , . . . , xn (n ≥ 2) is equal to 1, then x12 + · · · + xn2 ≥ n1 . Proof If x1  · · ·  xn  n1 , then x12 + · · · + xn2  n1 . Suppose that among the considered numbers there are at least two different numbers. Then among the numbers x1 , . . . , xn there are two numbers such that one of them is greater than n1 and the other one is less than n1 (Problem 1.10). Without loss of generality, one can assume that those numbers are x1 and x2 , and that x1 < n1 and x2 > n1 . Therefore, substituting x1 by n1 , and x2 by x1 + x2 − n1 , we obtain a new sequence of numbers n1 , x1 + x2 − n1 , x3 , . . . , xn such that their sum is again equal to 1. On the other 2  we have that 2  hand, x12 + x22 > n1 + x1 + x2 − n1 (see Problem 1.8), whence x12

+ ··· +

xn2

  2  1 1 2 > + x1 + x2 − + x32 + · · · + xn2 . n n

Repeating these steps a finite number of times, we obtain a sequence such that its all terms are equal to n1 , and the sum of their squares is less than the sum of the  2  2 squares of the numbers x1 , . . . , xn , that is, x12 + · · · + xn2 > n1 + · · · + n1  n1 . From the proof, it follows that equality holds if and only if x1  · · ·  xn  n1 .

Problems Prove the following inequalities (2.1–2.6). 2.1 AM-GM (arithmetic mean–geometric mean inequality): √ x1 + ··· + xn ≥ n x1 · · · xn , where n ≥ 2, xi > 0, i  1, . . . , n. n 2.2 QM-AM (quadratic mean–geometric mean inequality or RMS/root mean square  2.3 2.4 2.5

x12 + ··· + xn2 n

≥ x1 + ···n + xn . (1 − x1 ) ··· (1 − xn ) ≥ (n − 1)n , where n ≥ 2, xi > 0, i  1, . . . , n, and x1 ··· xn x1 + · · · + xn  1. 1 1 + · · · + 1+x ≥ 1+ √n xn1 ... xn , where n ≥ 2, x1 ≥ 1, . . . , xn ≥ 1. 1+x1 n 1 abc + bcd + cda + dab ≤ 27 + 176 abcd, where a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0, 27 inequality):

and a + b + c + d  1. 2.6 0 ≤ x y + yz + zx − 2x yz ≤

7 , 27

where x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z  1.

Problems

11

2.7 Among all triangles with no angle greater than 75º inscribed in a given circle, find the triangle such that its perimeter is (a) the greatest, (b) the smallest. 2.8 (a) Schur’s inequality: Prove that if for some numbers α and β one has the inequality [α f (a) + β f (b) ≤ f (αa + βb)]α f (a) + β f (b) ≥ f (αa + βb), where α ≥ 0, β ≥ 0, α + β  1, a, b are any numbers belonging to D( f )  I,1 and x1 , . . . , xn , y1 , . . . , yn ∈ I such that y1 ≥ . . . ≥ yn ,y1 ≤ x1 , y1 + y2 ≤ x1 + x2 , . . . , y1 + · · · + yn−1 ≤ x1 + · · · + xn−1 , y1 + · · · + yn  x1 + · · · + xn , then f (y1 ) + f (y2 ) +· · · + f (yn ) ≤ f (x1 )+ f (x2 ) +· · · + f (xn ) [ f (y1 ) + f (y2 ) + · · · + f (yn ) ≥ f (x1 ) + f (x2 ) + · · · + f (xn )]. (b) Popoviciu’s inequality: Prove that if for numbers α and β one has the inequality α f (a) + β f (b) ≥ f (αa + βb), where α, β ≥ 0, α + β  1, and a, b are any numbers belonging to the interval I , then for all numbers x, y, z from the interval I , one has the inequality f (x) + f (y) + f (z) + 3 f

x + y + z x + y y +z z + x  ≥2f +2f +2f . 3 2 2 2

2.9 Suppose that for numbers x1 , . . . , x1997 , the following conditions hold: √ (a) − √13 ≤ xi ≤ 3, i  1, . . . , 1997, √ (b) x1 + · · · + x1997  −318 3. 12 . Find the greatest possible value of the expression x112 + · · · + x1997

n−1 2

, where 2.10 Prove that cos α1 cos α2 · · · cos αn (tan α1 + · · · + tan αn ) ≤ (n−1) n−2 n 2 π n ≥ 2 and 0 ≤ αi < 2 , i  1, 2, . . . , n. n 2.11. Prove that xik (1 − xi ) ≤ ak , where k ≥ 2, k ∈ N, and ak  max(x k (1 − [0;1]

i1

x) + (1 − x)k x), xi ≥ 0, i  1, . . . , n, x1 + . . . + xn  1, n ≥ 2. 2.12. Prove the following inequalities: (a) 2(n − 1)(x1 x2 + x1 x3 + · · · + x1 xn + x2 x3 + · · · + x2 xn + · · · + xn−1 xn ) − n n−1 x1 x2 · · · xn ≤ n − 2, where n ≥ 2,x1 ≥ 0, . . . , xn ≥ 0 and x1 + x2 + · · · + xn  1, √ (b) x1 + x2 n+ ··· + xn − n x1 x2 · · · xn ≤ √ 2 √ √ 2 √ √ √ 2 √ 2 √ x − x2 ) +( x1 − x3 ) + ··· + ( x1 − xn ) + ··· + ( xn−1 − xn ) , where n ≥ 2, ≤ ( 1 x1 ≥ 0, . . . , xn ≥ 0.

1f

n

is any real-valued function, D is the domain of function f , D(f ) is denoted by I.

12

2 Sturm’s Method

(c) For n  4, the following inequality is called Turkevici’s inequality:  n 2 2 2 2 2 (n − 1)(x1 + x2 + · · · + xn ) + n x1 x2 · · · xn2 ≥ (x1 + x2 + · · · + xn )2 , where n ≥ 2, x1 ≥ 0, . . . , xn ≥ 0.

Proofs 2.1 Consider the numbers √n x1x1··· xn , . . . , √n x1xn··· xn , and note that their product is equal to 1. Then according to Example 2.1, we have that √n x1x1··· xn + · · · + √n x1xn··· xn ≥ n, or √ x1 + · · · + xn ≥ n x1 · · · xn . n

(2.1)

Note that in (2.1), the equality holds if x1  · · ·  xn . Inequality (2.1) is known as Cauchy’s inequality. 2.2 Consider the numbers x1 + x···1 + xn , . . . , x1 + x···n + xn , and note that their sum is equal to  2 2  1. According to Example 2.2, we have that x1 + x···1 + xn +· · ·+ x1 + x···n + xn ≥ n1 , or x12 + x22 + · · · + xn2  x1 + x2 + · · · + xn 2 ≥ . n n

(2.2)

This ends the proof. Note that in (2.2), equality holds if and only if x1  · · ·  xn . 2.3 Since x1 + · · · + xn  1, it follows by Problem 1.10 that there are two numbers such that one of them is not greater than n1 , and the other one is not less than n1 . Without loss of generality, one can assume that x1 ≤ n1 ≤ x2 . Let us substitute x1 by n1 , x2 by x1 + x2 − n1 . Then we obtain numbers n1 , x1 + x2 − n1 , x3 , . . . , xn such that (1 − n1 )(1 − x1 − x2 + n1 ) ··· (1−xn ) , since (1 − x1 )(1 − x2 ) (1 − x1 ) ··· (1 − xn ) ≥ ≥ 1 1 x1 ··· xn x1 x2 n ( x 1 + x 2 − n ) ··· x n 1 1 (1 − n )(1 − x1 − x2 + n ) (Problem 1.17) and (1 − x1 )(1 − x2 )  1 + 1 − (x1 + x2 ) . 1 1 x1 x2 x1 x2 n (x1 + x2 − n ) Repeating these steps a finite number of times, we obtain n numbers equal to 1 . For these numbers, the left-hand side of the inequality is equal to (n − 1)n , n (1−xn ) . and it is not greater than (1−xx11) ··· ··· xn √ n 2.4 Let us set x1 · · · xn  m. According to Problem 1.12, without loss of generality one can assume that x1 ≤ m ≤ x2 , and therefore, for numbers m, x1mx2 , x3 , . . . , xn , one has 1 1 1 1 1 + + ··· + ≥ , since x1 x2 + · · · + 1 + x1 1 + xn 1+m 1+ m 1 + xn

Proofs

13

1 + 1 +1x2 1 + x1 1 − x1 x2 . 1 + x1 + x2 + x1 x2

≥

1 1+m

+

1 1+

x1 x2 m

(Problem 1.17) and

After a finite number of steps, we deduce that 1 1  √n n . + ··· + 1 + m 1 + m 1 + x1 ··· xn 

1 1 + x1

1 1 + x1

+

1 1 + x2

+ ··· +

 1 + 1 1 + xn

≥

n

2.5 If a  b  c  d  14 , then one has equality. Let a < 41 < b. Consider the following two cases. (a) If c + d − 176 cd < 0, then 27 3    + b) A  ab c + d − 176 cd + cd(a + b) ≤ cd(a + b) ≤ c + d+(a  27 3 using Problem 2.1. cd ≥ 0, then note that (b) If c + d − 176 27       1 176 1 1 1 a+b− c+d − cd + cd + a+b− . A≤ 4 4 27 4 4

1 , 27

Hence, we must prove the inequality for numbers a1  41 , b1  a + b − 41 , c1  c, d1  d. In a similar way, either one can prove the inequality for numbers a1 , b1 , c1 , d1 , or it will be sufficient to prove the inequality for the case that among a1 , b1 , c1 , d1 , two numbers are equal to 41 . Continuing in this way, we obtain that it is sufficient to prove the inequality for numbers 1 1 1 1 , , , . Note that in this case, the inequality obviously holds. 4 4 4 4 2.6 Let x ≥ y ≥ z. Then y ≤ 21 , and therefore, 

0 ≤ y(x + z) + x z(1 − 2y) ≤ y

    1 1 1 1 + x +z− + x+z− (1 − 2y). 3 3 3 3

Therefore, if one substitutes the numbers x, y, z by the numbers 13 , y, x +z− 13 in the expression x y + yz + x z − 2x yz, then its value does not decrease. Continuing in a similar way, one can substitute the numbers 13 , y, x + z − 13 by the numbers 1 1 1 2 7 , , . Hence, we deduce that x y + yz + x z − 2x yz ≤ 19 + 19 + 19 − 27  27 . 3 3 3 2.7 According to the law of sines, we have that p  a + b + c  2R(sin α + sin β + sin γ ). (a) Let us prove that the greatest perimeter is that of an equilateral triangle. If a triangle is not equilateral, then without loss of generality one can assume that α > π3 > β. Now let us prove that

14

2 Sturm’s Method

sin α + sin β + sin(α + β) < sin

 π π + sin(α + β). + sin α + β − 3 3

We have that sin α + sin β − sin

 π α − π/3 α + π/3 π − sin α + β −  2 sin cos − 3 3 2 2 α − π/3 α + 2β − π/3 − 2 sin · cos  2 2   2π/3 − 2β α − π/3 α+β −2 sin   2 sin sin 2 4 2 α − π/3 β − π/3 α+β  4 sin sin sin < 0. 2 2 2

Therefore, the perimeter of a triangle with angles α, β, γ is less than the perimeter of a triangle with angles π3 , α + β − π3 , γ . If γ  π3 , then we obtain that the perimeter of a triangle with angles π3 , α + β − π3 , γ is less than the perimeter of a triangle with angles π3 , π3 , π3 . (b) Let us prove that the smallest perimeter is that of a triangle with angles ◦ ◦ ◦ 75 , 75 , 30 . It is sufficient to prove that the perimeter of a triangle with angles α, β, γ is not less than the perimeter of a triangle with angles ◦ ◦ 75 , α + β − 75 , γ .  ◦ ◦ Indeed, sin 75 + sin α + β − 75 ≤ sin α + sin β, since    ◦ ◦ ◦  ◦  sin α + sin β − sin 75 − sin α + β − 75  sin α − sin 75 + sin β − sin α + β − 75  ◦

 2 sin

◦

α + 75 α − 75 cos − 2 2 ◦

− 2 sin

◦

 4 sin ◦

◦

α + 2β − 75 α − 75 cos 2 2



◦

α − 75 β − 75 α+β sin sin ≥ 0, and 2 2 2 ◦

◦

0 < α ≤ 75 , 0 < β ≤ 75 .

2.8 (a) If xi  yi , i  1, 2, . . . , n, then f (x1 ) + · · · + f (xn )  f (y1 ) + · · · + f (yn ). Suppose that for some i, one has xi  yi . From the assumptions, it follows that if m is the smallest number such that xm  ym , then ym < xm . Let j be the greatest number such that y j < x j . Let k be the smallest number greater than j such that xk < yk . Note that there exists such a number k, for otherwise, we would have the following inequalities: x1 + · · · + x j−1 + x j > y1 + · · · + y j , x j+1 ≥ y j+1 , ··· ··· ··· xn ≥ yn .

Proofs

15

Summing these inequalities leads us to the contradiction x1 + · · · + xn > y1 + · · · + yn . Hence, it follows that x j > y j ≥ yk > xk .  δ Let δ  min x j − y j , yk − xk and λ  1 − x j −x . k Consider the numbers x1 , . . . , x j−1 , λx j + (1 − λ)xk  x ∗j , x j+1 , . . . , xk−1 , λxk + (1 − λ)x j  xk∗ , xk+1 , . . . , xn , and let us verify that for these numbers, the assumptions of the problem hold. We have x1 + · · · + xi ≥ y1 + · · · + yi , i  1, . . . , j − 1. The inequality y1 + · · · + y j ≤ x1 + · · · + xj−1 + λx j + (1 − λ)xk holds, since λx j + (1 − λ)xk  x j − δ ≥ x j − x j − y j  y j . Since x j+1 ≥ y j+1 , . . . , xk−1 ≥ yk−1 , it follows that x1 + . . . + xi ≥ y1 + . . . + yi , i  j + 1, . . . , k − 1.   For i ≥ k, we have x1 + · · · + x i  x1 + · · · + x j−1 + λx j + (1 − λ)xk + x j+1 + · · · + xk−1 + λxk + (1 − λ)x j + · · · + xi , and therefore, for the numbers x1 , . . . , x j−1 , x ∗j , x j+1 , . . . , xk−1 , xk∗ , xk+1 , . . . , xn , the assumptions of the problem again hold. On the other hand,     f (x1 ) + · · · + f x j−1 + f x j + · · · + f (xk ) + · · · + f (xn ) ≥         ≥ f (x1 ) + · · · + f x j−1 + f x ∗j + f x j+1 + · · · + f (xk−1 ) + f xk∗ + · · · + f (xn ),

      since f x ∗j  f λx j + (1 − λ)xk ≤ λ f x j + (1 − λ) f (xk ) and       ≤ λ f (xk ) + (1 − λ) f x j . Therefore, f xk∗   f λxk + (1 − λ)x j     f x ∗j + f xk∗ ≤ f x j + f (xk ). On the other hand, δ  x j − y j or δ  yk − xk , whence x ∗j  y j or xk∗  yk . Therefore, we have substituted the numbers x1 , x2 , . . . , x j , . . . , xn by the numbers x1 , . . . , x j−1 , x ∗j , x j+1 , . . . , xk−1 , xk∗ , xk+1 , . . . , xn . Hence, the assumptions of the problem again hold. On the other hand, the sum f (x1 ) + · · · + f (xn ) did not increase, and the number of indices i satisfying the condition xi  yi increased by 1. Therefore, after at most n − 1 steps, we obtain that f (x1 ) + · · · + f (xn ) ≥ f (y1 ) + · · · + f (yn ). , y3  y4  (b) Let x ≥ y ≥ z. Since for the numbers y1  y2  x+y 2 y+z x+y+z z+x , y  y  and x  x, x  x  x  , x  y, x6  z or 5 6 1 2 3 4 5 2 2 3 , x  z, the assumptions of Problem x1  x, x2  y, x3  x4  x5  x+y+z 6 3 2.8 (a) hold true, then it follows that f (x) + f (y) + f (z) + 3 f ( x+y+z 3 )  f (x 1 ) + f (x 2 ) + f (x 3 ) + f (x 4 ) + f (x 5 ) + f (x 6 ) ≥ y+z z+x ≥ f (y1 ) + f (y2 ) + f (y3 ) + f (y4 ) + f (y5 ) + f (y6 )  2 f ( x+y 2 ) + 2 f ( 2 ) + 2 f ( 2 ).

2.9 Let us first prove the following lemma. Lemma If a ≤ b and x > 0, then (a − x)12 + (b + x)12 > a 12 + b12 .

16

2 Sturm’s Method

Note that

    1 2 2 10 x b11 − a 11 + C12 x b + a 10 + · · · + 2x 12 > 0. (a − x)12 + (b + x)12 − a 12 − b12  C12

Set yi 

√ 3xi i  1, . . . , 1997. We have −1 ≤ yi ≤ 3,

(1)

y1 + · · · + y1997  −954,

(2)

y 12 + ··· +y 12

12 and x112 + · · · + x1997  1 36 1997 . If any two numbers among the numbers y1 , . . . , y1997 belong to (−1, 3), then according to the lemma, one can substitute these two numbers by numbers such that 12 one of them is equal to either −1 or 3. Then conditions (1), (2) hold, and y112 +· · ·+y1997 increases. 12 is the greatest possible if one substitutes the Therefore, the sum y112 + · · · + y1997 numbers y1 , . . . , y1997 by either −1, . . . , −1, 3, . . . , 3, or −1, . . . , −1, 3, . . . , 3, a, where a ∈ (−1, 3). Taking into consideration (2), we obtain that only the second case is possible, so + 1735, where k is the number of −1’s. Since a+2 ∈ Z and a ∈ (−1, 3), that k  a+2 4 4 we must have a  2. 12 is equal to Therefore, the greatest possible value of x112 + . . . + x1997 12 12 1736 + 260·3 + 2  189548. 36

2.10 Let

α1 + ··· +αn n

 ϕ (1). If α1  α2  · · ·  αn  ϕ, then

cos α1 cos α2 · · · cos αn (tan α1 + · · · + tan αn )  cosn ϕ · n · tan ϕ  n · sin ϕ · cos n−1 ϕ    cos 2 ϕ cos 2 ϕ 2 2 n−1  n sin ϕ · (cos ϕ) ··· ≤  n (n − 1)n−1 sin2 ϕ · n−1 n−1  ⎛ ⎞n  n−1 2ϕ 2ϕ   + · · · + cos sin2 ϕ + cos 2  n−1 n−1 n−1 ⎝ ⎠  n (n − 1)n−1 · 1  (n − 1) ≤ n (n − 1) n−2 n nn n 2

If αi  α j for some i and j (i  j), then among those numbers there are two numbers such that one of them is greater than ϕ and the other one is less than ϕ (see Problem 1.10). Let it be the numbers α1 and α2 such that α1 < ϕ < α2 . Thus, substituting α1 by ϕ and α2 by α1 + α2 − ϕ, we obtain a new sequence of numbers ϕ, α1 + α2 − ϕ, α3 , . . . , αn such that (1) holds. On the other hand, since

Proofs

17

cos α1 cos α2 

1 (cos(α1 + α2 ) + cos(α1 − α2 )) < 2 1 < (cos(α1 + α2 ) + cos(2ϕ − (α1 + α2 )))  2

 cos ϕ cos(α1 + α2 − ϕ), we have cos α1 cos α2 · · · cos αn (tan α1 + · · · + tan αn )   sin(α1 + α2 ) cos α3 · · · cos αn + cos α1 cos α2 cos α3 · · · cos αn (tan α3 + · · · + tan αn ) < < sin(α1 + α2 ) cos α3 · · · cos αn + + cos ϕ cos(α1 + α2 − ϕ) cos α3 · · · cos αn (tan α3 + · · · + tan αn )   cos ϕ cos(α1 + α2 − ϕ) cos α3 . . . cos αn (tan ϕ + tan(α1 + α2 − ϕ)+ + tan α3 + · · · + tan αn ).

Continuing in the same way for the numbers ϕ, α1 + α2 − ϕ, α3 , . . . , αn , we obtain a new sequence two of whose terms are equal to ϕ. Let us repeat these steps at most n − 1 times. Then we obtain a sequence n − 1 of whose terms are equal to ϕ, and the nth term is equal to nϕ − (n − 1)ϕ  ϕ. Hence, we have obtained cos α1 cos α2 · · · cos αn (tan α1 + · · · + tan αn ) < cosn ϕ · n · tan ϕ ≤

(n − 1) n

n−1 2

n−2 2

.

From the proof, it follows that equality holds if and only if α1  α2  · · ·  1 . αn  ϕ, where ϕ  arctan √n−1

2.11 Prove that if x ≥ 0, y ≥ 0, x + y ≤ 23 , and k ≥ 2, k ∈ N, then x k (1 − x) + y k (1 − y) ≤ (x + y)k (1 − x − y).

(2.6)

If x + y  0, then (2.6) holds, while if x + y  0, then xk yk · (1 − x) + · (1 − y) ≤ (x + y)k (x + y)k 



x x + y

2

 · (1 − x) +

y x + y

2 · (1 − y) 

(x + y)2 (1 − x − y) + x y(3(x + y) − 2) ≤ 1 − x − y. (x + y)2

Let xi+1 ≥ xi ≥ 0, i  1, . . . , n − 1, x1 + · · · + xn  1 and n ≥ 3. Then (n − 2)x1 + (n − 2)x2 ≤ (x3 + · · · + xn ) + (x3 + · · · + xn )  2 − 2x1 − 2x2 , and therefore, x1 + x2 ≤

2 2 ≤ . n 3

Therefore, if we substitute the numbers x1 , . . . , xn by 0, x1 + x2 , x3 , . . . , xn , then their sum will be equal to 1. Note that n  i1

xik (1 − xi ) ≤ (x1 + x2 )k (1 − x1 − x2 ) + x3k (1 − x3 ) + · · · + xnk (1 − xn ).

18

2 Sturm’s Method

Repeating these steps a finite number of times, we end up with the case n xik (1 − xi ) ≤ x k (1 − x) + (1 − x)k x, and therefore, of n  2, that is, n i1

i1

xik (1

− x i ) ≤ ak .

Remark Note that a1  max(2x(1 − x))  21 , a2  max(x(1 − x))  21 , [0;1]

[0;1]

1 1 a3  max(x(1 − x)(1 − 2x(1 − x)))  , a4  max(x(1 − x)(1 − 3x(1 − x)))  , [0;1] [0;1] 8 12 √ 5 10 − 14 a5  max(x(1 − x) − 4(x(1 − x))2 + 2(x(1 − x))3 )  [0;1] 27

(see Problem 12.12). 2.12 (a) Let x1 ≤ x2 ≤ · · · ≤ xn , n ≥ 3. Consider the following two cases. , then we have that If x2 · · · xn−1 ≥ 2(n−1) n n−1 A  2(n − 1)(x1 x2 + x1 x3 + · · · + x1 xn + x2 x3 + · · · + x2 xn + · · · + xn−1 xn ) − n n−1 x1 x2 · · · xn   2(n − 1)((x1 + xn )(x2 + · · · + xn−1 ) + x2 x3 + · · · + x2 xn−1 + · · · + xn−2 xn−1 )+ + x1 xn (2(n − 1) − n n−1 x2 · · · xn−1 ) ≤ 2(n − 1)(x(1 − x) + x2 x3 + · · · + x2 xn−1 + · · · + xn−2 xn−1 ),

where x  x2 + · · · + xn−1 .  x 2 x 2 +···+x 2 , and therefore, According to Problem 2.2, it follows that 2 n−2 n−1 ≥ n−2 n−3 2 x2 x3 + · · · + x2 xn−1 + · · · + xn−2 xn−1 ≤ 2(n−2) x , whence A ≤ 2(n − 1)(x(1 − x) +

n−3 2 n−1 n−1 x )  4(n − 2) · x · (1 − x) ≤ n − 2. 2(n − 2) 2(n − 2) 2(n − 2)

If x2 · · · xn−1 < 2(n−1) , then for x1  x2  · · ·  xn  n1 , we have A  n − 2. n n−1 Otherwise, if xi  n1 for some value of i, then x1 < n1 < xn . Substituting x1 by n1 and xn by x1 + xn − n1 , we see that the value of the expression A  2(n − 1)((x1 + xn )(x2 + · · · + xn−1 ) + x2 x3 + · · · + x2 xn−1 + · · · + xn−2 xn−1 )+ + x1 xn (2(n − 1) − n n−1 x2 · · · xn−1 )

increases (see Problem 1.17). Continuing in a similar way, either one can end the proof of the inequality or it will be sufficient to prove the inequality for x1  x2  · · ·  xn  n1 . (b) If y1 , y2 , . . . , yn ≥ 0 and y1 + y2 + · · · + yn > 0, then according to Problem 2.12 (a), for xi  y1 + y2 y+i ··· + yn , i  1, 2, . . . , n, we obtain that 2(n − 1)qn pnn−2 − n n−1 y1 · y2 · · · yn ≤ (n − 2) pnn , where pn  y1 + y2 + · · · + yn , qn  y1 y2 + y1 y3 + · · · + y1 yn + · · · + yn−1 yn . Therefore, for y1 · y2 · · · yn  1 we have qn ≤

(n − 2) pnn + n n−1 2(n − 1) pnn−2

.

(1)

Proofs

19

If x1  0, then we obtain the following inequality: x2 + · · · + xn ≤ n

√

x2 −

√

x3

2

+ ··· +

√

x2 −

√

xn

2

+ ··· + n

√

xn−1 −

√

xn

2

+ x2 + · · · + xn

,

√ √ √ √ or 2 x2 x3 + 2 x2 x4 + · · · + 2 x2 xn + · · · + 2 xn−1 xn ≤ (n − 2)(x2 + · · · + xn ). √ The last inequality can be proved using that 2 ab ≤ a + b(a, b ≥ 0). √ x If xi > 0, let yi  2n√x1 ·x2i ···xn , i  1, 2, . . . , n. Then y1 · y2 · · · yn  1, and p2 −2q

(n−1)( p2 −2q )−2q

(n−2) p2 +n

n n n n one needs to prove that n n n − 1 ≤ , or qn ≤ 2(n−1) . n The last inequality holds because by (1), it follows that (see Problem 2.1) (n−2) pn +n n−1 (n−2) pn2 +n . qn ≤ 2(n−1)n pn−2 ≤ 2(n−1) n

(c) For numbers y1  x12 , . . . , yn  xn2 , using the inequality of Problem 2.12 (b), we deduce that x12 + · · · + xn2 n

 2   (|x1 | − |x2 |)2 + · · · + (|x1 | − |xn |)2 + · · · + xn−1  − |xn | − n x12 · . . . · xn2 ≤ , n

 or (n − 1)(x12 + · · · + xn2 ) + n n x12 · · · xn2 ≥ (|x1 | + · · · + |xn |)2 ≥ (x1 + · · · + xn )2 (see Problem 1.13).

Problems for Independent Study Prove the following inequalities (1–4). 1. 2.

+ ··· +

1 1 + xn

a 2 + b2 + c2 + d 2 4

3

1 1 + x1 

3. 9 ≤ 4. (a) (b) (c)



≤

1+

n √ n x ··· x , 1 n

where n ≥ 2, 0 < xi ≤ 1, i  1, . . . , n.

≥ abc + abd +4 acd + bcd , where a > 0, b > 0, c > 0, d > 0. 2 2 2 x y + yz + zx ≤ 9 + x4 y z , where x + y + z  x yz and x > 0, y > 0, z > 0. 1 + x1 · · · 1 +xnxn ≥ (n + 1)n , where n ≥ 2, xi > 0, i  1, . . . , n, and x1 x1 + · · · + xn  1,  n 1+xn 1+x1 · · · 1−x ≥ nn −+ 11 , where n ≥ 2, xi > 0, i  1, . . . , n, and 1−x1 n x1 + · · · + xn  1, s , for (s − (n − 1)x1 ) · · · (s − (n − 1)xn ) ≤ x1 · · · xn , where 0 ≤ xi ≤ n−1 i  1, . . . , n, and

x1 + · · · + xn  s. 5. Prove that among all convex n-gons inscribed in a given circle, the greatest area it that of a regular n-gon. 6. Prove that among all convex n-gons inscribed in a given circle, the greatest perimeter is that of a regular n-gon. 7. Prove that among all convex polygons inscribed in a given circle, the greatest sum of the squares of the sides is that of an equilateral triangle.

20

2 Sturm’s Method

8. Let n ≥ 4 and xi > 0, i  1, . . . , n, be real numbers such that x1 +· · ·+ xn  1. Prove that for all λ (0 < λ < n), one has the inequality   1 1 n2 − λ − λx1 · · · xn ≤ + ··· + . x1 · · · xn x1 xn nn 9. Prove that for two triangles with angles α, β, γ and α1 , β1 , γ1 , one always has cos α1 cos β1 cos γ1 + + ≤ cot α + cot β + cot γ . sin α sin β sin γ 10. Prove that

(i 1 ,...,i n )

yia11 · · · yiann ≤

(i 1 ,...,i n )

yib11 · · · yibnn , (the summation is over all

permutations of the numbers (1, . . . , n)), where y1 > 0, . . . , yn > 0, and a1 ≥ . . . ≥ an , a1 ≤ b1 , a1 + a2 ≤ b1 + b2 , . . . , a1 + · · · + an−1 ≤ b1 + · · · + bn−1 , a1 + · · · + an  b1 + · · · + bn . 11. Prove that n(a1 b1 + · · · + an bn ) ≥ (a1 + · · · + an )(b1 + · · · + bn ) if given that from the condition ai < a < a j , it follows that bi ≤ b j , where a  a1 + a2 n+ ··· + an . 12. Let f be an odd function defined and decreasing on (−∞, +∞). Prove that f (a) f (b) + f (b) f (c) + f (c) f (a) ≤ 0, where a + b + c  0. 13. Suppose that for the numbers a1 , . . . , an (n ≥ 2), the following conditions hold: (a) a1 ≤ · · · ≤ an , (b) a1 + · · · + an  0, (c) |a1 | + · · · + |an |  S. Prove that an − a1 ≥ 2S . n 14. Let n be a given integer such that n ≥ 2. (a) Find the smallest constant C such that the inequality 

 xi x j (xi2 + x 2j ) ≤ C

1≤i< j≤n



4 xi

1≤i≤n

holds for all nonnegative numbers x1 , . . . , xn . (b) Find when the equality holds for the obtained constant C. 15. Prove that (a) (b)

1 + · · · + 1 + s1− xn ≤ 1, where s 1 + s − x1 x1 , . . . , xn > 0. 1 + a n + ··· + a n 1 + na1 ··· an n a1n + ··· + an−1 + na1 ··· an 1 n−2

 x1 + · · · + xn , x1 · · · xn  1 and + ··· +

1 a2n + ··· + ann + na1 ··· an

≤

1 , a1 ··· an

where a1 , . . . , an > 0. 16. Prove that 5(a 2 +b2 +c2 ) ≤ 6(a 3 +b3 +c3 )+1, where a, b, c > 0 and a +b+c  1.

Chapter 3

The HM-GM-AM-QM Inequalities

For solving some problems and in order to prove a large number of inequalities, one often needs to use various means and the relationships among them. For positive real numbers a and b, the expression 1 +2 1 is called their harmonic a b √ mean (HM), ab is called their geometric mean (GM), a +2 b is called their arithmetic  2 2 mean (AM), and a +2 b is called their quadratic mean (QM) or root mean square. One has the following relationship among the means. In this chapter we consider the so called the HM-GM-AM-QM inequalities, first for two positive numbers and afterward more generally for n arbitrary positive numbers. Lemma (HM-GM-AM-QM inequalities for two positive numbers) Let a and b be positive real numbers. Then  √ 2 a2 + b2 a +b ≤ . (3.1) ≤ ab ≤ 1 1 2 2 + b a Moreover, in (3.1) equality holds if and only if a  b. Proof Proofs of the inequalities (3.1) are given in Chapter 1 (Problems 1.2–1.4). Let us consider the following examples in order to see how these inequalities can be applied. Example 3.1 Prove that

a b

+

b c

+

c a

≥ 3, where a > 0, b > 0, c > 0.

Proof Without loss of generality one can assume that c ≥ a, c ≥ b. Using that a+b √ ≥ ab, 2

(3.2)

we deduce that © Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_3

21

22

3 The HM-GM-AM-QM Inequalities

a b + ≥ 2. b a Note that

b c

+

c a

−

b a

−1

(c − a)(c − b) ac

(3.3)

≥ 0, whence

b c b + − − 1 ≥ 0. c a a

(3.4)

Summing inequalities (3.3) and (3.4), we obtain ab + bc + ac ≥ 3. This ends the proof.    Example 3.2 Prove that b a+ c + c +b a + a +c b > 2, where a > 0, b > 0, c > 0. Proof Without loss of generality, one can assume that a ≥ b ≥ c. Thus 

       a b c a b+c b c b+c + +  + + + − ≥ b+c c +a a +b b+c a c +a a +b a       b c b+c b c b+c ≥ 2+ + − ≥ 2+ + −  c +a a +b a c +a 2a a       √ √ b c b c 1 1 + +  2+ √ − b+c ≥ 2+ √ − b+c  1 + ac 2 1 + bc 2 a a √ √ √ c  2+ √ b + c − 2c ≥ 2. 2a(b + c)

Equality a  b+c, b  a  c, which leads to a contradiction. Therefore,  holds if  c a b + c + a + a + b > 2. b+c This ends the proof.  Alternative proof Let us use the inequality b a+ c · 1 ≥ 1 + 2b + c  a +2ab + c , a   c b 2 2b 2 2c · 1 ≥  , · 1 ≥  . Summing these inequalia+c a+b+c a+b a+b+c 1 + a b+ c 1 + a +c b    ties, we obtain b a+ c + c +b a + a +c b ≥ 2. Equality holds if 1  b a+ c  a b+ c  a +c b ,    which leads to a contradiction. Thus, it follows that b a+ c + c +b a + a +c b > 2. This ends the proof. 

Generalization of inequalities (3.1). Note that (3.1) can be generalized for positive numbers a1 , . . . , an . √ a12 + ··· + an2 a + ··· + a n are called the harThe expressions 1 + ··· + 1 , n a1 · · · an , 1 n n , n a1

an

monic mean (HM), geometric mean (GM), arithmetic mean (AM), and quadratic mean (QM) or root mean square of the numbers a1 , . . . , an Lemma (HM-GM-AM-QM inequalities for n arbitrary positive numbers) Let a1 , . . . , an be positive real numbers. Then

3 The HM-GM-AM-QM Inequalities

1 a1

n + ··· +

1 an

23

√ a1 + · · · + an ≤ ≤ n a1 · · · an ≤ n



a12 + · · · + an2 . n

(3.5)

Moreover, in (3.5), inequality holds if and only if a1  · · ·  an . Proof See Example 3.3, Problems 2.1 and 2.2. Example 3.3 (HM-GM) Prove that 0, . . . , an > 0. Proof Using the inequality 1 , i ai 1 √ n a ··· a . 1 n

1 a1

n + ··· +

x1 + ··· + xn n

≥

≤

1 an

√ n

√ n

a1 · · · an , where n ≥ 2, a1 >

x1 · · · xn for positive numbers xi 

1, . . . , n (see Problem 2.1), we obtain the following inequality:

This ends the proof. √ Remark a1 + ···n + an ≥ n a1 · · · an ≥

1 a1

n + ··· +



1 an

1 a1

+ ··· + n

1 an

≥

, whence

1 1 + ··· + (a1 + · · · + an ) a1 an

 ≥ n2 .

(3.6)

Let us consider the following examples in order to see how these inequalities can be applied. Example 3.4 Solve the following system:  ⎧   1 ⎪ ⎪ , 1 + x1 + · · · + 1 + x100  100 1 + ⎨ 100  ⎪   ⎪ ⎩ 1 − x + · · · + 1 − x  100 1 − 1 . 1 100 100 Solution. Using the inequality a1 + · · · + an ≤ n



a12 + · · · + an2 n

(3.7)

√ √ (see Problem√ 2.2) for the numbers 1 + x1 , . . . , 1 + x100 and √ 1 − x1 , . . . , 1 − x100 , we deduce that  √ √ 1 + x1 + · · · + 1 + x100 100 + (x1 + · · · + x100 ) ≤ 100 100  √ √ 1 − x1 + · · · + 1 − x100 100 − (x1 + · · · + x100 ) and ≤ . (3.8) 100 100

24

3 The HM-GM-AM-QM Inequalities

From the obtained inequalities and the given system, we deduce that 

1 ≤ 1+ 100



x1 + · · · + x100 , and 1+ 100



1 ≤ 1− 100



1−

x1 + · · · + x100 . 100

Therefore, x1 + · · · + x100  1, and thus it follows that inequality (3.8) is an equality. √ √ Hence, we obtain that 1 + x1  · · ·  1 + x100 , or x1  · · ·  x100  0.01. It is not difficult to verify that these numbers satisfy the given system. The relationship between the arithmetic mean and the geometric mean helps in solving problems the involve finding maximum and minimum values. Remark 1 Consider all n-tuples (a1 , . . . , an ) of positive numbers having the same geometric mean. Prove that among those n-tuples, the minimum arithmetic mean has the n-tuple with a1  · · ·  an . Remark 2 Consider all n-tuples (a1 , . . . , an ) of positive numbers having the same arithmetic mean. Prove that among those n-tuples, the minimum geometric mean has the n-tuple with a1  · · ·  an . Example 3.5 Among all triangles with a given area, find the triangle that has the minimum perimeter. Proof Let us denote the side lengths of that triangle by a, b, c, and the perimeter by 2p. By Heron’s formula and  the inequality of Problem 2.1, we obtain S  3  √  p(p − a)(p − b)(p − c) ≤ p (p − a) + (p −3 b) + (p − c)  93 p2 , and therefore,  √ 2p ≥ 2 3 3 · S. Equality holds if a  b  c. Therefore, among all triangles with a given area, the one with the minimum perimeter is an equilateral triangle. Example 3.6 Among the triangles with perimeter equal to 2p, find the triangle with the maximum area. Proof Denote the area of that triangle by S, and the side lengths by a, b, c. √  By Heron’s formula, S  p · (p − a)(p − b)(p − c). Let us find the maximum value of the expression (p − a)(p − b)(p − c). Since (p − a) + (p − b) + (p − c)  p is a constant, it follows from Remark 2 that the product (p − a)(p − b)(p − c) attains its maximum value when p − a  p − b  p − c. Therefore, a  b  c, and so among all triangles with perimeter equal to 2p, the one with maximum area is an equilateral triangle. This ends the proof.

Problems

25

Problems Prove the following inequalities (3.1–3.28). 3.1. (a + b)(b + c)(c + a) ≥ 8 abc, where a > 0, b > 0, c > 0. 3.2. (a + b + c − d )(b + c + d − a)(c + d + a − b)(d + a + b − c) ≤ (a + b)(b + c)(c + d )(d + a), where a > 0, b > 0, c > 0, d > 0. 3.3. (a) Schur’s inequality: a3 + b3 + c3 + 3abc ≥ a2 b + ab2 + b2 c + c2 b + a2 c + ac2 , where a >0, b > 0,  c > 0.   1 + c4b 1 + a4c > 25, where a > 0, b > 0, c > 0. (b) 1 + b4a +c +a +b − 1) log a 3.4. log(a < , where a > 1. log a log(a + 1) 3.5. Schur’s inequality (alternative form): abc ≥ (a + b − c)(a + c − b)(b + c − a), where a > 0, b > 0, c > 0. 1 , if x + y  1. 3.6. x8 + y8 ≥ 128   2  1 2 3.7. a + a + b + 1b ≥ 12.5, if a > 0, b > 0 and a + b  1.  2 2  2 n2 + 1 3.8. x1 + x11 + · · · + xn + x1n ≥ ( n ) , if n ≥ 2, x1 > 0, . . . , xn > 0 and x1 + · · · + xn  1. 3.9. a4 + b4 + c4√≥ abc (a + b + c). xy  1. 3.10. x√2 + y2 ≥ 2 √2(x − y), if√ √ √ √ 3.11. 6a1 + 1 + 6a2 + 1 + 6a3 + 1 + 6a4 + 1 + 6a5 + 1 ≤ 55, if a1 > · · + a5  0, . . . , a5 > 0 and √ a1 + ·√ √1. 3.12. 6a + 4b +5c ≥ 5 ab + 3 bc + 7 ca, where a ≥ 0, b ≥ 0, c ≥ 0. 3.13. 2 a4 + b4 + 17 > 16 ab. 3.14. (a) b + c1− a + a −1b + c + a + b1 − c ≥ a1 + 1b + 1c , where a, b, c are the side lengths of some triangle. √ (b) √2b2 + a2c2 − a2 + √2c2 + b2a2 − b2 + √2a2 + c2b2 − c2 ≥ 3, where a, b, c are the side lengths of some triangle.  n+1 3.15. 1n++nb ≥ bn , where n ∈ N, b > 0. 1 n  n+1  , where n ∈ N, 3.16. (a) 1 + 1n < 1 + n +1 1     1 n+1 1 n+2 > 1 + n+1 , where n ∈ N, (b) 1 + n    n−1 n  n  m m m 1 m−1 m < 1+ n < 1 + m− , where m > 1, n > 1, and (c) 1 + n − 1 n m ∈ N, n ∈ N. n  3.17. n! < n +2 1 , where n  2, 3, 4, . . . . 1

3.18. (a) n(n + 1) n < n + Sn , where Sn  11 + · · · + 1n , n  2, 3, 4, . . . , 1 (b) n − Sn > (n − 1)n 1−n , where Sn  11 + · · · + 1n , n  3, 4, . . . .   3.19. (qn − 1) qn+1 + 1 ≥ 2nqn (q − 1), where q > 1, n ∈ N.

26

3 The HM-GM-AM-QM Inequalities

3.20. (a) a2 + b2 + c2 + d 2 + ab + ac + ad + bc + bd + cd ≥ 10, where a > 0, b > 0, c > 0, d > 0, and abcd  1.      abc 3 (b) a − 1 + 1b b − 1 + 1c c − 1 + a1 ≤ 12√+ abc , where a, b, c > 0.       1 1 1 (c) a + b − t b + c − t c + a − t ≤ (a + b + c) a1 + 1b + 1c (1 − t)2 + 4 − 3t, where a, b, c, t > 0 and abc  1. √ √ 3.21. n n a1 · · · an −(n − 1) n−1 a1 · · · an−1 ≤ an , where ai > 0, i  1, . . . , n, n  3, 4, . . ..   √ n a1 · · · an + n b1 · · · bn + . . . + n k1 · · · kn 3.22.  ≤ n (a1 + b1 + · · · + k1 )(a2 + b2 + · · · + k2 ) · · · (an + bn + · · · + kn ), where a1 > 0, . . . , an > 0, b1 > 0, . . . , bn > 0, . . . , k1 > 0, . . . , kn > 0. √ √ 3.23. a1 + a1 a2 + · · · + n a1 · · · an ≤ e(a1 + · · · + an ), where n ≥ 2, a1 ≥ 0, . . . , an ≥ 0. 3.24. (a) nak − kan ≤ n − k, where n > k, n, k ∈ N, a > 0. x2 x3 xn+1 (b) x12 + x22 + · · · + xn n ≥ x1 + · · · + xn , where n ≥ 2, n ∈ N, x1  1 3 min(x1 , . . . , xn ) > 0. 3.25.

ax1 −x2 x1 + x2

+

ax2 −x3 x2 + x3

+ ··· +

axn −x1 xn + x1

≥

2

n2 n 

xi

, where a > 0, xi > 0, i  1, . . . , n.

i1

√ p

√ 3.26. x1 + 1 + · · · + p xn + 1 ≤ n + 1, where n ≥ 2, x1 > 0, . . . , xn > 0, x1 + · · · + xn  p, p ∈ N, p ≥ 2. 3.27. (a) (1 + α)n ≥ 1 + n α, where α ≥ −1, n ∈ N (Bernoulli’s inequality). (b) (1 + α)n > 1 + n α, where α  0, α ≥ −1, n ∈ N, n > 1. √ 3 3 , 16 k m xk (1 − xm ) ≤ k ·m1+ k , where 0 ≤ x ≤ 1, k, m ∈ N, (k + m) m √ y x z + + ≥ 3 2 3 , where x, y, z > 0 and x2 + y2 + z 2 1 − x2 1 − y2 1 − z2 √ 1 + 1 −1 y + 1 −1 z ≥ 9 + 23 3 , where x, y, z > 0 and x2 + y2 + z 2 1−x

3.28. (a) cos3 t sin t ≤ (b) (c) (d)

3.29. Find the minimum value of the function f (x)  n ∈ N, n > 1.

1 √ n 1+x

 1,  1.

+ √n 11− x in [0, 1), where

3.30. Find the minimum value of the function f (x)  axm + xbn in (0, ∞), where a > 0, b > 0, m, n ∈ N. 3.31. Find in [a, b] (0  < a < b) a point x0 such that the function f (x)  (x − a)2 b2 − x2 attains its maximum value in a, b at x 0 . 3.32. Find the greatest possible value of the product xyz given x > 0, y > 0, z > 0, √ and 2x + 3y + π z  1. 3.33. Find the maximum and minimum values of the function y  ax2x+ b , where a > 0, b > 0.

Problems

27

3.34. (a) Find the maximum value of the function y  (b) Find the maximum value of the function y 

√ 5 x2 + 6x + 8 + 12 . x+3  2 3 2 2 (x + 1) (x + 3) . 3x2 + 4

3.35. Suppose the sum of the six edges of a triangular pyramid (tetrahedron) PABC is equal to S, and  APB   BPC   CPA  900 . Find among such pyramids one with the greatest volume. 3.36. Solve the system of equations  x + y  2, xy − z 2  1. 3.37. Solve the system of equations 

x + y + z  3, x2 + y2 + z 2  3.

3.38. Given numbers a, b, c, d , e such that  a + b + c + d + e  8, a2 + b2 + c2 + d 2 + e2  16, find the greatest possible value of e. 3.39. Find the minimum value of the expression x4 ≤ x5 ≤ x6 ≤ 1000.

x1 x2

+

x3 x4

+

x5 x6

if 1 ≤ x1 ≤ x2 ≤ x3 ≤

3.40. Solve the equation x4 + y4 + 2  4xy. 3.41. Find all integer solutions of the equation

xy z

+

xz y

+

yz x

 3.

3.42. Prove that β

β

(a) x1α + · · · + xnα ≥ x1 + · · · + xn , where n ≥ 2, x1 > 0, . . . , xn > 0, α > β ≥ 0, and x1 · · · xn  1, β β (b) x1α + · · · + xnα ≥ x1 + · · · + xn , where n ≥ 2, x1 > 0, . . . , xn > 0, α ≥ (n − 1)|β|, and x1 · · · xn  1, 3.43. Prove that x2 y +y2 z +z 2 x ≤

4 , where x 27

≥ 0, y ≥ 0, z ≥ 0, and x +y +z  1.

3.44. Prove that b + 11++bc + 11++cdc + and abcd  1, + 11++bc + 11++cdc + (b) 11++ab a b and abcd  1.

(a)

1+a 1 + ab

1+d 1 + da

≥ 4, where a > 0, b > 0, c > 0, d > 0,

1 + da 1+d

≥ 4, where a > 0, b > 0, c > 0, d > 0,

28

3 The HM-GM-AM-QM Inequalities

 3.45. Prove that 2ST > 3(S + T )(S(bd + df + fb) + T (ac + ce + ea)), where 0 < a < b < c < d < e < f and a + c + e  S, b + d + f  T . 3.46. Prove that √ √ a+

ab +

3

abc + 4

√ 4 abcd

0, c > 0, d > 0.

≤

 4 a·

a+b 2

·

a+b+c 3

·

a+b+c+d , 4

where a > 0, b >

3.47. Prove that a12 + (ab)6 + (abc)4 + (abcd )3 ≤ 1, 43(a12 + b12 + c12 + d 12 ).

Proofs 3.1. Let us use the inequality√(3.2) for positive √ numbers a√and b, b and c, c and a. ≥ ca, and on multiplying Then we obtain a +2 b ≥ ab, b +2 c ≥ bc, c+a 2 these inequalities, we deduce that (a + b)(b + c)(c + a) ≥ 8abc. 3.2. Note that three factors on the left-hand side of the given inequality are positive. If only one factor on the left-hand side is not positive, then the proof is obvious. Consider the case in which all four factors are positive. In this case, according to inequality (3.2) for the numbers a + b + c − d and b + c + d − a, a + b + c − d and d + a + b − c, b + c + d − a and c + d + a − b, c + d + a − b and d + a + b − c, we obtain the following inequalities:  (a + b + c − d ) + (b + c + d − a)  b + c, (a + b + c − d )(b + c + d − a) ≤ 2   (a + b + c − d )(d + a + b − c) ≤ a + b, (b + c + d − a)(c + d + a − b) ≤ c + d ,  (c + d + a − b)(d + a + b − c) ≤ a + d ,

and on multiplying these inequalities, we obtain the required inequality. 3.3. (a) The given inequality is equivalent to the following inequality: abc ≥ (a + b − c)(a + c − b)(b + c − a). See Problem 3.5. (b) The given inequality is equivalent to the following inequality: 7abc + a3 + b3 + c3 > a2 b + b2 a + c2 a + a2 c + b2 c + c2 b, and the proof follows from the inequality of Problem 3.3(a). 3.4. The given inequality is equivalent to the following inequality: log(a − 1) log(a + 1) < log2 a. If log(a − 1) ≤ 0, then log(a − 1) log(a + 1) ≤ 0 < log2 a. Otherwise, if log(a − 1) > 0, then using inequality (3.2), that   we deduce  2 log(a − 1) log(a + 1) ≤ log(a − 1) + log(a + 1)  log a2 − 1 < log a2 . This inequality is equivalent to the inequality log(a − 1) log(a + 1) < log2 a. Alternative proof. The given inequality is equivalent to the following inequality: log a log a + 1 − 1) log a 1 − log(a > 1 − log(a , or loga −a 1 > log(a +a 1) . This proof follows from the log a + 1) inequalities a −a 1 > a +a 1 > 1 and a + 1 > a > 1.

Proofs

29

3.5. Since a > 0, b > 0, c > 0, at least two factors on the right-hand side of the inequality are positive. If only one factor is not positive, then the proof is obvious. Consider the case in which all three factors on the right-hand side are positive. √ Since x +2 y ≥ xy for nonnegative numbers, it follows that   (a + b − c) + (a + c − b)  a, (a + b − c)(a + c − b) ≤ (a + b − c)(b + c − a) ≤ b, 2  (a + c − b)(b + c − a) ≤ c.

On multiplying these inequalities, we obtain the required inequality. 3.6. We have x8 + y8 ≥ 2



x4 + y4 2

2

≥

x2 + y2 2

4 ≥

3.7. Since a + b  1, we deduce from (3.2) that inequality (3.7), we obtain  2  2  a+ a + a1 + b + 1b ≥ 2

1 a

+b+ 2

1 b



2 

1 ab

 x + y 8 1 .  2 128 ≥ 4. Hence, according to

1 1 + ab 2

2

≥

1+4 2

2

25 . 4



3.8. Since x1 + · · · + xn  1, we deduce from (3.6) that x11 + · · · + x1n ≥ n2 . Using inequality (3.7), we obtain  2  



x1 + x11 + · · · + xn + x1n 1 2 1 2 x1 + + · · · + xn + ≥n  x1 xn n  2 2  2

1 + x11 + · · · + x1n 1 + n2 1 + n2 . n ≥n  n n n 3.9. According to inequality (3.2), we have a4 +b4 ≥ 2a2 b2 , b4 +c4 ≥ 2b2 c2 , c4 + a4 ≥ 2c2 a2 . Summing these inequalities, we obtain a4 + b4 + c4 ≥ a2 b2 + b2 c2 + c2 a2 .

(3.9)

According to inequality (3.2), we have a2 b2 + b2 c2 ≥ 2ab2 c, b2 c2 + c2 a2 ≥ 2abc2 , c2 a2 + a2 b2 ≥ 2a2 bc. Summing these inequalities, we deduce that a2 b2 + b2 c2 + c2 a2 ≥ abc(a + b + c).

(3.10)

30

3 The HM-GM-AM-QM Inequalities

From inequalities (3.9) and (3.10) it follows that a4 + b4 + c4 ≥ abc(a + b + c). 3.10. Using that x2 + y2  (x − y)2 + 2xy and xy  1, we deduce that x2 + y2  √ 2 √ 2 ≥ 2 2(x − y). (x − y)2 + 3.11. According to inequality (3.2), for every value of λ, one has the following inequalities: √ (6ai + 1) + λ2 ≥ 2λ 6ai + 1, i  1, . . . , 5. Summing these inequalities and using that a1 + · · · + a5  1, for λ > 0 we n √  + 5λ2 ≥ 6ai + 1. obtain 11 2λ i1  5 √ √  Taking λ  11 , it follows that 55 ≥ 6ai + 1. 5 i1

3.12. According (3.2), we have √ √ √to inequality 5 ab + 7 ac + 3 bc ≤ 5(a2+ b) + 7(a2+ c) +

 6a + 4b + 5c.   4 4 2 2 2 a4 + b4 + 17 ≥ 4a2 b2 + 17 > 3.13. Since  2 2a + b ≥ 2a b , it follows that 4 a b + 4 ≥ 16ab, and therefore 2 a4 + b4 + 17 > 16ab. 3(b + c) 2

3.14. (a) Using inequality (3.6) for the numbers and 1 , c

1 , a −1b + c a+b−c 1 1 b+c−a + a+b−c ≥ 1b , 2

and 1 a−b+c

1 , a+b−c + a + b1 − c 2

1 b+c−a

and

we obtain that ≥

1 1 , a−b+c b+c−a 1 1 b+c−a + a−b+c ≥ 2

1 . a

Summing these inequalities, we obtain the required inequality. (b) According to the inequality of Problem 1.2, we have a b c  +  +   2b2 + 2c2 − a2 2c2 + 2a2 − b2 2a2 + 2b2 − c2   √ a2 b2 c2  3  +  +  ≥ 3a2 (2b2 + 2c2 − a2 ) 3b2 (2c2 + 2a2 − b2 ) 3c2 (2a2 + 2b2 − c2 ) ⎛ ⎞ √ √ a2 b2 c2 ⎠  3, + + ≥ 3⎝ 2 2 2 2 2 2 2 2 2 2 2 2 3a + (2b + 2c − a ) 2

and therefore,

√

a 2b2 +2c2 −a2

3b + (2c + 2a − b ) 2

+

√

b c + √2a2 +2b ≥ 3. 2 −c2 2c2 +2a2 −b2 √  n+1 n+1 bn , and thus it follows that 1n++nb 1

√

3.15. One has 1 + b + · · · + b ≥ (n + 1)

3c + (2a + 2b − c ) 2

≥

n

bn . 3.16. (a) Using

the

inequality

of

Problem 2.1 for the  numbers  n 1+ 1 +...+ 1+ 1 +1 (1 + 1/n), . . . , (1 + 1/n), 1, we obtain ( n ) n + (1 n ) > n+1 1 + 1n ,    nn+1  n  > 1 + 1n . or 1 + n +1 1

Proofs

31

(b) Using

the inequality of Problem 2.1 for the numbers 1 1 , 1, we obtain 1− ,...,1 − n + 1 n + 1     n+1  n+1  n+1  n+2 1 n+2 1 − n+2 > 1 − n +1 1 or 1 + 1n > 1 + n +1 1 . (c) Using the inequality of Problem 2.1 for the numbers  n−1 n   m m m m ,...,1 + 1+ , 1, we obtain 1 + n − < 1 + mn m . 1 n − 1  n − 1  n−1

Using the inequality of Problem 2.1 for the n  n   m m 1 m−1 . 1 + , . . . , 1 + , 1, we obtain 1 + mn m < 1 + m − n n n    m−1

3.17. Using the inequality of Problem 2.1, we obtain n+1 . 2

√ √ n n!  n 1 · · · n <

numbers

1+ ··· +n n



 √ 3.18. (a) Sn +n  n+ 11 +· · ·+ 1n  2+ 23 +· · ·+ n +n 1 > n· n 2 · 23 · · · n +n 1  n· n n + 1.  1 (b) n − Sn  21 + · · · + n −n 1 > (n − 1) · n−1 21 · · · n −n 1  (n − 1)n 1−n .   3.19. We have qn − 1  (q − 1) qn−1 + · · · + 1 and q > 1. Therefore we obtain an inequality equivalent to the given inequality:  n−1   q + · · · + 1 qn+1 + 1 ≥ 2nqn .

(3.11)

According to Problem  2.1 for the casen−1n > 1, we have n+1 qn−1 + · · · + 1 ≥ n · n qn−1 · · · 1  nq 2 and qn+1 + 1 ≥ 2q 2 . Multiplying these inequalities, we obtain (3.11).  10 3.20. (a) a2 + b2 + c2 + d 2 + ab + ac + ad + bc + bd + cd ≥ 10 · (abcd )5  10. (b) Since a, b, c > 0, at least two factors on the left-hand side are positive (see Problem 1.9(a)). If only one factor on the left-hand side is nonpositive, then the proof is obvious. Consider the case in which all three factors on the left-hand side are positive. In this case, note that







 1 1 1 1 3 + 3abc  b a − 1 + +c b−1+ +a c−1+ + bc a − 1 + + b c a b



 1 1 + ac b − 1 + + ab c − 1 + . c a

According to Problem 2.1, we deduce that 





 1 2 1 2 1 2 3 b−1+ c−1+ 3 + 3abc ≥ 6 a3 b3 c3 a − 1 + . b c a

32

3 The HM-GM-AM-QM Inequalities

This is equivalent to the given inequality. (c) Define a + b + c  A and a1 + 1b + 1c  B. Then one needs to prove that

t 3 + (AB − A − B)t 2 +

 a b c + + + A + B − 2AB t + AB + 2 − A − B ≥ 0. (1) c a b

For t  1, inequality (1) holds, since for t  1 it is equivalent to Problem 3.2(b) for abc  1. Therefore, for t  1, from (1) it follows that ac + ab + bc ≥ A + B − 3.  √ 3 1 Since t > 0,A  a + b + c ≥ 3 abc  3, B  a1 + 1b + 1c ≥ 3 3 abc  3, and AB + 2 − A − B  (A − 1)(B − 1) + 1 > 0, it follows that a b c t 3 + (AB − A − B)t 2 + ( + + + A + B − 2AB)t + AB + 2 − A − B ≥ c a b ≥ t 3 + (AB − A − B)t 2 + (2A + 2B − 2AB − 3)t + AB + 2 − A − B   (t − 1)2 (t + AB + 2 − A − B) ≥ 0, 3.21. an + n·

√ √ a1 · · · an−1 + · · · + n−1 a1 · · · an−1 ≥  

n−1

 

an ·

√

n−1

n−1

a1 · · · an−1 · · · 

 n−1 √ (n − 1) n−1 a1 · · · an−1 ≤ an .

3.22.

n

 n

√

n−1

√ a1 · · · an−1 , and therefore n n a1 · · · an − 

a1 a2 an · ··· + a1 + b1 + · · · + k1 a2 + b2 + · · · + k2 an + bn + · · · + kn  b2 bn b1 + n · ··· + ···+ a1 + b1 + · · · + k1 a2 + b2 + · · · + k2 an + bn + . . . + kn  k1 k2 kn + n · ··· ≤ a1 + b1 + · · · + k1 a2 + b2 + · · · + k2 an + bn + · · · + kn 

a1 a2 an 1 + + ··· + ≤ + n a1 + b1 + · · · + k1 a2 + b2 + · · · + k2 an + bn + · · · + kn

 b1 b2 bn 1 + ···+ + + ··· + + n a1 + b1 + · · · + k1 a2 + b2 + · · · + k2 an + bn + · · · + kn 

1 k1 k2 kn +  + + ... + n a1 + b1 + · · · + k1 a2 + b2 + · · · + k2 an + bn + · · · + kn 

1 a1 b1 k1  + + + ··· + n a1 + b1 + · · · + k1 a1 + b1 + · · · + k1 a1 + b1 + · · · + k1 

1 a2 b2 k2 + + ···+ + + ··· + n a2 + b2 + · · · + k2 a2 + b2 + · · · + k2 a2 + b2 + · · · + k2 

1 an bn kn +  1. + + ... + n an + bn + · · · + kn an + bn + · · · + kn an + bn + · · · + kn

Proofs

33

3.23. Note that 1 a1 + 1·2 2 a2 + a2 ≥ 2·3 ...

a1 ≥

1 1 a1 + · · · + a1 , 2·3 n(n + 1) 2 2 a2 + · · · + a2 , 3·4 n(n + 1)

an ≥ n(nn+ 1) an , and therefore, + 2a2 + ··· + a1 + a2 + · · · + an ≥ a21 + a1 2·3

a1 + 2a2 + ··· + kak k(k + 1)

+ ··· +

a1 + 2a2 + ... + nan . n(n + 1)

2 + ··· + kak ≥ (a1 · · · ak ) k , k  1, . . . , n. Let us prove that e a1 + 2a k(k + 1) According to Problem 2.1, we have 1

a1 + 2a2 + · · · + kak ≥ k(k + 1)

√ √ k k 1 a1 · 2a2 · · · kak k!  (a1 · · · ak ) k . k +1 k +1

In order to complete the proof, one needs to prove that √ k  k k! > 1e or k! > k +e 1 (see Problem 7.18). k +1 Hence, we deduce that e(a1 + · · · + an ) ≥

n n   1 a1 + 2a2 + · · · + kak e ≥ (a1 · · · ak ) k . k(k + 1) k1

k1

Now let us prove that it is not possible to substitute e in the given inequality n  1 by a number smaller than e. Indeed, let c(a1 + · · · + an ) ≥ (a1 · · · ak ) k . k1

Then taking ak  k1 , k  1, . . . , n, it follows that c

n  1 k1

k

≥

n  k1

1 (k!)

1 k

>

n  k8

k·

e √ k

k

(3.12)

(here we have used the inequality from Problem 14.18).

 k  On the other hand, according to Problem 7.17, we have 1 + k2 > 1+    2 √ k 2 > k, whence k < 1 + k2 , and therefore, k k2 + k(k−1) 2 k n  k8

 e e  . > √  k 2 k· k k8 k 1 + n

k

(3.13)

34

3 The HM-GM-AM-QM Inequalities

From (3.12) and (3.13), it follows that n n   1  1√ < c e , or 2 k k 1 + k k8 k1 √2 n n n    1 1 k
k8 n  k8

1 k

− e·

√ αn · e − 2e ·

k8 k 1 + n √  √2 k k k8 n  1 √ k k k1 n  1 k k1

k

< c

n  k1

k1

1 , k

or n 

< c, where αn 

k8 n  k1

7 

1 k

1−

1 k

k1 n  k1

1 k 1 k

.

Passing to the limit, we obtain e ≤ c (see Problems 7.15 and 7.16). 3.24. (a) Note that n − k + kan  n



1 + · · · + 1 + a n + · · · + an n−k

≥

n ·

k

1 · · · 1 · an · · · an  nak , k

whence n − k ≥ nak − kan .  k k (b) Note that yxk−1 + (k − 1)y ≥ k k yxk−1 · yk−1  kx, where k ∈ N, k ≥ 2, whence xk yk−1

≥ kx − (k − 1)y.

(1)

Using inequality (1), we obtain xn+1 x12 x23 + 2 + · · · + nn x2 x3 x1 ≥ (2x1 − x2 ) + (3x2 − 2x3 ) + · · · + (nxn−1 − (n − 1)xn )+ + ((n + 1)xn − nx1 )  2(x1 + · · · + xn ) − nx1 ≥ x1 + · · · + xn , whence

x12 x2

+

x23 x32

+ ··· +

xnn+1 x1n

≥ x1 + · · · + xn .

x2 −x3 xn −x1 ax1 −x2 + ax2 + x3 + · · · + axn + x1 x1 + x2 n √ ≥ (x1 + x2 )+ ···n +(xn + x1 ) n (x1 + x2 )·(x2 + x3 ) ··· (xn +x1 )

3.25. Note that

n

≥ n· 

2

n2 n 

xi

 n

ax1 −x2 x1 + x2

·

ax2 −x3 x2 + x3

···

axn −x1 xn +x1



.

i1

3.26. Using the inequality of Problem 2.1 for the numbers xi + 1 and 1, 1, . . . , 1,    p−1 √  where i  1, . . . , n, we obtain p xi + 1  p (xi + 1) · 1 · ·  · 1 ≤ p−1  √ √ (xi + 1)+(p − 1)·1 xi x1 p p + ··· + x  1 + , whence + 1 + · · · + x + 1 ≤ 1 + 1 n p p p  n 1 + xpn  n + x1 +···+x  n + 1. p

Proofs

35

3.27. (a) Since 1+α ≥ 0, it is enough to prove the inequality for the case 1+nα ≥ 0. According to Problem 2.1, for n > 1, we obtain n (1 + nα) · 1 · · 1 ≤  · n−1

1 + n α + 1 + ··· + 1 n

 1 + α. Therefore, 1 + nα ≤ (1 + α)n . (b) See the proof of Problem 3.27(a). 2

 27 · cos3 3.28. (a) cos6 t sin2 t 4

2 2 2 cos t cos t cos t 2 3 + 3 + 3 +1−cos t 27  4 and therefore cos3 t sin t ≤ (b) We have

t

√ 3 3 . 16

  1 − cos2 t   27 , whence cos3 t · sin t  ≤ 44 cos2 t 3

·

·

cos2 t 3

1   1    m xk 1 − xm  xkm (1 − xm )m  mxk · · · mxk (k − kxm · · · k − kxm ) m · ≤



1 m

m k

·k

·

mxk + · · · + mxk + (k − kxm ) + · · · + (k − kxm ) m+k

 m+k

≤ √ 3 3 , 16

1 m

mk ·k

≤

k

km ·m

m



k

(m + k) m +1

(c) According to Problem 3.28(b), it follows that x(1 − x2 ) ≤

√ 2 3 , 9

.

whence

x y2 z2 y z x2 + + + +  1 − x2 1 − y2 1 − z 2 x(1 − x2 ) y(1 − y2 ) z(1 − z 2 ) √  3 3 9  2 2 2 . ≥ √ x +y +z  2 2 3 Therefore, (d) We have

x 1 − x2

+

y 1 − y2

+

z 1 − z2

≥

√ 3 3 . 2

1 1+y 1+z 1 1 1 x 1 1 1+x + +  + + + + + +  1−x 1−y 1−z 1 − x2 1 − y2 1 − z 2 1 − x2 1 − y2 1 − z 2 1 − x2 √ z 9 x y z y + ≥ + + + ≥ 4.5 + 1.5 3, + 1 − y2 1 − z2 1 − x2 + 1 − y2 + 1 − z 2 1 − x2 1 − y2 1 − z 2

(see Problem 3.28b). Therefore,

1 1−x

+

1 1−y

3.29. From inequality (3.2), it follows that 

+

1 1−z

√ ≥ 4.5 + 1.5 3.

2 f (x)  (1 + x)− n + (1 − x)− n ≥ 2 (1 + x)− n (1 − x)− n  2n√1−x ≥ 2, 2 Since x ∈ [0; 1). On the other hand, f (0)  2. Thus the minimum value of the function f (x) is equal to 2. 1

1

1

1

3.30. Let us represent the function f (x) as f (x)  (m + n) ·

n·

axm n

+m· m+n

b mxn

.

36

3 The HM-GM-AM-QM Inequalities

By Problem 2.1, we  obtain    b m m+n axm n m+n a n bm ·  + n) , where equality holds f (x) ≥ (m + n) (m n mxn nn mm if

axm n



b . mxn

Therefore, x 

 m+n

bn . am

 n m Answer. min f (x)  f (x0 )  (m + n) m + n nan mb m , where x0  (0,+∞)

 m+n

bn . am

3.31. Let us represent the function f (x) as α(b − x)· β(b + x) , where α > 0, β > 0. f (x)  (x − a)· (x − a)·αβ Using Problem 2.1, we deduce that  4 4 (x − a) · (x − a) · α(b − x) · β(b + x) ≤ (x − a) + (x − a) + α(b − x) + β(b + x)   (2 − α + β)x + (α + β)b − 2a.

Note that the right-hand side does not depend on x if α − β  2, and equality holds if x − a  α(b − x)  β(b + x). Thus, it follows that α  x−a , β  x−a . b−x b+x Hence, from the equation α − β  2, we deduce that 2x2 − ax − b2  0. This √ 2 2 equation has only one positive root, x0  a+ a4 +8b . One can easily prove that x0 ∈ [a, b]. Therefore,f (x) attains its maximum value in [a, b] at the point x0 . √ 2 2 Answer. x0  a+ a4 +8b . 3.32. Let us represent the √product xyz as xyz  2√13π · 2x · 3y · π z, and according to Problem 2.1, for the numbers 3  √ √ √ 2x, 3y, π z, we have xyz  2√13π · 2x · 3y · π z ≤ 2√13π · 2x+ 33y+πz  √ √ 1 √ , where inequality holds if 2x  3y  π z. From 2x + 3y + π z  1, 54 3π it follows that x  16 , y  Answer. 54√1 3π .

√

3 ,z 9



1 . 3π

3.33. Note that f (0)  0, and if x < 0, then f (x) < 0. On the other hand, if x > 0, then f (x) > 0. Therefore, the function f (x) attains its maximum value in (0, +∞), and its minimum value in (−∞, 0). First proof. From inequality (3.2) for the numbers ax2 and b, we have √ ax2 + b ≥ |x| · ab, 2 where equality holds if ax2  b. From inequality (3.14), it follows that

x ax2 +b

≤

√1 . 2 ab

(3.14)

Proofs

37

Therefore, the maximum value of the function f (x) is equal to 2√1ab , and it  attains this value at the point x  ab . Since f (x) is an odd function, its minimum value is equal to − 2√1ab , which the function f (x) attains at the point  x  − ab . Alternative proof. In (0, +∞), the function f (x) coincides with the function g(x)  ax 1+ b . This function attains its maximum value at the point where the x

function h(x)  ax + bx attains its minimum value. Since the product ax · bx  ab is constant, the sum ax + bx attains its minimum  value when ax  bx , i.e., when x  ab . Therefore,in (0, +∞), the function f (x) attains its maximum value at the

point x  ab , and it is equal to 2√1ab . One can find the minimum value of the function f (x) similarly as in the first proof. √ 5 (x + 2)(x + 4) + 12 , and thus if x ≥ −2, then x+3 √ 25x + 50 + x + 4 + 12 (25x + 50)(x + 4) + 12 2 ≤  13, and y  13  x+3 x+3 23 + 4, i.e., if x  − 12 . If x ≤ −4, then y < 0.

3.34. (a) We have y 

if 25x + 50  y x Therefore, the maximum value of the function y is equal to 13. Answer. 13. (b) We have  3 y 

   x2 + 1 x2 + 1 25 x2 + 65 · 3x2 + 4 + 1 + 25 x2   3 3x2 + 4

5 · 2

x2

+ 1 + x2

Note that y 

4 15

·

≤

3

5 2

 3

5 , 2

+

6 5

 

3

 

3

5 · 2

    3 x2 + 1 x2 + 1 25 x2 + 65 3x2 + 4

5 12x2 + 16 4   ·  · 2 15 3x2 + 4 15

 3

5 . 2

if x2 + 1  25 x2 + 65 , i.e., if x2  13 .

Therefore, the maximum value of the function y is equal to  4 Answer. 15 · 3 25 .

≤

4 15

·

 3

5 . 2

 x2 + y2 + 3.35. Let PA  x, PB  y, PC  z. Thus, it follows that S  x + y + z + √  1 2 2 2 2 2 2 y + z + z + x and VPABC  6 xyz. Since a + b ≥ 2ab, it follows that √ √ √ √ √  √ √ √ √ S ≥ x + y + z + 2 xy + yz + xz ≥ 3 3 xyz + 3 2 3 xy · yz · xz. 3  Therefore, V ≤ 16 · 3 1 +S √2 , and if x  y  z  3 1 +S √2 , we have ( ) ( ) 3  1 S√ V  6 · 3 1 + 2 . Then the volume of the pyramid with edge lengths ( ) x  y  z  3 1 +S √2 is maximal. ( )

38

3 The HM-GM-AM-QM Inequalities

3.36. From inequality (3.2), it follows that 1 ≥ z 2 + 1, and therefore, z  0, x  y  1. 2  2 2 2 3.37. Note that the inequality x + y3 + z ≥ x + 3y + z becomes an equality, since x + y + z  3 and x2 + y2 + z 2  3. Therefore, x  y  z  1 (see the proof of Problem 2.2). Answer. (1, 1, 1). 3.38. We have

  2 2 2 ≥ a + b +4 c + d (Problem 2.2). It follows that 16−e ≥ 8−e , 4 4 whence 5e − 16e ≤ 0, and therefore, 0 ≤ e ≤ 3.2. If a  b  c  d  1, 2, then e  3.2. Thus, it follows that the maximum value of e is equal to 3.2. Answer. 3.2. a2 + b2 + c2 + d 2 4 2

3.39. We have   1 x1 x3 x5 1 x2 x4 x2 x4 3 1 + + ≥ + + ≥3 · · 33 ≥ 0.3. x2 x4 x6 x2 x4 x6 x2 x4 x6 x6 If x1  1, x2  10, x3  10, x4  102 , x5  102 , x6  103 , then xx21 + xx43 + x5  0.3. x6 Therefore, the minimum value of the expression xx21 + xx43 + xx56 is equal to 0.3. Answer. 0.3. 3.40. From Problem 2.1, for the numbers x 4 , y4 , 1, 1, it follows that  4 4 4 4 x + y + 1 + 1 ≥ 4 x · y4 · 1 · 1  4|x| · |y| ≥ 4xy, whence x4 + y4 + 2 ≥ 4xy, and therefore, x4  y4  1. Answer. (1, 1), (−1, −1). 3.41. We have (xy)2 + (yz)2 + (zx)2  3xyz, whence xyz > 0. On the other hand, xyz ∈ Z, and hence xyz ≥ 1. From Problem 2.1, it follows √ that 3xyz  (xy)2 + (yz)2 + (zx)2 ≥ 3xyz · 3 xyz ≥ 3xyz, and hence xyz  1, and the inequality becomes an equality. It follows that (xy)2  (yz)2  (zx)2 . Therefore, x2  y2  z 2  1. Answer. (1, 1, 1), (1, −1, −1), (−1, 1, −1), (−1, −1, 1). 3.42. (a) Note that for x > 0, α > β ≥ 0, we have xα − xβ ≥ xα−β − 1, since (xα−β − 1)(xβ − 1) ≥ 0.

Proofs

39

Therefore, β

β

x1α + · · · + xnα − (x1 + · · · + xnβ )  (x1α − x1 ) + · · · + (xnα − xnβ ) ≥ α−β

≥ (x1

α−β

− 1) + · · · + (xnα−β − 1)  x1

 n α−β α−β + · · · + xnα−β − n ≥ n x1 · · · xn − n  0.

β

β

It follows that x1α + · · · + xnα ≥ x1 + · · · + xn . (b) If β ≥ 0, then from Problem 3.42(a), it follows that x1α + · · · + xnα ≥ β β x1 + · · · + xn . If β < 0, then β

x1 + · · · + xnβ 

1 −β x1

+ ··· +

1 −β xn

−β

 x2

−β(n−1)

−β(n−1)

−β

· · · xn−β + · · · + x1

−β

· · · xn−1 ≤

−β(n−1)

−β(n−1) x + · · · + xn−1 + · · · + xn + ··· + 1 n−1 n−1 −β(n−1)  x1 + · · · + xn−β(n−1) ≤ x1α + · · · + xnα .

≤

x2

β

β

Therefore, x1α + · · · + xnα ≥ x1 + · · · + xn . 3.43. Proof. We have 4 4 4 (3 − 3z + z 2 ), z 2 + xy > 27 (3 − 3x + x2 ), x2 + yz > 27 (3 − 3y + y2 ). y2 + xz > 27 2 2 2 Summing these inequalities, we deduce that 23(x +y +z )+27(xy+yz +zx) > 24, which leads to a contradiction, since 23  23(x + y + z)2 ≥ 23(x2 + y2 + z 2 ) + 27(xy + yz + zx). 4 (3 − 3x + x2 ). Then Let z 2 + xy ≤ 27 x2 y + y2 z + z 2 x  x(z 2 + xy) + y2 z ≤

 4 0, 5y + 0, 5y + z 3 4  x(3 − 3x + x2 ) + 4 , 27 3 27

4 and therefore, x2 y + y2 z + z 2 x ≤ 27 . Alternative proof. Without loss of generality one can assume that max(x, y, z)  x, in which case x2 y +y2 z +z 2 x ≤ x2 y +xyz +0.5z 2 x +0.5zx2  0.5x(x + z)(2y + z). According to Problem 2.1 (AM-GM inequality), we obtain

x2 y + y2 z + z 2 x ≤ 0.5x(x + z)(2y + z) 

4 x + (x + z) + (2y + z) 3 ,  ≤ 0.5 3 27 and therefore, x2 y + y2 z + z 2 x ≤

4 . 27

40

3 The HM-GM-AM-QM Inequalities

3.44 (a) Note that 1+a 1+b 1+c 1+d cd + acd ad + adb 1+c 1+d + + +  + + +  1 + ab 1 + bc 1 + cd 1 + da cd + abcd ad + abcd 1 + cd 1 + da  cd (1 + ab) c(1 + ad ) d (1 + ab) 1+ +1+ ≥2+2  4, 1 + cd 1 + da 1 + cd

whence

1+a 1+ab

1+b 1+bc

+

+

1+c 1+cd

1+d 1+da

+

≥ 4.

(b) Note that 1 + ab 1 + bc 1 + cd 1 + da cd + abcd 1 + bc 1 + cd bc + abcd + + +  + + + ≥ 1+a 1+b 1+c 1+d cd + acd 1+b 1+c bc + bcd 4 4b 4 + (1 + bc)  (1 + cd ) + ≥ (1 + cd ) cd + acd + 1 + c 1 + b + bc + bcd bcd + abcd + b + bc 4(b(1 + cd ) + 1 + bc) 4   4, + (1 + bc) 1 + b + bc + bcd 1 + b + bc + bcd

and therefore,

1 + ab 1+a

+

1 + bc 1+b

+

1 + cd 1+c

+

1 + da 1+d

≥ 4.

3.45. Note that (c − b)(c − d ) + (e − f )(e − d ) + (e − f )(c − b) < 0, and therefore, (bd +df +fb)−(ac+ce+ea) < (c+e)(b+d +f −a−c−e), or τ −σ < s(T −S), where τ  bd + df + fb, σ  ac + ce + ea, s  c + e. We have Sτ + T σ  S(τ − σ )+ (S + T )σ < Ss(T − S) + (S + T )(ce + as) ≤   2 ≤ Ss(T − S) + (S + T ) s4 + as  s 2ST − 34 (S + T )s . It fol    3 3 lows that (S + T )(Sτ + T σ ) < (S + T )s · 2ST − 43 (S + T )s ≤ 4 4    1 3 (S + T )s + 2ST − 43 (S + T )s  ST . 2 4  Therefore, 3(S + T )(S(bd + df + fb) + T (ac + ce + ea)) < 2ST . 3.46. Note that a +  4 a·

√ √ √ 3 4 ab + abc + abcd a+b 2

 

4

1·



+

·

a+b+c 3

·

a+b+c+d 4

3a 4a 2a · · + a +b a +b+c a +b+c +d

+ + +

 4

1·

3b 4b 2a · · + a +b a +b+c a +b+c +d

2b 2b 3a 3b 3c 2b · · · · · · a +b a +b a +b a +b+c a +b+c a +b+c a

3c 4d 1 2a 3a 2b 4 · · ≤ 1+ + + 1· a +b a +b+c a +b+c +d 4 a +b a +b+c

 2a 3b 4b 1 1+ + + + 4 a +b a +b+c a +b+c +d

1 6b 3a 3b 3c 12c 3+ + + + + 12 a +b a +b+c a +b+c a +b+c a +b+c +d

 2b 3c 4d 1 1+ + +  4. 4 a +b a +b+c a +b+c +d 12

1·1·1·

 +



Therefore, a +

√

ab +

4c 4c 4c · · + +b+c +d a +b+c +d a +b+c +d  4a + a +b+c +d

 +

 √ √ 3 4 a +b a +b+c a +b+c +d abc + abcd 4 ≤ a· · · . 4 2 3 4

3.47. Without loss of generality one can assume that a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0.

Proofs

41

Let x, y, z be arbitrary positive numbers. Then according to Problem 2.1 (AM-GM inequality), we obtain 1 1 a12 + (ab)6 + (abc)4 + (abcd )3  a12 + 6 (xa · b)6 + 4 8 (xya · yb · c)4 + x x y 1 1 + 3 6 9 (xyza · yzb · zc · d )3 ≤ a12 + 6 (x12 a12 + b12 )+ x y z 2x 1 + 4 8 (x12 y12 a12 + y12 b12 + c12 )+ 3x y 1 + 3 6 9 (x12 y12 z 12 a12 + y12 z 12 b12 + z 12 c12 + d 12 )  A(a12 + b12 + c12 + d 12 ). 4x y z

We take the numbers x, y, z such that 1+

x8 y4 x9 y6 z 3 1 y4 y6 z 3 x6 + +  + + 2 3 4 2x6 3x4 4x3 1 z3 1  + 3 6   A. 4 8 3 3x y 4x y 4x y6 z 9

Therefore, x12 

1



, √ 2 3 3 A( A(A − 1) − 0,5)

1 −

1 , A

y12



1 −

√ 1 , 2 A(A − 1)

z 12



1 −

and

  3  2 256 3 A 3 A A(A − 1) − 0, 5 − 1  1. 27

(1)

Consider the following function:

 3 √ 2 3 256 f (t)  27 t 3 t t(t − 1) − 0, 5 − 1 − 1 in [1, 42; 1, 43]. Note that 3    2 256 3 · 1, 42 3 1, 42 1, 42 · 0, 42 − 0, 5 − 1 − 1 < 27 3

 256 256 3 < · 1, 42 3 1, 42 · 0, 27232 − 1 − 1 < · 1, 42(3 · 0, 4723 − 1)3 − 1 < 0 and 27 27 3   2  256 3 f (1, 43)  · 1, 43 3 1, 43 1, 43 · 0, 43 − 0, 5 − 1 − 1 > 27 3   256 256 > · 1, 43 3 3 1, 43 · 0, 08 − 1 − 1 > · 1, 43(3 · 0, 48 − 1)3 − 1 > 0. 27 27

f (1, 42) 

Therefore, there exists a number A such that 1, 42 < A < 1, 43 and f (A)  0. Then (1) holds. It follows that x > 0, y > 0, z > 0 (A > 1.42) and a12 + (ab)6 +(abc)4 +(abcd )3 ≤ A(a12 +b12 +c12 +d 12 ) ≤ 1, 43(a12 +b12 +c12 +d 12 ).

42

3 The HM-GM-AM-QM Inequalities

Problems for Independent Study Prove the following inequalities (1–16, 20–26). p

q

1. ab ≤ ap + bq , if p1 + q1  1, a > 0, b > 0, p > 0, q > 0 where p and q are rational nnumbers.  2. 1 + 1n > 2, where n ∈ N. n 3. (1 + a1 ) · · · (1 + an ) ≤ 1 + 1!S + · · · + Sn! , where n ≥ 2, S  a1 + · · · + an , ai > 0, n.   i 1, . . . , 4. 1 + a1 1 + 1b 1 + 1c ≥ 64, where a > 0, b > 0, c > 0 and a + b + c  1. √ √ n n a2n−k + a2n+k ≥ 3a − 1, where n ≥ 2, a > 0, n > k, n, k ∈ N. 5. n(n+1) n 6. ana(a−−11) ≥ n + 1 − a 2 , where a > 0, a  1. n+1 1 ≥ (n  + 1)an , where a > 0.  7. na √ +√ √ √ √ √ √ √ √ 8. k + k +1 k + 1 + k + 2 ··· n+ n+1 ≥ n− k n+ √ k − 1 + 2, where n > k, n, k ∈ N. 9. aa21 + · · · + aan−1 + aan1 ≥ n, where ai > 0, i  1, . . . , n. n 1 10. an+1 + a1 (a2 −a1 )(a3 −a ≥ n + 2, where 0 < ak < ak+1 , k  1, . . . , n. 2 )···(an+1 −an ) x 1 11. 1 + 2 ≤ √1 − x , where 0 ≤ x < 1.

2 π . 12. sin(2α) < 3α−α 3 , where 0 < α <  a 4  b 4  c 4  d 4  e 4 2 a b c d e 13. b + c + d + e + a ≥ b + c + d + e + a , where abcde  0.  1999  b 1999  c 1999  d 1999 14. ab + c + d + a ≥ ab + bc + dc + da , where a > 0, b > 0, c> 0,d >  0.   √ 15. (a) a1 a+3 a2 + a2 a+4 a3 + · · · + an−1a1+ an + an a+2 a1 ≥ n 2, where n > 2 and a1 > 0, . . . , an > 0. √ (b) 1 +xx2 + 1 +yy2 + 1 +z z2 ≤ 3 4 3 , where x2 + y2 + z 2  1.    n 16. a12 − 1 · · · a12 − 1 ≥ n2 − 1 , where n ≥ 2, a1 > 0, . . . , an > 0, n 1 and a1 + · · · + an  1. 17. Find the maximum and minimum values of the expression (1 + u)(1 + v)(1 + w) 7 7 7 , 0 < v ≤ 16 , 0 < w ≤ 16 , and u + v + w  1. if 0 < u ≤ 16 18. Find the maximum value of the expression xp yq if x + y  a, x > 0, y > 0, and p, q ∈ N . 19. Find the maximum value of the expression a + 2c if for all x, one has ax2 + bx + c ≤ √1 1− x2 , where |x|< 1. √ Hint. Take x  √12 and x  − √12 . Then it follows that a + 2c ≤ 2 2. √ √ 1 Prove that if a  2, b  0, c  √12 , then 2x2 + √12 ≤ √1−x for |x|< 1. 2      a b c a+b+c 20. 1 + b 1 + c 1 + a ≥ 2 1 + √3 , where a > 0, b > 0, c > 0. abc     Hint. 1 + ab 1 + bc 1 + ac  2 + ab + ab + bc + bc + ac + ac and 13 ab + 13 ab + 13 bc ≥  3 a2  √3 a . bc abc

Problems for Independent Study

43

· 11 −+ aa22 · · · 11 −+ aan+1 ≥ nn+1 , where −1 < a1 , a2 , . . . , an+1 < 1 and a1 + a2 + n+1 · · · + an+1 ≥ n − 1. Hint. We have that 1+ai ≥ (1−a1 )+ · · · +(1−ai−1 )+(1−ai+1 )+ · · · +(1−an+1 ). 22. (a + b)3 (b + c)3 (c + d )3 (d + a)3 ≥ 16a2 b2 c2 d 2 (a + b + c + d )4 , where a > 0, b > 0, c > 0,d > 0. Hint. We have (a + b + c + d )2  (a + b)(b + c) + (a + b)(d + a) + (b + c)(c + d ) + (c + d )(d + a).  2 2 2  2 2 2     23. 1 + ab + 1 + bc + 1 + ac 1 + ab + 1 + bc + 1 + ac ≥ a+b  b+c c+a 2 4 c + a + b ,where a > 0, b > 0, c > 0.  2  2  2  2 Hint. We have 1 + ab + 1 + bc + 1 + ac ≥ 13 1 + ab + 1 + bc + 1 + ac .

21.

1 + a1 1 − a1

24. (a2 + bc)3 (b2 + ac)3 (c2 + ab)3 ≥ 64(a3 + b3 )(b3 + c3 )(c3 + a3 )a3 b3 c3 , where a > 0, b > 0, c > 0. Hint. We have (a2 + bc)(b2 + ac)  c(a3 + b3 ) + ab(ab + c2 ). √ √ 3 25. (a) a + ab + abc ≤ 43 (a + b + c), where a > 0, b > 0, c > 0.  √ √ 3 (b) a + ab + abc ≤ 3 3 a · a +2 b · a + 3b + c , where a > 0, b > 0, c > 0. 5

5

5

26. (ab) 4 + (bc) 4 + (ca) 4 ≤

√ 3 , 9

where a > 0, b > 0, c > 0 and a + b + c  1.

Chapter 4

The Cauchy–Bunyakovsky–Schwarz Inequality

Historical origins. The Cauchy–Bunyakovsky–Schwarz inequality, also known as the Cauchy–Schwarz inequality, is one of the most important inequalities in mathematics. The inequality for sums was published in 1821 by the French mathematician Augustin-Louis Cauchy, born 21 August 1789 in Paris, France, died 23 May 1857 in Sceaux, one of the wealthy suburbs of Paris. Despite the fact that the influential French mathematician Pierre-Simon Laplace and the Italian–French mathematician Joseph-Louis Lagrange were Cauchy’s father’s friends and that on Lagrange’s advice, Cauchy was enrolled in the École Centrale du Panthéon in Paris, France, Cauchy never obtained a doctorate in mathematics. Besides being a prolific writer, Cauchy was awarded the title of baron (in return for his services to King Charles X of France, who asked Cauchy to be tutor to his grandson, the duke of Bordeaux). The names of Cauchy, Laplace, and Lagrange are among the 72 names inscribed on the Eiffel Tower. Cauchy was an advisor of 2 doctoral students, including a well-known Russian mathematician Viktor Bunyakovsky. The corresponding inequality for integrals was first proved in 1859 by Viktor Bunyakovsky, born 16 December 1804 in Bar, Russian Empire (now Bar, Ukraine), died 12 December 1889 in Saint Petersburg, Russian Empire (now Russia). In 1824, Bunyakovsky received his bachelor’s degree from the University of Paris (the Sorbonne), Paris, France, and continuing his studies, he wrote three doctoral dissertations under Cauchy’s supervision and obtained his doctorate in 1825. The modern proof of the integral inequality was given in 1888 by the Prussian mathematician Karl Hermann Amandus Schwarz, born 25 January 1843 in Hermsdorf, Prussia (now Jerzmanowa, Poland), died 30 November 1921 in Berlin, Germany. He obtained his doctorate from the University of Berlin in 1864 under the supervision of the well-known mathematicians Ernst Kummer and Karl Weierstrass. Schwarz was the advisor to 20 doctoral students.

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_4

45

46

4 The Cauchy–Bunyakovsky–Schwarz Inequality

There are several generalizations of this inequality, one of which is known as Hölder’s inequality, named after the German mathematician Otto Ludwig Hölder, born 22 December in Stuttgart, Germany, died 29 August 1937 in Leipzig, Germany. In 1877, Hölder began his studies at the University of Berlin, where he was a student of the well-known mathematicians Leopold Kronecker, Ernst Kummer, and Karl Weierstrass, and he obtained his doctorate under the supervision of Paul du BoisReymond from the University of Tübingen in 1882. Hölder was the advisor of 58 doctoral students. In this chapter we study some inequalities whose proofs involve the Cauchy–Bunyakovsky–Schwarz inequality. But first, let us prove the simplest case of the Cauchy–Bunyakovsky–Schwarz inequality for arbitrary numbers a1 , a2 , b1 , b2 .  1 , b2 ).  Consider the vectors a (a1 , a2 ) and b(b      It is known that a · b  a1 b1 + a2 b2  | a | · b cos a , b .  We have Let us estimate the absolute values of the scalar product a · b.                  a | · b. a | · b · cos a , b  ≤ | a · b  | On the other hand,           a · b  |a1 b1 + a2 b2 | ≤ | a | · b  a12 + a22 · b12 + b22 , or     (a1 b1 + a2 b2 )2 ≤ a12 + a22 b12 + b22 .

(4.1)

Inequality (4.1) is the simplest formulation of the Cauchy–Bunyakovsky–Schwarz inequality for arbitrary numbers a1 , a2 , b1 , b2 . Note that in (4.1), equality holds if and only if a1 b2 − a2 b1  0. The following generalization of inequality (4.1) for arbitrary numbers a1 , . . . , an , b1 , . . . , bn , is called the Cauchy–Bunyakovsky–Schwarz inequality:    2 a1 + · · · + an2 b12 + · · · + bn2 ≥ (a1 b1 + · · · + an bn )2 .

(4.2)

First, let us prove inequality (4.2) for the numbers a1 ≥ 0, . . . , an ≥ 0, b1 ≥ 0, . . . , bn ≥ 0.    Proof Let xk  a12 + · · · + ak2 b12 + · · · + bk2 , where k  1, . . . , n. In this case,   2  2 2 xk+1  a12 + · · · + ak2 + ak+1 b1 + · · · + bk2 + bk+1      

2

2 2 2 2 2 2 2

 a1 + · · · + ak + ak+1 b1 + · · · + bk + bk+1 ≥ 

2   2 2 2 2 a1 + · · · + ak · b1 + · · · + bk + ak+1 · bk+1  xk + ak+1 bk+1 . ≥

4 The Cauchy–Bunyakovsky–Schwarz Inequality

47

Therefore, we obtain xk+1 ≥ xk + ak+1 bk+1 , where k  1, . . . , n − 1. Summing these inequalities, we deduce that    a12 + · · · + an2 b12 + · · · + bn2 ≥ a1 b1 + · · · + an bn , or  2   a1 + · · · + an2 b12 + · · · + bn2 ≥ (a1 b1 + · · · + an bn )2 . Now let us consider the case in which a1 , . . . , an , b1 , . . . , bn are arbitrary real numbers. In this case,       2 a1 + · · · + an2 b12 + · · · + bn2  |a1 |2 + · · · + |an |2 |b1 |2 + · · · + |bn |2 ≥ ≥ (|a1 b1 | + · · · + |an bn |)2 ≥ |a1 b1 + · · · + an bn |2  (a1 b1 + · · · + an bn )2 . This ends the proof. Alternative proof. n   2   2  2 2 2 2 ai b j − bi a j ≥ 0. a1 + · · · + an b1 + · · · + bn − (a1 b1 + · · · + an bn )  i; j1 i≥ j

This ends the proof. Let us consider the following examples. Example 4.1 Prove that sin α · sin β + cos α + cos β ≤ 2. Proof According to the Cauchy–Bunyakovsky–Schwarz inequality, we have sin α · sin β + cos α · 1 + 1 · cos β ≤

  sin2 α + cos2 α + 12 · sin2 β + cos2 β + 12  2.

This ends the proof. Example 4.2 Prove that a 3 + b3 > a 2 + b2 if a > 0, b > 0, and a 2 + b2 > a + b. Proof According to the Cauchy–Bunyakovsky–Schwarz inequality, we have             3  2 3 2 3 2 1 2 1 2 3 1 3 1 2 a2 + b2 ≥ a2 ·a2 +b2 ·b2  a 2 + b2 . a + b3 (a + b)  a 2 + b 2 3

3

2

2

+b Since aa 2 +b ≥ aa+b > 1, it follows that a 3 + b3 > a 2 + b2 . +b2 This ends the proof.

48

4 The Cauchy–Bunyakovsky–Schwarz Inequality

Problems Prove the following inequalities (4.1–4.19, 4.22, 4.23). 4.1. a 2 + b2 + c2 ≥ 14 if a + 2b + 3c ≥ 14.    4.2. ab + 1 − a 2 1 − b2 ≤ 1 if |a| ≤ 1, |b| ≤ 1 . √ √ √ c(a − c) + √c(b − c) ≤ ab if a > c, b > c, c > 0. 4.3. √ 4.4. a a 2 + c2 + b b2 + c2 ≤ a 2 + b2 + c2 . 4.5. √1ab + √1bc + √1ca ≤ a1 + b1 + 1c , where a > 0, b > 0, c > 0.  √  √ √ 4.6. a(a + c − b) + b(a + b − c) + c(b + c − a) ≤ a 2 + b2 + c2 (a + b + c), where a, b, c are of a triangle.  the side lengths  1 1 4.7. (a1 + · · · + an ) a1 + · · · + an ≥ n 2 , where a1 > 0, . . . , an > 0. 2  a 2 +···+a 2 n 4.8. 1 n n ≥ a1 +···+a . n 2  1. 4.9. a1 a2 + a2 a3 + · · · + a9 a10 + a10 a1 ≥ −1 if a12 + · · · + a10 4 4 3 3 4.10. x + y ≥ x y + x y . 2  3  4.11. |a1 |3 + · · · + |an |3 ≤ a12 + · · · + an2 .       3 a 2 + b2 + c2 + x 2 + y 2 + z 2 + 6 a 2 + b2 + c2 x 2 + y 2 + z 2 4.12. ≥ (a + b + c + x + y + z)2 . 4.13. a 2 + b2 + c2 ≥ab + bc + ca.    4.14. (a1 + · · · + an ) a17 + · · · + an7 ≥ a13 + · · · + an3 a15 + · · · + an5 , where a1 > 0, . . .√, an > 0.√ √ a + 1 + 2a − 3 + 50 − 3a ≤ 12, where 23 ≤ a ≤ 50 , 4.15. (a) 3 (b) a + b + c ≤ abc + 2, where a 2 + b2 + c2  2, (c) 2(a + b + c) − abc ≤ 10, where a 2 + b2 + c2  9, (d) 1 + abc 3min(a, where a 2 + b2 + c2  9.

≥

b, c), n n n  k+1  −1  k ≥n ai ai ai , where k, n ∈ N and a1 > 0, . . . , an > 0. 4.16. i1 i1 i1 √ 4.17. a+b+c ≥ 3 abc, where a > 0, b > 0, c > 0. 3 k  k a +···+a k n , where k, n ∈ N and a1 ≥ 0, . . . , an ≥ 0. 4.18. 1 n n ≥ a1 +···+a n    1 1 4.19. 1 + sin α 1 + cos α > 5, where 0 < α < π2 . 4.20. Find the distance from a point A(x0 , y0 ) to the line defined by the equation ax + by + c  0 (a 2 + b2  0). 2 √ 4.21. Find the smallest possible value of the expression (u − v)2 + 2 − u 2 − v9 √ if 0 < u < 2, v > 0.  2  2 n 2 + · · · + x1 +···+x ≤ 4 x12 + · · · + xn2 . 4.22. x12 + x1 +x 2 n This inequality is a particular case of Hardy’s inequality p   n  n   p p a1 +···+ak p ≤ p−1 · ak , where p > 1, ai ≥ 0, i  1, . . . , n. k k1 k1   2 n 1 1 , where a1 > 0, . . . , an > 0, + . . . + < 4 + . . . + 4.23. (a) a11 + a1 +a a1 +...+an a1 an 2

Problems

49

(b)

1 a1

2 n + a1 +a + · · · + a1 +···+a <2 2 n



1 a1

+ ··· +

1 an

 , where a1 > 0, . . . , an > 0.

Proofs    4.1. From (4.2) we have that a 2 + b2 + c2 12 + 22 + 32 ≥ (a + 2b + 3c)2 ≥ 142 , and therefore, a 2 + b2 + c2 ≥ 14.  2 √ √ √ 2 2 4.2. We have a · b + 1−a · 1−b ≤ a2 + 1 − a2 ·  2 √ b2 + 1 − b2  1. 4.3. We have 

 √ √ √ √ c(a − c) + c(b − c)  c · a − c + b − c · c ≤  2 √ √ 2 √ 2 √ 2 √ c + b−c · a − c + c  ab. ≤

4.4. Note that √ √ a · a 2 + c2 + b2 + c2 · b ≤

4.5. √1a · √1b + 4.6. Note that

√1 b

·

√1 c

+

√1 c

 a2 +

√

b2 + c2

2

 ·

 a 2 + b2 + c2 .   1 · √1a ≤ + b1 + 1c b1 + a

1 c

√ 2 a 2 + c2 + b2

+

1 a





1 a

+

1 b

+ 1c .

√ √ √ a(a + c − b) + b(a + b − c) + c(b + c − a)         a(a + c − b) (a + c − b) + b(a + b − c) (a + b − c) + c(b + c − a) (b + c − a)   ≤ (a(a + c − b) + b(a + b − c) + c(b + c − a)) · (a + c − b) + (a + b − c) + (b + c − a)    a 2 + b2 + c2 (a + b + c).

4.7. We have 

2

2 

2 √  √ 2  √ 1 1 1 1 √ 2 a1 + · · · + an + ··· + √ a1 · √ + · · · + an · √  n2 . ≥ √ a1 an a1 an

⎛ ⎞   2 4.8. We have a1 + · · · + an2 ⎝12 + · · · + 12 ⎠ ≥ (a1 + · · · + an )2 , whence n

a12 + · · · + an2  a1 + · · · + an 2 . ≥ n n

50

4 The Cauchy–Bunyakovsky–Schwarz Inequality

4.9. We have   2  2 2 |a1 a2 + . . . + a9 a10 + a10 a1 | ≤ a12 + . . . + a92 + a10 a2 + . . . + a10 + a12  1, hence a1 a2 + · · · + a9 a10 + a10 a1 ≥ −1. 4.10. We have     x 4 + y 4  x 4 + y 4 · x 4 + y 4 ≥ x 4 + y 4 · 2x 2 y 2   2  2  x 2 + y 2 · (x y)2 + (x y)2 ≥ x 3 y + x y 3 .  4.11. We have

 2    2      3   a1  + · · · + an3   |a1 | · a12 + · · · + |an | · an2 ≤ a12 + · · · + an2 · a14 + · · · + an4   2  3 ≤ a12 + · · · + an2 a12 + · · · + an2  a12 + · · · + an2 .

4.12. Note that the left-hand side of the inequality 2 √  3 a 2 + b2 + c2 + x 2 + y 2 + z 2  B. According to the inequality of Problem 4.8, we have

is

equal

to

⎛

⎞2

2   2 a + b + c x + y + z ⎠  (a + b + c + x + y + z)2 . B ≥ 3⎝ 3 + 3 3 3 4.13. ab + bc + ca ≤ 4.14. We have

   a 2 + b2 + c2 b2 + c2 + a 2  a 2 + b2 + c2 .



 2     √  √ 2   5 2 2 5 5 a1 + · · · + an a1 + · · · + an5 (a1 + · · · + an ) a1 + · · · + an   2 ≥ a13 + · · · + an3 .

   ≥ In a similar way, we deduce that a17 + · · · + an7 a13 + · · · + an3   5 2 a1 + · · · + an5 .   Multiplying these inequalities, we obtain (a1 + · · · + an ) a17 + · · · + an7 ≥    3 a1 + · · · + an3 a15 + · · · + an5 . 4.15. (a) We have 1·

√ √ √ α + 1 + 1 · 2α − 3 + 1 · 50 − 3α ≤  2 √ 2 √ 2   √  12. 12 + 12 + 12 α+1 + 2α − 3 + 50 − 3α ≤

(b) We have a · (1 − bc) + (b + c) · 1 ≤

  (a 2 + (b + c)2 )((1 − bc)2 + 12 )  4(1 − b2 c2 ) + 2(1 + bc)b2 c2 ≤ 2,

Proofs

51 2

2

since bc ≤ b +c ≤ 1. 2 2 2 2 2 (c) Let a  max(a , b2 , c2 ); hence a 2 ≥ 3 and x  bc ≤ b +c ≤ 3, and 2 therefore,  a · (2 − bc) + (b + c) · 2 ≤ (a 2 + (b + c)2 )((2 − bc)2 + 22 )   100 + (2x − 7)(x + 2)2 ≤ 10. (d) Let a ≤ b ≤ c. If 0 ≤ a ≤ 1, then a(b − a)(c − a) ≥ 0 and 9  a 2 + b2 + c2 ≤ bc + b2 + c2 ≤ (b + c)2 . Hence 1 + abc ≥ 1 − a 3 + a 2 (b + c) ≥ (1 − a) · 3a + a 2 · 3  3a, whence 1 + abc ≥ 3a. If a > 1 and c ≥ 2, then 1 + abc ≥ 1 + 2a 2 > 3a, and thus 1 + abc > 3a. If a > 1 and c < 2, then a 2 + b2 > 5, whence b2 > 2, 5 and 1 + abc ≥ 1 + ab2 > 1 + 2, 5a > 0, 5c + 2, 5a ≥ 3a, and therefore, 1 + abc > 3a. If a < 0 and bc ≤ 3, then 1 > 0 ≥ a(3−bc), and therefore, 1+abc > 3a. 2 , If a < 0 and bc > 3, since 9  a 2 + b2 + c2 ≥ a 2 + 2bc, or bc ≤ 9−a 2 it follows that −a(bc − 3)  |a|(bc − 3) ≤ |a|

(|a| − 1)2 (|a| + 2) 3 − a2 1− ≤ 1, 2 2

whence 1 + abc ≥ 3a. n  4.16. Define aiS  A S . From inequality (4.2), it follows that i1



Ak+1 · Ak−1

n   k+1 2 ai 2  i1

 n   n 2   k−1 2  k+1 k−1 2 2 2 ai ai · ai  A2k . ≥ i1

i1

In a similar way, we obtain Ak · Ak−2 ≥ A2k−1 , Ak−1 · Ak−3 ≥ A2k−2 , . . . , A2 · A0 ≥ A21 , A1 · A−1 ≥ A20 . Multiplying these inequalities, we deduce that Ak+1 · A−1 ≥ Ak · A0 , or  n  n  n    −1 k+1 ai ai aik . ≥ i1

i1

i1

4.17. We have (a + b + c)(b + c + a)  ≥



√ 2 √ 2 √ 2 √ 2 √ 2 √ 2 ≥ a + b + c b + c + a

√ √ √ 2 ab + bc + ca ,

4 √ √ √ 2  √ √ √ √ 2 √ √ 4 4 4 ab + bc + ca c+ a+ b ≥ abc + abc + abc , (a + b + c)(1 + 1 + 1) ≥

√ √ √ 2 a+ b+ c .

Multiplying these inequalities, we obtain that (a + b + c)3 ≥ 27abc, or

52

4 The Cauchy–Bunyakovsky–Schwarz Inequality a+b+c √ 3 ≥ abc. 3

4.18. We have that Ak · Ak−2 ≥ A2k−1 , Ak−1 · Ak−3 ≥ A2k−2 , . . . , A2 · A0 ≥ A21 , therefore multiplying these inequalities, we obtain Ak · A0 ≥ Ak−1 · A1 , Ak−1 · A0 ≥ Ak−2 · A1 , . . . , A2 · A0 ≥ A1 · A1 , A1 · A0 ≥ A0 · A1 .

Multiplying these inequalities, we deduce that Ak · Ak0 ≥ A0 · Ak1 , or

Here Ak 

n  i1

 a + . . . + a k a1k + . . . + ank 1 n ≥ . n n

aik .

4.19. We have 

2 

2 

2 1 1 1 2 2  1 + √ ≥ 1+ √ 1 + √ cos α sin α sin α cos α ⎛ ⎞2 ⎛ ⎞2  √ 2 1 1 ⎠ ≥ ⎝1 +  ⎠  1 + 2 > 5.  ⎝1 +  sin 2α 2

1 2

4.20. Let M(x, y) be a point on the line defined by the equation ax + by + c  0. Let us estimate the minimum value of the distance AM   (x − x0 )2 + (y − y0 )2 .    According to inequality (4.1), we have a 2 + b2 (x − x0 )2 + (y − y0 )2 ≥ 2 (a(x − x√ 0 ) + b(y − y 0 )) , whence a 2 + b2 · (x − x0 )2 + (y − y0 )2 ≥ |ax + by − ax0 − by0 |, or  2 0 +c| . (x − x0 ) + (y − y0 )2 ≥ |ax√0a+by 2 +b2 0 0  y−y . In the last inequality, equality holds if x−x a b Solving the following system: ⎧ ⎨ ax + by + c  0, y − y0 x − x0 ⎩  . a b

0 +c 0 +c We obtain x  −a axa0 +by + x0 , y  −b axa0 +by + y0 . 2 +b2 2 +b2 Therefore, the distance from point A to the given line is equal to 4.21. We have

|ax0 +by0 +c| √ . a 2 +b2

 2 9 2 9 9 2 − u2 −  2 − 2 uv + 2 − u 2 + v2 + ≥ v v v   2 2  2   √ 2 √ 2 9 9 81 √ + v2 + 2 − u2  v2 + v2 + 2 − 2 ≥ 2 · 9 − 2  8. ≥ 2 − 2 u 2 + v v v

(u − v)2 +

Proofs

53

2  If u  1, v  3, then (u − v)2 + 2 − u 2 − v9  8; hence the minimum value of the given expression is equal to 8. 4.22. We have



1 1 1 1 1 , + a12 √ − √ + · · · + a12 √ − √ a12 ≥ a12 1 − √ n n+1 2 2 3

√ √ √ 1 1 1 1 1 1 , + a22 2 √ − √ + · · · + a22 2 √ − √ a22 ≥ a22 2 √ − √ n n+1 2 3 3 4 .......... ......... ......... .......

√ 1 1 an2 ≥ an2 n √ − √ . n n+1

Therefore,





√  1 1 1 + a12 + a22 2 √ − √ + ···+ a12 + a22 + · · · + an2 ≥ a12 1 − √ 2 2 3

  √ √ 1 1 + a12 + a22 2 + · · · + an2 n √ − √ . n n+1

Let us prove that √ √ √  1  a 1 + · · · + a k 2 k +1− k 2 a1 + · · · + ak2 k ≥ . √ √ 4 k k· k+1

(4.3)

According to inequality (4.2), it follows that

 √  1 a12 + · · · + ak2 k 1 + · · · + √ ≥ (a1 + · · · + ak )2 . k √

√

4 k+1− k In order to prove inequality (4.3), it is sufficient to prove that ( √k √k+1 ) k 2 >

1 + ··· +

√1 . k

We have

1 2 1 2 1 + √ + ··· + √ ≤ 1 + √  + ··· + √ √ k k+ k−1 2 2+1  √  √ √ √  1 + 2 2 − 1 + · · · + 2 k − k − 1  2 k − 1. Let us prove that √

2 k−1<

Since that

√ 2k √ k+1+ k

4k 2


√

√  √ k+1− k 4k k .  √ √ √ k k+1 k + 1 + k2 + k

√ 2 √ √ k − 1 < k + k 2 + k − 2 k + 1, it follows

54

4 The Cauchy–Bunyakovsky–Schwarz Inequality

√ √ √ < k + k 2 + k − 2 k + 1, or k+1+ k √ √ √  √ 2k k + 1 − k < k + k 2 + k − 2 k + 1, or    √ √ √  √ 2 k k + 1 + k 2 + k − k + 1 + k 2 + k < 4k k. 2k

√

Hence we deduce that √ 2 k−1<

√ 4k k . √ k + 1 + k2 + k

Now let us prove that on the right-hand side, it is impossible to choose a multiplier smaller than 4. 2 2     n 2 +· · ·+ x1 +···+x ≤ c x12 + x22 + · · · + xn2 Suppose the inequality x12 + x1 +x 2 n holds for all x1 , . . . , xn . Taking xi  √1i , i  1, . . . , n, we deduce that  1+

√1 2

1+



2 + ··· +

2

1+

√1 2

+ ··· +

√1 n

n

2

1 1 . ≤ c 1 + + ··· + 2 n

According to the inequality of Problem 14.16, we have 2 √ √ n     1/1 1+···+1/n n 1 1 √2 − 2 , whence 4 − 8 < c 1 + + · · · + < c. k 2 n 1+1/2+···+1/n k

k1

Passing to the1 limit,1we

obtain √ +···+ √ lim 4 − 8 11+1 1 +···+n 1 n ≤ lim c. Since

n→∞

2

n

n→∞

1 √ +···+ n √1 n 1 1 1+ 21 +···+ n1

lim  0, we must have 4 ≤ c. n→∞ 4.23. (a) The proof follows straightforwardly from the proof of part (b). (b) We have





1 1 1 1 1 1 1 1 1 1− 2 + , > − 2 + ··· + − a1 a1 2 a1 2 2 3 a1 n 2 (n + 1)2







2 2 2 1 2 1 2 1 1 1 2 1 1 1 n2 1 1 ,..., . > − 2 + − 2 + ··· + − > − a2 a2 2 2 3 a 2 32 4 a2 n 2 an an n 2 (n + 1)2 (n + 1)2

Therefore, 2

1 1 1 + + ··· + a1 a2 an

Let us prove that



2

1 k2 + ··· + a1 ak





1 1 1 +2 1− + a1 a1 22   1 1 n2 +2 − + ··· + a1 an n2 >2

1 1 − k2 (k + 1)2

>

22 a2



1 1 − 22 32

1 . 2 (n + 1)

k , a1 + · · · + ak

+ ···

Proofs

55

or equivalently, (a1 + · · · + ak )

1 k2 + ··· + a1 ak

>

k 3 (k + 1)2 . 2(2k + 1)

We have ⎛ ⎞2 

1 k 3 (k + 1)2 k2 1 k 2 (k + 1)2 k2 > . ≥ ⎝ a1 · + ··· + + ··· + · ak ⎠  (a1 + · · · + ak ) a1 ak a1 ak 4 2(2k + 1)

It follows that





n 1 1 1 1 k2 1 > 2 + ··· + 2 + ··· + − a1 an a1 ak k2 (k + 1)2 k1 >

n  k1

k . a1 + · · · + ak

Now let us prove that on the right-hand side, it is  choose a multiplier  impossible to n 1 1 . ≤ c + · · · + smaller than 2. Indeed, let a11 + · · · + a1 +···+a a1 an n −2 Assume that ak  k, k  1, . . . , n, and thus 2xn+1 ≤ c, where xn  1 + xn   xn+1 1 1 · · · + n . Therefore, 2 lim xn − xn ≤ c, or 2 ≤ c. This ends the proof.

1 2

+

n→∞

Problems for Independent Study Prove the following inequalities (1–16). 1. (sin α1 + · · · + sin αn )2 + (cos α1 + · · · + cos αn )2 ≤ n 2 . √ n ≥ n a1 · · · an , where n ≥ 2, a1 > 0, . . . , an > 0. 2. a1 +···+a √ n √ √ √ 3. a1 b1 + · · · + an bn ≤ a1 + · · · + an · b1 + · · · + bn , where ai ≥ 0, bi ≥ 0, i  1, . . . , n. 2 2 + ≤ 4. (x (x1 y3 − x3 y1 )2  12y2 −2 x2 y12) 2 + 2 (x2 2 y3 − x3 y2 ) x1 + x2 + x3 y1 + y2 + y3 .

2 n

n n √   bi , where xi > 0, ai > 0, bi > 0 i  5. ai bi ≤ ai xi xi i1

1, .n. . , n. n   6. xi yi i1

i1

xi yi

i1

i1

≥

n 

i1

2 xi

, where xi > 0,

yi > 0 i  1, . . . , n.

   a 2 + b2 + c2 x 2 + y 2 + z 2 ≥ 23 (a + b + c)(x + y + z).    8. ( p1 q1 − p2 q2 − · · · − pn qn )2 ≥ p12 − p22 − · · · − pn2 q12 − q22 − · · · − qn2 , if p12 ≥ p22 + · · · + pn2 , q12 ≥ q22 + · · · + qn2 . 7. ax + by + cz +

56

9.

4 The Cauchy–Bunyakovsky–Schwarz Inequality

   √ 2 + x y + y 2 y 2 + yz + z 2 + 2 + yz + z 2 z 2 + zx + x 2 + x y  √ + z 2 + zx + x 2 x 2 + x y + y 2 ≥ (x + y + z)2 . Hint. We have 

   √ 2  √ 2 2   x z 2 3x 3z



2 2 2 2 x + x y + y y + yz + z  + + ≥ y+ y+ 2 2 2 2  z  3x z x  y+ + ≥ y+ . 2 2 4 

10. a1 (b1 + a2 ) + a2 (b2 + a3 ) + · · · + an (bn + a1 ) < 1, where n ≥ 3, a1 , . . . , an > 0 2 + bn2  1. and a1 + · · · + a n  1, b1 + · · ·   x+y 2  y+z 2  2 √ 1 − 2 + 1 − 2 + 1 − z+x ≥ 6, where x, y, z ≥ 0, x 2 + y 2 + 11. 2 z 2  1.    a  c b abc 12. + + ≥ 2 1 + (a+b)(b+c)(c+a) , where a, b, c > 0. c+a  b+c a+b  √ 13. a + (b − c)2 + b + (c − a)2 + c + (a − b)2 ≥ 3, where a, b, c ≥ 0 and a+ b + c  1.   √ a+b b+c c+a − ab+ − bc+ − ca ≥ 2, where a, b, c ≥ 0 and a+b+c  2. 14. 2 2 2 √ √ √ √ √ √ 1 − x y 1 − yz+ 1 − yz 1 − zx+ 1 − zx 1 − x y ≥ 2, where x, y, z ≥ 15. 0 and x 2 + y 2 + z 2  1. √ √ √ √ 16. x 1 − yz + y 1 − zx + z 1 − x y ≥ 2 3 2 , where x, y, z ≥ 0 and x + y + z  1. 17. Find the distance from apoint A(x0 , y0 , z 0) to the plane defined by the equation ax + by + cz + d  0 a 2 + b2 + c2  0 . 18. Prove (a) the following identity: (a1 c1 + · · · + an cn ) (b1 d1 + · · · + bn dn ) − (a1 d1 + · · · + an dn ) (b1 c1 + · · · + bn cn )  (ai bk − ak bi ) (ci dk − ck di ).  1≤i
(b) (a1 c1 + · · · + an cn )(b1 d1 + · · · + bn dn ) ≥ (a1 d1 + · · · + an dn )(b1 c1 + · · · + bn cn ),where bi di > 0 (i  1, . . . , n) or bi di < 0 (i  1, . . . , n) and a1 ≤ ab22 ≤ . . . ≤ abnn , dc11 ≤ dc22 ≤ . . . ≤ dcnn . b1 19. Find the maximum and minimum values of the expression   x 2 + y 2 + (x − 2)2 + (y − 1)2   . x 2 + (y − 1)2 + (x − 2)2 + y 2

Problems for Independent Study

57

20. Find the minimum value of the expression



1 1 1 1 + − 1 + − 1 , x n an y n bn where x, y, a, b > 0, x + y  1, a + b  1. Hint. We have







1 1 1 1 b(1 + a + · · · + a n−1 ) a(1 + b + · · · + bn−1 ) 1 1 + −1 + −1  + + ≥ x n an y n bn xn an yn bn ⎛  n−1 ⎞2 ⎛ ⎞2 

n

n 1 + √ab + · · · + √ab n−1 )(1 + b + · · · + bn−1 ) 1 1 (1 + a + · · · + a ⎟ ⎜ ⎠ ≥⎝ √ ≥⎝ √ + + ⎠ √ n−1 xy a n−1 bn−1 xy ab ≥ (2n+1 − 1)2 .

Chapter 5

Change of Variables Method

The change of variables method is a basic technique used to simplify problems in which the original variable or variables are replaced with other variables in order to express the given problem in new variables in such a way that the problem becomes simpler or easier to prove. In order to prove inequalities, one of the most useful techniques is the change of variables method. Depending on the problem, one can deal with the single-variable case or the multiple-variables case. Both cases can be treated using this method. Let us explain this method by considering the following examples. Example 5.1 Prove that − 21 ≤

(x+y)(1−x y)

(1+x 2 )(1+y 2 )

≤ 21 .

Proof Let us perform the following change of variables: x  tan α, y  tan β. α+tan β)(1−tan α tan β) (x+y)(1−x y)  (tan 1+tan 2α (1+x 2 )(1+y 2 ) ( )(1+tan2 β ) 1 β) cos (α + β)  2 sin 2 (α + β) , therefore

It then follows that



sin(α+β) cos(α+β) cos α cos β · cos α cos β 1 · 1 cos2 α cos2 β



 sin (α + − 21 ≤ 21 sin 2 (α + β) ≤ This completes the proof.        Example 5.2 Prove that a 2 1 + b4 + b2 1 + a 4 ≤ 1 + a 4 1 + b4 . 1 . 2

Proof Let us perform the following of a 2  tan α, b2 tan β.   change  variables:  2 2 2 Then tan α 1 + tan β + tan β 1 + tan α ≤ 1 + tan α 1 + tan2 β . This inequality is equivalent to the following inequality: sin 2α + sin 2β ≤ 2. This ends the proof. Example 5.3 Prove that if a > 0, b > 0, c > 0, a +b+c  1, then a 3 +b3 +c3 ≥ 19 .

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_5

59

60

5 Change of Variables Method

Proof Let us perform the following change of variables: a  x + 13 , b  y + 13 , c  1 1 1 z + 13 . Then a 3  x 3 + x 2 + x3 + 27 , b3  y 3 + y 2 + 3y + 27 , c3  z 3 + z 2 + 3z + 27 , and x+y+z 1 3 3 3 2 2 2 2 therefore, a + b + c  x (1 + x) + y (1 + y) + z (1 + z) + 3 + 9  x (1 + x) + +y 2 (1 + y) + z 2 (1 + z) + 19 ≥ 19 , as x > − 13 , y > − 13 , z > − 13 . This ends the proof. Example 5.4 Prove that if a > 0, b > 0, c > 0, and abc  1, then 1 1 1 + 1+b+c + 1+a+c ≤ 1. 1+a+b √ √ √ Proof Let us perform the following change of variables: 3 a  x, 3 b  y, 3 c  z. x yz x yz 1 z Then x yz  1 and 1+a+b  x yz+x 3 +y 3 ≤ x yz+x y(x+y)  x+y+z . In a similar way, we y 1 x 1 ≤ x+y+z and 1+a+c ≤ x+y+z . Therefore, we deduce that obtain that 1+b+c y 1 1 1 x z + 1+b+c + 1+a+c ≤ x+y+z + x+y+z + x+y+z  1. 1+a+b This ends the proof. Example 5.5 Prove that if x1 > 0, . . . , xn > 0 and ··· +

xnn n

≥1+

1 2

1 x1

+···+

1 xn

 n, then x1 +

x22 2

+

+ · · · + n1 .

Proof Let us perform the following change  of variables: xi  1+ yi , i  1, . . . , n, where yi > −1. We have (x1 + · · · + xn ) x11 + · · · + x1n ≥ n 2 , and hence y1 + · · · + yn ≥ 0. According to Bernoulli’s inequality (Problem 3.27a), we deduce that x2 xn x1 + 22 + · · · + nn ≥ 1 + 21 + · · · + n1 + y1 + y2 + · · · + yn ≥ 1 + 21 + · · · + n1 . Therefore, x1 +

x22 2

+ ··· +

xnn n

≥1+

1 2

+ · · · + n1 . This ends the proof.

Example 5.6 Prove that if a1 ≥ 1, . . . , an ≥ 1 and n ≥ 2, k ≥ 2, k ∈ N, then n−1 √ √ k √ k k a1 − 1 + · · · + k an − 1 ≤ (n−1) a1 · · · · · an . n−k n k √ Proof Let us perform the following change of variables: bi  k (n − 1)(ai − 1), k ≥ 1, . . . , bnk ≥ 1. Then from where i  1, 2, . . . , n and b1k < 1, . . . , blk < 1,bl+1 Problems 10.6,  11.14,  it follows  that k     bk −1 b −1 bk −1 bk −1 · · · · · 1 + ln 1 + l+1n · · · · · 1 + nn ≥ 1 + 1n    k k k k b +···+b −l b +···+b −(n−l) ≥ 1+ 1 n l 1 + l+1 nn  ⎞ ⎛ k k k⎞⎛ k k k b1 +...+bl +1 + · · · + 1 · · + 1 k +bl+1 +···+bnk 1 + · 

 ⎟⎜ ⎟  b1 +b2 +···+bn k ⎜ n−l l ⎝ . ⎠⎝ ⎠≥ n n n bk −1

i , i  1, 2, . . . , n, and thus it follows that Note that 1 + i n  (n−1)a n √ √ k  √ k  n−1 n n−1( k a1 −1+···+ k an −1) a · · · · · a ≥ , whence 1 n n n n−1 √ √ √ (n−1) k k a1 . . . an ≥ k a1 − 1 + · · · + k an − 1. n−k

n

k

This ends the proof.

Problems

61

Problems Prove the following inequalities (5.1–5.5, 5.8–5.12, 5.16–5.20). √ √ 2 2 5.1. 4 √ ≤ a 2 + b2 + ab √ + 4 − a · 9 − b ≤ 19, where 0 ≤ a ≤ 2 and 0 ≤ b ≤ 3. 5.2. n√ m − 1 + m√ n − 1 ≤ mn, where m ≥ 1, n ≥ 1. m 2 − n2 +  2mn − n 2 ≥ m, where m > n > 0. 5.3. √  √ 5.4. x > x − 1 + x x − 1 , where x ≥ 1.  2 5.5. Prove that 1 + √12 + ... + √1n ≥ n n+1 , wher e n ∈ N 5.6. Prove that among seven arbitrary numbers one can find two numbers x and y √ x−y 3 < . such that 0 ≤ 1+x y 3 5.7. Prove the inequality of Problem 4.3. √ √ √ ≤ √1+a|a−c| + √1+b|b−c| . 5.8. √1+a|a−b| 2 · 1+b2 2 · 1+c2 2 · 1+c2 √ √ n 5.9. (a) √ Huygens’s inequality: (a1 + b1 ) · · · (an + bn ) ≥ n a1 · · · · · an + n b1 · · · bn , where ai > 0, bi > 0, i  1, . . . , n, +···+an )(b1 +···+bn ) bn b1 + · · · + aann+b ≤ (a(a11+···+a , where ai > (b) Milne’s inequality: aa11+b 1 n n )+(b1 +···+bn ) 0, bi > 0, i  1, . . . , n. 5.10. (x +x )(y +y8 )−(z +z )2 ≤ x y 1−z 2 + x y 1−z 2 , where x1 > 0, x2 > 0, and x1 y1 −z 12 > 1

5.11.

5.12. 5.13. 5.14.

2

1

2

1

2

1 1

1

2 2

2

0, x2 y2 − z 22 > 0. √ √ √ √ a − 1 + b − 1 + c − 1 ≤ √23 abc, where a ≥ 1, b ≥ 1, c ≥ 1, (a) √ √ √ √ √ √ a − 1 + b − 1 + c − 1 + d − 1 ≤ 3 4 3 abcd, where a ≥ 1, b ≥ (b) 1, c ≥ 1, d ≥ 1. k  a1k +···+ank n ≥ a1 +···+a , where k ∈ N and a1 > 0, . . . , an > 0. n n Let f (x) be a given differentiable function on [a, a + 4] . Prove that there exists a point x0 (x0 ∈ [a, a + 4]) such that f  (x0 ) < 1 + f 2 (x0 ) . Let f (x) be a function defined in [a, b] such that f (x) > 0, x ∈ (a, b), and f (a)  f (b)  0, f (x) + f n (x) > 0 for all x ∈ [a, b] . Prove that b − a > π.    2 2 2 a 2 −b2 ≥ a +b − a+b . 2 2 2 n 3 3 x1 3+· · ·+xn ≤3 3 , where x1 +· · ·+xn  0 and xi ∈ [−1, 1] , x + · · · + x  ≤ 2n, where x1 + · · · + xn  0 and xi n 1

5.15. Prove that if a ≥ 21 , b ≥ 21 , then 5.16. (a) (b) 5.17.

x 1+x 2

1, 2, . . . , n. y z + 1+y 2 + 1+z 2 ≤

√ 3 3 , where x 2 + y 2 + z 2  1. 4 √ 3 2 √ b √ c + ≤ , where a, b, c 2 b2 +c2√ c2 +a 2√ √

i  1, . . . , n, ∈ [−2.2] , i 

> 0. 5.18. 1 < √a 2a+b2 + √ √ √ √ √ 5.19. 1 − a + 1 − b+ 1 − c+ 1 − d ≥ a + b+ c+ d, where a, b, c, d > 0, a 2 + b2 + c2 + d 2 1.  √ √ √ 2 √ √ 2 √ √ 2 3 , where 5.20. a+b+c c − abc ≤ max a − b , b − , c − a 3 a > 0, b > 0, c > 0.

62

5 Change of Variables Method

Proofs   5.1. Let a  2 cos α, b  3 cos β, where α, β ∈ 0, π2 . It follows that

  4 − a 2 · 9 − b2  4 cos2 α + 9 cos2 β + 6 cos α cos β + 6 sin α sin β  .  4 cos2 α + 9 cos2 β + 6 cos (α − β) ≤ 4 + 9 + 6  19.

a 2 + b2 + ab +

On the other hand, 6 sin α sin β ≥ 6 sin2 α or 6 sin α sin β ≥ 6 sin2 β. If 6 sin α sin β ≥ 6 sin2 α, then we have that 4 cos2 α + 9 cos2 β + 6 cos α cos β + 6 sin α sin β ≥ 4 cos2 α + 6 sin2 α + 9 cos2 β + 6 cos α cos β ≥ 4. On the other hand, if 6 sin α sin β ≥ 6 sin2 β, then 4 cos2 α + 9 cos2 β + 6 cos α cos β + 6 sin α sin β√≥ 6. √ Therefore, in both cases we obtain a 2 + b2+ ab+ 4 − a 2 · 9 − b2 ≥ 4. 5.2. Let m  cos12 α , n  cos12 β , where α, β ∈ 0, π2 . Then √ √ n m−1+m n−1 mn

1 2

sin α · cos α+

sin β · cos β

1 2

2β cos α  cos β  sin 2α+sin ≤ 1. 1 2 cos2 α cos2 β √ √ Therefore, n m − 1 + m n − 1 ≤ mn. 5.3. Let mn  sin α, where α ∈ 0; π2 . It follows that √ √ √  m cos α+m 2 sin α−sin2 α m 2 −n 2 + 2mn−n 2   cos α + 2 sin α − sin2 α  m m  √  cos α + sin α + sin α (1 − sin α) ≥ cos α + sin α ≥ cos2 α + sin2 α  1.

√ √ 2 Therefore, m 2 − n 2 + 2mn  π− n ≥ m. 1 5.4. Let x  cos2 α , where α ∈ 0, 2 . Then the given inequality can be rewritten as   1 1 − cos α 1 > tan α + , or 1 > sin α cos α + cos α (1 − cos α). 2 cos α cos α cos α (5.1) √ From inequality (3.2), it follows that cos α (1 − cos α) ≤ 21 and sin α cos α ≤ 1 , where equality does not hold simultaneously in both inequalities. 2 Summing these inequalities, we obtain inequality (5.1).  5.5. First, let us prove that

√1 k

+

√ 1 n+1−k

≥2

2 , n+1

where k  1, . . . , n.

Let k  (n + 1) sin α, where 0 < α < π2 . It follows that   1  1 2 √ 1 √ 1 √1 √2 ≥ +  + cos α sin 2α n+1· sin α n+1· cos α n+1 sin α n+1 2

+ · · · + √1n   2 + · · · + √1n ≥ n n+1 .

Since 1 + √1 2

√1 2

1 2



√1 1

+

√1 n



+ ··· +

1 2



√1 n

 2 ≥ 2 n+1 .  + √11 , we have 1 +

  5.6. Let xk  tan αk , where − π2 < αk < π2 , k  1, . . . , 7. Let us divide − π2 , π2 into six equal parts. According to Dirichlet’s principle, at least two among the

Proofs

63

numbers α1 , . . . , α7 belong to the same interval, and therefore, 0 ≤ αi − α j < π . It follows that 6 √   xi − x j 3 π 0≤  tan αi − α j < tan  . 1 + xi x j 6 3   5.7. Define a  cosc2 α , b  cosc2 β , where α, β ∈ 0, π2 . After several similar transformations, we get the inequality sin (α + β) ≤ 1. The last inequality obviously holds. 5.8. Set a  tan α, b  tan β, c  tan γ . Then the given inequality can be rewritten as |sin (α − β)| ≤ |sin (α − γ )| + |sin (β − γ )| . That can be proved in the following way: |sin (α − β)|  |sin (α − γ ) cos (γ − β) + sin (γ − β) cos (α − γ )| ≤ ≤ |sin (α − γ ) cos (γ − β)| + |sin (γ − β) cos (α − γ )| ≤ |sin (α − γ )| + |sin (γ − β)| , therefore |sin (α − β)| ≤ |sin (α − γ )| + |sin (β − γ )| . 5.9. (a) Let abii  tan αi , i  1, . . . , n. After several similar transformations, we obtain   n sin2 α1 · · · · · sin2 αn + n cos2 α1 · · · · · cos2 αn ≤ 1. From Problem 2.1, it follows that   sin2 α1 + · · · + sin2 αn cos2 α1 + · · · + cos2 αn n n sin2 α1 · · · sin2 αn + cos2 α1 · · · cos2 αn ≤ +  n n n    1 sin2 αi + cos2 αi  1.  n i1



 tan αi , i  1, . . . , n. Hence, we obtain the inequality tan α1 bn2 tan2 αn (b1 tan2 α1 +···+bn tan2 αn )(b1 +···+bn ) , which is + · · · + bn tan ≤ 2 α +b b1 tan2 α1 +b1 n n (b1 tan2 α1 +···+bn tan2 αn )+(b1 +···+bn ) equivalent to   

(b) Let b12

n 

ai bi

2

n 

bi sin αi ≤ 2

i1



n 



n 

bi tan2 αi

i1

n 

n  i1

bi

i1 bi cos2 αi





n 



n 



bi tan αi bi . i1  i1 sin α j sin αi The factor bi b j cos on the left-hand side is not greater than 2 α + cos2 α j j  2  sin2 αi the factor bi b j tan αi + tan2 α j on the right-hand side, that is, cos 2α + j bi sin αi

i1

sin2 α j cos2 αi

2

 i12

≤

, or

bi cos2 αi

2

2

≤ tan2 αi + tan2 α j . This holds because we have that   2 sin2 α j (sin2 αi −sin2 α j )2 ≥ 0. sin αi tan2 αi + tan2 α j − cos  2 α + cos2 α cos2 αi · cos2 α j j i

64

5 Change of Variables Method

√ √ 5.10. Let z 1  x1 y1 · sin α, z 2  x2 y2 · sin β. Then the given inequality can be rewritten as A ≤

8 x1 y1 cos2 α + x2 y2 cos2 β + x1 y2 cos2 α + x2 y1 cos2 β +

√

x1 y2 sin α −

√

y1 x2 sin β

2 ≤

1 1 + . x1 y1 cos2 α x2 y2 cos2 β

Note that A≤

x1 y1

cos2

α + x2 y2

cos2

8 . β + x1 y2 cos2 α + x2 y1 cos2 β

(5.2)

On the other hand,    1 1 x1 y1 cos2 α + x2 y2 cos2 β + x1 y2 cos2 α + x2 y1 cos2 β x1 y1 cos  2 α + x y cos2 β 2 2       2 2 2 2 cos α x2 y2 cos β β cos α  2 + xx21 yy21 cos + yy21 + yy21 + xx21 cos ≥ 8, and there+ xx21 cos 2 β + x y cos2 α 2β cos2 α 1 1 fore, we obtain 8 x1 y1 α + x2 y2 β + x1 y2 cos2 α + x2 y1 cos2 β 1 1 + . ≤ x1 y1 cos2 α x2 y2 cos2 β cos2

cos2

(5.3)

From (5.2) and (5.3) follows the required inequality. 5.11. Let a  cos12 α , b  cos12 β , c  cos12 γ , d  cos12 ϕ , where α, β, γ , ϕ ∈  π 0; 2 . (a) The given inequality can be rewritten as 2 tan α + tan β + tan γ ≤ √3 cos α cos , or β cos γ

cos α cos β cos γ (tan α + tan β + tan γ ) ≤ √23 . Since cos α cos β cos γ (tan α + tan β + tan γ )  sin (α + β) cos γ + cos α cos β cos γ   B, it follows from inequality (4.1)  2   sin (α + β) + cos2 α cos2 β cos2 γ + sin2 γ  that B ≤  2 2 2 sin (α + β) + cos α cos β. Now let us prove sin2 (α + β) + cos2 α cos2 β ≤ 43 . Indeed, 

2 1 (cos (α − β) + cos (α + β))  sin2 (α + β) + 2 1 1 1 1 2 + cos (α − β) + cos (α − β) cos (α + β) + cos2 (α + β) ≤ sin2 (α + β) + + 4 2 4 4 3 5 1 1 1 + |cos (α + β)| + cos2 (α + β)  − cos2 (α + β) + + |cos (α + β)|  2 4 4 4 2   3 4 1 2 4 |cos (α + β)| − − + ≤ 4 3 3 3

sin2 (α + β) +

Proofs

65

(b) The given inequality can be rewritten as√ M  3 3 cos α cos β cos γ cos ϕ (tan α + tan β + tan γ + tan ϕ) ≤ 4 . One can easily prove that M  sin (α + β) cos γ cos ϕ + sin (γ + ϕ) cos α cos β.  ϕ 2 According to inequality (3.2), we have that M ≤ sin (α + β) cos γ +cos + 2 2  cos α+cos β sin (γ + ϕ) . 2   cos x+cos y x+y Since  cos 2 cos x−y ≤ cos x+y , where x, y ∈ 0; π2 , we 2 2 2 must have + sin (γ + ϕ) cos2 α+β  M ≤ sin (α + β) cos2 γ +ϕ 2  2  γ +ϕ γ +ϕ γ +ϕ α+β α+β   2 cos 2 cos 2 sin 2 cos 2 + sin 2 cos α+β 2 +ϕ cos γ +ϕ sin α+β+γ ≤ 2 cos2  2 cos α+β 2 2 2 α+β+γ +ϕ 3 α+β+γ +ϕ sin .  4 cos 4 4

Since cos3 t sin t ≤

√ 3 3 16

α+β+γ +ϕ 4

sin

α+β+γ +ϕ 2

(see Problem 3.28), we have M ≤



√ 3 3 . 4

n  a, ai  a+xi , i  1, . . . , n. Then from the condition a+xi > 5.12. Let a1 +...+a n 0, it follows that xai > −1. Hence, according to Bernoulli’s inequality, we have k  1 + xai ≥ 1 + k xai , i  1, . . . , n, and therefore, (a + xi )k ≥ a k + ka k−1 xi . Summing these inequalities, we obtain (a + x1 )k + · · · + (a + xn )k ≥ na k + ka k−1 (x1 + . . . + xn )  na k , since  k a k +...+a k n x1 +. . .+xn  a1 +. . .+an −na  0. Thus, we deduce that 1 n n ≥ a1 +...+a . n 5.13. Proof by contradiction. Assume that for all x ∈ [a, a + 4] , f  (x) ≥ 1 + f 2 (x) . Let arctan f (x)  g (x) . Then − π2 < g (x) < π2 and f (x)  tan(g(x)).  ≥ 1 + f 2 (x)  cos21g(x) , and hence g  (x) ≥ 1. It follows that f  (x)  cosg2(x) g(x) Therefore, g (x) − x is a nondecreasing function. Hence,

g (b) − b ≥ g (a) − a, where b  a + 4 or g (b) − g (a) ≥ b − a.

(5.4)

Since  π π , we have g (b) − g (a) < π. g (b) , g (a) ∈ − ; 2 2

(5.5)

From (5.4)–(5.5) and the condition b − a  4 it follows that π > 4; this leads to a contradiction. Therefore, there exists x0 such that x0 ∈ [a, a + 4] and f  (x0 ) < 1 + f 2 (x0 ) . 5.14. Without loss of generality one can assume that a  0 (otherwise, consider the function f (x − a)). Then we need to prove that b > π. We proceed with a proof by contradiction argument.   Assume that  b ≤ π. Consider the function F (x)  − πb f (x) cos πb x + f  (x) sin πb x in [0, b] . Note that F (0)  F (b)  0; Hence by Rolle’s theorem, there exists c ∈ (0, b) such that F  (c)  0.

66

5 Change of Variables Method

On the other hand, we  have    2 F  (c)  πb2 f (c) sin πb c + f  (c) sin πb c   2         πb2 f (c) + f  (c) sin πb c ≥ f (c) + f  (c) sin πb c > 0. It follows that F  (c) > 0, which leads to a contradiction with F  (c)  0. Therefore, b > π. 5.15. Without loss of generality one can assume that a ≥ b. Let a  b tan α, where π ≤ α < π2 . Then the given inequality can be rewritten as 4 2 2α 1 α ≥ b √2 cos − b sin2α+cos , or b4 4cos cos4 α cos α α   √ b3 cos2 2α ≥ 2 cos3 α 2 − sin α − cos α . In order to finish the proof, it is sufficient to prove that √  for α in π π  1 2 3 , , one has 8 cos 2α ≥ 2 cos α 2 − sin α − cos α , which is 4 2 equivalent to the following inequality: 21 cos2 (α + π/4) sin2 (α + π/4) ≥ √ 2  2 cos3 α (1 − sin (α + π/4)) , or √ 1 1 − sin2 (α + π/4) sin2 (α + π/4) ≥ 2 2 cos3 α (1 − sin (α + π/4)) . 2 This inequality holds because for the and right  difference between its left√ hand sides we have 21 (1 − sin (α + π/4)) sin2 (α + π/4) + sin3 (α + π/4) − 4 2 cos3 α ≥ 0, √       since 1 ≥ sin α + π4 and sin2 α + π4 ≥ 2 cos2 α, 1 + sin α + π4 ≥ 2 2 ·   cos α π4 ≤ α < π2 . 5.16. (a) Let xi  sin αi , i  1, . . . , n. We have n  n    3 sin αi − 4 sin3 αi  sin 3αi , and therefore, i1 n  i1

xi 

n  i1

sin αi 

1 3

n  i1

i1

sin 3αi ≤ n3 .

(b) Let xi  2 sin αi , i  1, 2, . . . , n. We have n 

n 

(3 sin αi − 4 sin3 αi ) 

i1

sin 3αi , and therefore, n  n    3   x + x 3 + . . . + x 3   2  sin 3αi  ≤ 2 |sin 3αi | ≤ 2n. n 1 2   i1 i1   It follows that x13 + x23 + . . . + xn3  ≤ 2n. 5.17. Without loss of generality one can assume that x, y, z ≥ 0. Indeed, note that y |y| |x| |z| x z 2 2 2 + 1+y 2 + 1+z 2 ≤ 1+|x|2 + 1+|y|2 + 1+|z|2 and |x| +|y| +|z|  1. 1+x 2  π Let x  tan α, y  tan β, z  tan γ , where α, β, γ ∈ 0, 4 . We have x y + yz + x z ≤ x 2 + y2 + z2 (Problem 4.13), whence tan α tan β + tan β tan γ + tan γ tan α ≤ 1.   tan α+tan β ≤ 1, or tan(α + β) ≤ tan π2 − γ . Hence, tan γ · 1−tan α tan β i1

Proofs

67

It follows that α + β ≤

π 2

y π x . Note that 1+x 2 + 1+y 2 2 √ 2α+2β+2γ ≤ 23 sin π3  3 4 3 . 3

− γ , that is, α + β + γ ≤

 21 (sin 2α + sin 2β + sin 2γ ) ≤ 21 · 3 sin 5.18. Note that a b c √ a + √b2b+c2 + √c2c+a 2 > a+b + b+c + c+a > a 2 +b2 √ a a 2 +b2 2 2

a a+b+c

+

b a+b+c

+

c a+b+c

+ 1+zz 2

 1,

√ b + √c2c+a 2 > b2 +c2 2 2 2 2

+ 1. and therefore, 2 2 Set a + b  m , b + c  n , c + a  k 2 , where m, n, k > 0. We have (m + n)2 > m 2 + n 2 > k 2 , whence m + n > k, and in a similar way, we obtain n + k > m, m + k > n. Let an acute triangle with side lengths m, n, k have angles α, β, γ . Then using the law √ law of cosines, √ this inequality √ can be rewritten as √ of sines and the sin α sin 2γ + sin β sin 2α + sin γ sin 2β ≤ 3 sin α sin β sin γ . Therefore, from inequality (4.2), it follows that   √ sin α sin 2γ + sin β sin 2α + sin γ sin 2β  ≤ (sin2 α + sin2 β + sin2 γ )(sin 2α + sin 2β + sin 2γ )   4 sin α sin β sin γ (2 + 0, 25 cos2 (α − β) − (cos γ − 0, 5 cos(α − β))2  ≤ 3 sin α sin β sin γ . It follows √ √ √ √ that sin α sin 2γ + sin β sin 2α + sin γ sin 2β ≤ 3 sin α sin β sin γ . 5.19. Let us prove that for an arbitrary number a, where 0 ≤ a ≤ 1, the following inequality holds: √ √ √ 2 (1 − 4a 2 ). 1−a ≥ a+ 2   Let a  sin2 α, where α ∈ 0, π2 . Then the given inequality can be rewritten as √ 2 cos 2α(2 − cos 2α). (1) cos α − sin α ≥ 4 √ If 0 ≤ α ≤ π4 , then cos α−sin α ≥ 0, cos α+sin α ≤ 2, and 2−cos 2α ≤ 2. Therefore, inequality (1) holds. If π4 ≤ α ≤ π2 , then on setting α  π4 + β, the inequality (1) can be rewritten as cos2 β cos β2 ≥ sin β2 . In order to prove this inequality it is sufficient to prove that cos2 β(1 + cos β) ≥ sin β. Since 0 ≤ β ≤ π4 , we have cos2 β(1 + cos β) ≥ √

√

+ 22 ) > 22 ≥ sin β. It follows that cos2 β(1 + cos β) > sin β. We have √ √ √ √ √ √ √ √ √ 1 − a + 1 − b + 1 − c + 1 − d ≥ a + b + c + d + 22 (1 − 4a 2 + 1 − 4b2 + 1 − 4c2 + 1 − 4d 2 ), 1 (1 2

68

5 Change of Variables Method

whence √

1−a+

√

1−b+

√ √ √ √ √ √ 1 − c + 1 − d ≥ a + b + c + d.

5.20. Without loss of generality one can assume that a ≥ b ≥ c. Let a  x 3 , b  y 3 , c  z 3 . Therefore, x ≥ y ≥ z > 0. Thus, the given inequality can be rewritten as We have

x 3 +y 3 +z 3 −x yz 3

≤

√

x3 −

√ 2 z3 .

  x 3 + y3 + z3 1 − x yz  (x + y + z) (x − z)2 + (x − y)2 + (y − z)2 ≤ 3 6  √ 2 √ 2 √ √ √ 2  √ 2  x  x + xz ≤ x + xz + z  ≤ x− z x− z (x − z)2 + (x − y + y − z)2  2    2  2 x 3 + y3 + z3 x 3 − z 3 , whence x 3 − z3 .  − x yz ≤ 3

This ends the proof.

Problems for Independent Study Prove the following inequalities (2–27). + b2  1. Prove that 1. Given that a 2√ √ √ (a) |a + b| ≤ 2, (b) |a − b| ≤ 2, (c) |ab| ≤ 21 , (d) |a 2 b + ab2 | ≤ 22 .      2. x y − 1 − x 2 1 − y 2  ≤ 1, where |x| ≤ 1, |y| ≤ 1.   √  2 2 2 3. 1 − x + 1 − y ≤ 2 1 − x+y , where |x| ≤ 1, |y| ≤ 1. 2   4. x + x1 arccot x > 1, where x > 0.    1  1 2 5. cos α+cos − 1 ≤ − 1 − 1 , where π3 ≤ α < π2 , π3 ≤ β < β cos α cos β 6. 7. 8. 9.

n(a1 +···+an ) an a1 + · · · + 1−a ≥ n−(a , where 0 ≤ a1 < 1, . . . , 0 ≤ an < 1. 1−a1 n 1 +···+an ) 1 1 3 √1 √ √ + 1+b2 + 1+c2 ≤ 2 , where a, b, c > 0 and a + b + c  abc. 1+a 2 |x−y| |y−z| |x−z| + 1+a|y−z| ≥ 1+a|x−z| , where a > 0. 1+a|x−y| 2x (1−x 2 ) 2y (1−y 2 ) 2z (1−z 2 ) y x z + + ≤ 1+x 2 + 1+y 2 + 1+z 2 , where x > 0, y > 0, (1+x 2 )2 (1+y 2 )2 (1+z 2 )2

π . 2

z>0

and x y + yz + zx  1. √ √ √ √ √ √  √  1 a1 − a2 + 2 a2 − a3 + · · · + 10. √a1+ a2 + · · · + an ≤ √ √ n an − an+1 , where a1 ≥ . . . ≥ an ≥ an+1  0. 1 1 1 11. 1 +···+ − 1 +···+ 1 1 ≥ n , where a1 > 0, . . . , an > 0. 1+a1 1+an a1 an √ 12. a + b + c − 2 abc ≥ ab + bc + ca − 2abc, where 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, 0 ≤ c√≤ 1. √ √ √ a (1 − b) (1 − c) + b (1 − a) (1 − c) + c (1 − a) (1 − b) ≤ 1 + abc, 13. where 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, 0 ≤ c ≤ 1.

Problems for Independent Study

69

14. ((x + y)(y +z)(x +z))2 ≥ x yz(2x + y +z)(2y +z +x)(2z +x + y), where x, y, z ≥ 0. Hint. If x 2 + y 2 + z 2  0, then without loss of generality one can assume that x + y + z  1. Let x  tan α tan β, y  tan β tan γ , z  tan α tan γ , where α + β + γ  π2 . Then one needs to prove that ((1 − x)(1 − y)(1 − z))2 ≥ x yz(1 + x)(1 + y)(1 + z). < 41 , where 0 < a < 1, 0 < b < 1. 15. ab(1−a)(1−b) (1−ab)2   Hint. Let a  sin2 α, b  sin2 β, where α, β ∈ 0, π2 . 16. max(a1 , . . . , an ) ≥ 2, where n > 3, a1 + · · · + an ≥ n, a12 + · · · + an2 ≥ n 2 . Hint. Let ai  2 − bi and bi > 0, i  1, . . . , n. We have that b1 + · · · + bn ≤ n, b12 + · · · + bn2 − 4(b1 + · · · + bn ) ≥ n(n − 4), and therefore, (b − 4)(b1 + · · · + bn ) ≥ n(n − 4), where b  max(b   1 , . . . , bn ). 2 √ √ √ (an −an−1 )2 n −an−1 ) 17. a1 + 4(n−2) + · · · + an−2 + (a4(n−2) + an−1 + an ≤ n, where n ≥ 3, a1 , . . . , an ≥ 0, and a1 + a2 + · · · + an  1. √ √ Hint. an−1  x + t, an  x − t. Then x ≤ √12 , and the left-hand side of the  inequality is not greater than (n − 2)(1 − 2x 2 ) + 2x.  18. 2 (x 2 − 1)(y 2 − 1) ≤  2(x − 1)(y − 1) + 1, where 0 ≤ x, y ≤ 1. 3 3 3 19. a + b + c − 3abc ≤ (a 2 + b2 + c2 )3 . 1 1 + · · · + n−1+x ≤ 1, where x1 , . . . , xn > 0 and x1 · · · · · xn  1. 20. n−1+x 1 n Hint. Let xi  yin , i  1, . . . , n, where yi > 0. Then y1 · · · · · yn  1 and yn

(n−1)y n−1

i i n − 1 + xi  n − 1 + y1 ·····y ≥ n − 1 + y n−1 +···+y n−1 . −yin−1 n n 1 y x+y x 21. √1−x + √1−y ≥ √ x+y , where 0 ≤ x < 1, 0 ≤ y < 1.

22.

1− 2 √ √ x +···+ x √ x1 √ xn + · · · + ≥ 1√n−1 n , 1−xn 1−x1 x1 + . . . + xn  1.

where n ≥ 2, n ∈ N, x1 , . . . , xn > 0 and

Hint. Take 1 − xi √ai , i  1, . . . , n. 23. √ x 2 + √4xy2 +1 ≤ 22 , where 0 ≤ x, y ≤ 0.5. 4y +1

Hint. Let 2x  tan α, 2y  tan β, where 0 ≤ α, β ≤ π4 . 24. 0 ≤ ab+bc+ca−abc ≤ 2, where a > 0, b > 0, c > 0 and a 2 +b2 +c2 +abc  4. 25. a + b + c ≤ 3, where a > 0, b√> 0, c > 0, and a 2 + b2 + c2 + abc  4. 26. (x − 1)(y − 1)(z − 1) ≤ 6 3 − 10, where x > 0, y > 0, z > 0, and x + y + z  x yz.   Hint. Let x  tan α, y  tan β, z  tan γ , where α, β, γ ∈ 0, π2 . Then β−sin β α + β + γ  π and β > π4 . Note that (tan α − 1)(tan γ − 1)  2 + cos ≤ cos α cos γ 2+

cos

cos β−sin β . cos(α+γ − π3 )

π 3

70

5 Change of Variables Method

  3 z+x ≤ 5(x+y+z)+9 , where x > 0, y > 0, z > 0, and x yz  1. + 3 y+z + 2x 2y 8   4    3 2 −ab+b2 3  (a+b) , Hint. 3 21 a 3 b3 a 3 + b3 ≤ 21 (a + b) ab+ab+ab+a 4 8

 27.

3

x+y 2z

where a > 0, b > 0. 28. Prove that among four arbitrary numbers there are two numbers a and b such √ > 21 . that √ 1+ab 1+a 2 · 1+b2 29. Given that x + y + z  0 and x 2 + y 2 + z 2  6. Find all possible values of the expression x 2 y + y 2 z + z 2 x.  √ 1− 1−h 2n 30. Let (h n ) be a sequence such that h 1  21 and h n+1  , n  1, 2, . . .. 2 Prove that h 1 + · · · + h n ≤ 1.03.

Chapter 6

Using Symmetry and Homogeneity

In order to prove certain inequalities, one often needs to use the symmetry and homogeneity of mathematical expressions. Definition 1 A mathematical expression is called symmetric with respect to the set of variables x1 , x2 , . . . , xn , n ∈ N, if its value remains unchanged under every permutation of variables x1 , x2 , . . . , xn . Remark For purposes of simplicity and brevity, if a mathematical expression is symmetric with respect to the set of variables x1 , x2 , . . . , xn , then hereinafter we shall call it symmetric. Definition 2 A mathematical expression is called homogeneous of degree k with respect to the set of variables x1 , x2 , . . . , xn , n ∈ N if for every positive λ, on replacing the variables x1 , x2 , . . . , xn by the variables λx1 , λx2 , . . . , λxn , the value of the given mathematical expression is multiplied by λk . Remark For purposes of simplicity and brevity, if a mathematical expression is homogeneous of degree k with respect to the set of variables x1 , x2 , . . . , xn , then hereinafter we shall call it homogeneous. The symmetry and homogeneity of mathematical expressions allow us to deduce additional conditions for variables. For example, if a mathematical expression is symmetric, then an additional condition for variables can be assuming without loss of generality that x1 ≤ x2 ≤ · · · ≤ xn , and if it is homogeneous, then such a condition can be assuming without loss of generality that x1 ·x2 · · · xn  1 or x1k +x2k +· · ·+xnk  1, where k ∈ N. In order to demonstrate how these additional conditions can be obtained, let us consider the following examples of symmetric or/and homogeneous mathematical expressions.

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_6

71

72

6 Using Symmetry and Homogeneity

Example 6.1 Prove that x(x − z)2 + y(y − z)2 ≥ (x − z)(y − z)(x + y − z),

(6.1)

where x ≥ 0, y ≥ 0, z ≥ 0. Proof Note that (6.1) can be rewritten as x3 + y3 + z 3 − x2 y − x2 z − y2 x − y2 z − z 2 x − z 2 y + 3xyz ≥ 0, whence we obtain a symmetric inequality with respect to the variables x, y, z. Therefore, without loss of generality, one can assume that x ≥ z ≥ y. It follows that x(x − z)2 + y(y − z)2 ≥ 0 ≥ (x − z)(y − z)(x + y − z), whence x(x − z)2 + y(y − z)2 ≥ (x − z)(y − z)(x + y − z). This ends the proof. Example 6.2 Hölder’s inequality (particular  case):    Prove that (a1 b1 c1 + · · · + an bn cn )3 ≤ a13 + · · · + an3 b31 + · · · + b3n c13 + · · · + cn3 , where ai > 0, bi > 0, ci > 0, i  1, . . . , n. Proof Note (that if we substitute the variables a1 , . . . , an by the variables λa1 , . . . , λan , where λ is an arbitrary positive number, then we obtain an equivalent inequality. Therefore, we can choose λ such that (λa1 )3 + · · · + (λan )3  1. Without loss of generality one can assume that a13 + · · · + an3  b31 + · · · + b3n  3 c1 + · · · + cn3  1. Hence in order to complete the proof, it is sufficient to prove that a1 b1 c1 + · · · + an bn cn ≤ 1. a3 +b3 +c3 We have ai bi ci ≤ i 3i i , i  1, . . . , n (Problem 2.1). Summing these inequalities, we obtain a1 b1 c1 + · · · + an bn cn ≤ 1. This ends the proof.  a  b  c Example 6.3 Prove that b+c + a+c + a+b > 2, where a > 0, b > 0, c > 0. Proof Without loss of generality one can assume that a + b + c  1. Note that  a ≥ 2a, 1−a

(1)

and the equality in (1) holds if a  21 . It follows that       a b c a b c + +  + + > 2a + 2b + 2c  2. b+c a+c a+b 1−a 1−b 1−c This ends the proof.

Problems

73

Problems Prove the following inequalities. 6.1. abc ≥ (a + b − c)(b + c − a)(a + c −b), where a >  0, b > 0, c > 0. 6.2. (a1 b1 + · · · + an bn )2 ≤ a12 + · · · + an2 b21 + · · · + b2n .  n 6.3. (a + b)2 · . . . · (an + bn )2 ≥ an+1 + bn+1 , where a > 0, b > 0. α  β  β β 6.4. a1α + · · · + anα ≤ a1 + · · · + an , where 0 < β < α, a1 > 0, . . . , an > 0. a b + c+a + c ≥ 23 , where a > 0, b > 0, c > 0. 6.5. Nesbitt’s inequality: b+c   a+b   a b c d + a+c+d + a+b+d + a+b+c > 2, where a > 0, b > 0, c > 0, d > 6.6. b+c+d 0.   +bcd ab+ac+ad +bc+bd +cd 6.7. 3 abc+abd +acd ≤ , where a > 0, b > 0, c > 0, d > 0. 4 6 √ √ √ 3 6.8. 2 ab + bc + ac ≤ 3 (b + c)(c + a)(a + b), where a > 0, b > 0, c > 0. 2      6.9. 8 x3 + y3 + z 3 ≥ 9 x2 + yz y2 + xz z 2 + xy , where x > 0, y > 0, z > 0. 6.10. (a) 4a3 +4b3 +4c3 +15abc ≥ 1, where a ≥ 0, b ≥ 0, c ≥ 0 and a +b+c  1,   d , where a ≥ 0, b ≥ 0, c ≥ 0 and (b) a3 + b3 + c3 + abcd ≥ min 41 , 19 + 27 a + b + c  1.   √ a12 +···+an2 1 n ≥ + 1 − 1n n a1 · · · an , where n ≥ 2, ai > 0, i  6.11. a1 +···+a n n n 1, . . . , n. 6.12. Turkevici’s inequality: a4 + b4 + c4 + d 4 + 2abcd ≥ a2 b2 + a2 c2 + a2 d 2 + b2 c2 + b2 d 2 + c2 d 2 , where a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0. a13 a3 + · · · + bnn ≥ 1, where ai > 0, bi > 0, i  1, . . . , n, and (a12 + · · · + an2 )3  b1 b21 + · · · + b2n . + b+a + c+b , where a > 0, b > 0, c > 0. 6.14. ab + bc + ac ≥ a+c b+c c+a   a+b n a1 an n 6.15. + · · · + an +λan1 ···an ≥ √1+λ , where n ≥ 2, a1 > 0, . . . , an > 0, an +λa1 ···an 6.13.

1

n

and λ ≥ n2 − 1.  √ 6.16. ( k 2 − 1)(a1 + · · · + an ) < k 2a1k + · · · + 2n ank , where k ∈ N, k ≥ 2, a1 > 0, . . . , an > 0. 6.17. 3(x2 y +y2 z +z 2 x)(xy2 +yz 2 +zx2 ) ≥ xyz(x +y +z)3 , where x > 0, y > 0, z > 0. 6.18. (x1 + · · · + xn + y1 + · · · + yn )2 ≥ 4n(x1 y1 + · · · + xn yn ), where x1 ≤ · · · ≤ xn ≤ y1 ≤ · · · ≤ yn .

74

6 Using Symmetry and Homogeneity

Proofs 6.1. Without loss of generality one can assume that a ≤ b ≤ c. Note that the given inequality is equivalent to the following inequality: (b − c)2 (b + c − a) ≥ a(a − b)(c − a). The last inequality can be proved in the following way: (b − c)2 (b + c − a) ≥ 0 ≥ a(a − b)(c − a). 6.2. If a12 + · · · + an2  0 and b21 + · · · + b2n  0, then without loss of generality one can assume that a12 + · · · + an2  1, b21 + · · · + b2n  1. a2 +b2 a2 +b2 We have that − i 2 i ≤ ai bi ≤ i 2 i , i  1, . . . , n. Summing these inequalities we obtain −1 ≤ a1 b1 + · · · + an bn ≤ 1; hence    (a1 b1 + · · · + an bn )2 ≤ 1  a12 + · · · + an2 b21 + · · · + b2n . If a12 + · · · + an2  0 or b21 + · · · + b2n  0, then the proof is obvious. 6.3. Without loss of generality one can assume that an+1 + bn+1  1. Therefore, 0 < a < 1, 0 < b < 1 and ai + bi ≥ an+1 + bn+1  1, i  1, . . . , n.  n Thus, it follows that (a + b)2 · · · (an + bn )2 ≥ 1  an+1 + bn+1 . β β 6.4. Without loss of generality one can assume that a1 +· · ·+an  1, and therefore, β 0 < ai ≤ 1, i  1, . . . , n. Since α > β, we obtain that aiα ≤ ai , i  1, . . . , n. β  β β It follows that a1α + · · · + anα < a1 + · · · + an  1, and hence a1α + · · · + anα ≤ α  β β 1  a1 + · · · + an . 6.5. Without loss of generality one can assume that a + b + c  1. x ≥ 9x−1 . Therefore, Note that if 0 < x < 1, then 1−x 4 a b c a b c 9a − 1 9b − 1 9c − 1 3 + +  + + ≥ + +  . b+c c+a a+b 1−a 1−b 1−c 4 4 4 2

6.6. Without loss of generality one can assume that a + b + c + d  1. We have     a b c d + + + b+c+d a+c+d a+b+d a+b+c     a b c d + + + > 2a + 2b + 2c + 2d  2.  1−a 1−b 1−c 1−d 6.7. Without loss of generality one can assume that d  1. From Problems 2.1 and 2.2, it follows that

Proofs

75

 3



abc + ab + bc + ac ≤ 4



 

ab+bc+ac 3 3 + ab + bc + ac  3

ab + bc + ac + a + b + c ≥ 6

4 ab + bc + ac +

√

 A and

3(ab + bc + ac)

6

 B.

Let us prove that B ≥ A. Let ab + bc + ac  3t 2 , where t > 0. Then the inequality B ≥ A is equivalent to the inequality (t − 1)2 (t + 2) ≥ 0. 6.8. Without loss of generality one can assume that ab + bc + ac  1.

(1)

We need to prove that 8 (b + c)(a + b)(a + c) ≥ √ . 3 3   Let a  tan α, b  tan β, c  tan γ , where α, β, γ ∈ 0, π2 . From (1) it follows that α +β +γ  π2 , and therefore, (b+c)(a +b)(a +c)  cos α cos1 β cos γ . In order to end the proof, it is sufficient to prove that √  π π 3 3 , if α, β, γ ∈ 0, and α + β + γ  . cos α cos β cos γ ≤ 8 2 2 Note that 1 1 cos α cos β cos γ  (cos(α + β) + cos(α − β)) cos γ ≤ (1 + sin γ ) cos γ 2 2   1 1  (1 + sin γ )3 (1 − sin γ )  √ (1 + sin γ )3 (3 − 3 sin γ ) 2 2 3 √ 

1 + sin γ + 1 + sin γ + 1 + sin γ + 3 − 3 sin γ 2 1 3 3 ≤ √  . 4 8 2 3

6.9. Without loss of generality one can assume that x3 + y3 + z 3  1.

(1)

Then we have     A  x2 + yz y2 + xz z 2 + xy  2(xyz)2 + xyz + x3 y3 + y3 z 3 + z 3 x3 . Since x3 + y3 + z 3  1, it follows that xyz ≤ 13 and x3 y3 + y3 z 3 + z 3 x3 ≤ (x3 +y3 +z3 )2  1 , and therefore, A ≤ 2 + 1 + 1  8 . 3 3 9 3 3 9

76

6 Using Symmetry and Homogeneity

6.10. (a) Note that the inequality 4a3 +4b3 +4c3 +15abc ≥ (a + b + c)3 is equivalent to the inequality a3 + b3 + c3 + 3abc ≥ a2 b + a2 c + b2 a + b2 c + c2 a + c2 b (see Example 6.1). (b) If d ≥

15 , 4

then

a3 + b3 + c3 + abcd ≥ a3 + b3 + c3 +

 1 15 1 3 4a + 4b3 + 4c3 + 15abc ≥ . abc  4 4 4

3  If d < 15 , then according to 6.10(a) and the inequality abc ≤ a+b+c  4 3   1 15 3 3 3 − 3 ≥ , we have 27 a + b + c − 27abc), and therefore, (1 27  4  d 27 a3 + b3 + c3 − 3 ≥ d (1 − 27abc), or a3 + b3 + c3 + abcd ≥ 19 + 27 . 6.11. Without loss of generality one can assume that a1 a2 · · · an  1. We need to prove that

   2 1  2 a1 + · · · + an2 − √ (n − 1) a12 + · · · + an2 + 2a1 a2 + · · · + 2a1 an−1 + 2a1 an + · · · + 2an−1 an ≥ (n − 1)2 , or n n

2  1 (n − 1) √ a12 + · · · + an2 − 1 + 2a1 a2 + · · · + 2a1 an−1 + 2a1 an + · · · + 2an−1 an ≥ n(n − 1). n 1−

The proof of this inequality follows from Problem 4.1: a1 a2 + a1 a3 + · · · + a1 an−1 + a1 an + · · · + an−1 an ≥ Cn2

 (a1 a2 . . . an )n−1  Cn2 .

Cn2

6.12. Without loss of generality one can assume that a ≥ b ≥ c ≥ d ≥ 0. We have A  a4 + b4 + c4 + d 4 + 2abcd − a2 b2 − a2 c2 − a2 d 2 − b2 c2 − b2 d 2 − c2 d 2  (a2 − b2 )(a2 − c2 − d 2 ) + 2bcd (a − b) + (b2 + cd − c2 − d 2 )2 + 2cd (c − d )2    (a − b) a(a2 − d 2 ) − b(c − d )2 + a(a − b)(ab − c2 ) + (b2 + cd − c2 − d 2 )2 + 2cd (c − d )2 .

Since a − b ≥ 0, a(a2 − d 2 ) ≥ b(a − d )2 ≥ b(c − d )2 and ab ≥ c2 , it follows that √ A ≥ 0. 6.13. Let 3 bi  ci , i  1, . . . , n. Then we have c16 + · · · + cn6  (a12 + · · · + an2 )3 . a3 a3 We need to prove that c31 + · · · + c3n ≥ 1. 1

n

Without loss of generality one can assume that a12 + · · · + an2  1, and hence c16 + · · · + cn6  1. From inequality (8.4) from Chapter 8, it follows that a13 c13

+ ··· +

a4 (a12 + · · · + an2 )2 an3 a4 1  13 + · · · + n3 ≥  ≥ 1. 3 cn an cn a1 c1 a1 c13 + · · · + an cn3 a1 c13 + · · · + an cn3

Then from inequality (4.2) from Chapter 4, it follows that 1  (a12 + · · · + an2 )(c16 + · · · + cn6 ) ≥ (a1 c13 + · · · + an cn3 )2 , whence a3 a3 1 ≥ a1 c13 + · · · + an cn3 , and hence c31 + · · · + c3n ≥ 1. n 1 6.14. Without loss of generality one can assume that max(a, b, c)  a. Then a ≥ b ≥ c or a ≥ c ≥ b.

Proofs

77

If a ≥ b ≥ c, then we have a b c a+c b+a c+b c(a − b) a(b − c) b(a − c) + + − − −  + − b c a b+c c+a a+b b(b + c) c(a + c) a(b + a) c(a − b) a(b − c) b(a − c) + −  0, ≥ a(b + a) a(a + b) a(b + a) whence a+c b+a c+b a b c + + ≥ + + . b c a b+c c+a a+b If a ≥ c ≥ b, then we have a b c a+c b+a c+b c(a − b) a(c − b) b(a − c) + + − − −  − − b c a b+c c+a a+b b(b + c) c(a + c) a(b + a) c(a − b) a(c − b) b(a − c) − −  0, ≥ b(b + c) b(b + c) b(b + c) and therefore, a b c a+c b+a c+b + + ≥ + + . b c a b+c c+a a+b 6.15. Without loss of generality one can assume that a1 · . . . · an  1. Then  n the a1 expression on the left-hand side of the given inequality is equal to an +λ + 1  n an · · · + an +λ . n By the Cauchy–Bunyakovsky–Schwarz inequality and inequality (8.4) from Chapter 8, we have

a1n + ··· + a1n + λ



n 2 a12

n 2 an2

ann   + ··· +  n n ann + λ an2 (ann + λ) a12 (a1n + λ)

n

n n 2 n 2 a12 + · · · + an2 a12 + · · · + an2 ≥  . ≥  n n n n   a12 (a1n + λ) + · · · + an2 (ann + λ) a12 + · · · + an2 a1n + · · · + ann + nλ

In order to complete the proof, it is sufficient to prove that n (1 + λ)(b1 + · · · + bn )3 ≥ n2 (b31 + · · · + b3n ) + n3 λ, where bi  ai3 > 0, i  1, . . . , n, and b1 · . . . · bn  1. Note that (b1 + · · · + bn )3  (b21 + · · · + b2n + 2b1 b2 + · · · + 2b1 bn + · · · + 2bn−1 bn )(b1 + · · · + bn )

78

6 Using Symmetry and Homogeneity  2  ≥ (b21 + · · · + b2n + n(n − 1) (b1 · . . . · bn )n−1 n(n−1) )(b1 + · · · + bn )  (b21 + · · · + b2n + n(n − 1))(b1 + · · · + bn ) ≥ b31 + · · · + b3n + b21 b2 + · · · + b21 bn + · · · + b2n−1 bn + n2 (n − 1) ≥ b31 + · · · + b3n + n(n − 1) + n2 (n − 1)  b31 + · · · + b3n + n3 − n,

whence (b1 + · · · + bn )3 ≥ b31 + · · · + b3n + n3 − n. Thus (1 + λ)(b1 + · · · + bn )3 − n2 (b31 + · · · + b3n ) − n3 λ ≥ (1 + λ)(b31 + · · · + b3n + n3 − n) − n2 (b31 + · · · + b3n ) − n3 λ ≥ (λ − (n2 − 1))(b31 + · · · + b3n ) + (1 + λ)(n3 − n) − n3 λ ≥ (λ − (n2 − 1))n + (1 + λ)(n3 − n) − n3 λ  0,

and therefore, (1 + λ)(b1 + · · · + bn )3 ≥ n2 (b31 + · · · + b3n ) + n3 λ. 6.16. Without loss of generality one can assume that 2a1k + · · · + 2n ank  1, whence 2i aik < 1.     Therefore, a1 +. . .+an < and thus

1 √ k 2

+. . .+

1 √ k 2

n

<

1 √ k 2

+. . .+

1 √ k 2

n

+. . . 

1 √ , k 2−1

 √ k ( 2 − 1)(a1 + · · · + an ) < 1  k 2a1k + · · · + 2n ank . 6.17. According to Example 6.2 from Chapter 6, we have 3(x2 y + y2 z + z 2 x)(xy2 + yz 2 + zx2 ) ≥ xyz(x + y + z)3 ,   3(x2 y + y2 z + z 2 x)(xy2 + yz 2 + zx2 )  (13 + 13 + 13 )  ·

 3  3  3 3 3 3 zx2 + xy2 + yz 2

 ≥

3 2 x y

3 +

 3     3 2 3 3 2 y z + z x

 √ 3 xyz(x + y + z) 3  xyz(x + y + z)3 ,

and therefore, 3(x2 y + y2 z + z 2 x)(xy2 + yz 2 + zx2 ) ≥ xyz(x + y + z)3 . 6.18. Note that if we substitute the numbers x1 , . . . , xn , y1 , . . . , yn by the numbers x1 + x, . . . , xn + x, y1 + x, . . . , yn + x, where x is an arbitrary number, then we obtain an equivalent inequality. Hence, we can choose the number x such that x1 + x + · · · + xn + x + y1 + x + · · · + yn + x  0. Therefore, without loss of generality one can assume that x1 + · · · + xn + y1 + · · · + yn  0. Note that the given inequality is equivalent to the inequality x1 y1 + · · · + xn yn ≤ 0.

(1)

If x1 ≤ · · · ≤ xn ≤ 0 ≤ y1 ≤ · · · ≤ yn , then x1 y1 ≤ 0, . . . , xn yn ≤ 0, and hence (1) holds.

Proofs

79

In order to complete the proof, it is sufficient to prove (1) if x1 ≤ · · · ≤ xk ≤ 0 ≤ xk+1 ≤ · · · ≤ xn ≤ y1 ≤ · · · ≤ yn (such a number k exists, for otherwise, we could substitute the numbers x1 , . . . , xn , y1 , . . . , yn by the numbers −yn , . . . , −y1 , −xn , . . . , −x1 ). Hence, we have that x1 y1 + · · · + xn yn  y1 (x1 − xn ) + · · · + yk (xk − xn ) + xk+1 yk+1 + · · · + xn−1 yn−1 + xn (−x1 − . . . − xn − yk+1 − . . . − yn−1 ) ≤ xn (x1 − xn ) + · · · + xn (xk − xn ) + xk+1 yk+1 + · · · + xn−1 yn−1 + xn (−x1 − . . . − xn − yk+1 − . . . − yn−1 )  −kxn2 − xn (xk+1 + · · · + xn ) + yk+1 (xk+1 − xn ) + . . . + yn−1 (xn−1 − xn ) ≤ 0,

whence x1 y1 + · · · + xn yn ≤ 0.

Problems for Independent Study Prove the following inequalities. z−ln y y−ln x z−ln x < ln z−x < ln y−x , where 0 < x < y < z. 1. ln z−y 2. ab bc cd d a ≥ ba cb d c ad , where 0 < a ≤ b ≤ c ≤ d . x1 xn n + · · · + S−x ≥ n−1 , where n ≥ 2, S  x1 + · · · + xn , x1 > 0, . . . , xn > 0. 3. S−x 1 n 1 3 3 3 4. a + b + c + 6abc ≥ 4 (a + b + c)3 , where a ≥ 0, b ≥ 0, c ≥ 0. 5. a2 (2b + 2c − a) + b2 (2a + 2c − b) + c2 (2a + 2b − c) ≥ 9abc, where a, b, c are the side lengths of some √ triangle. √ √ 6. (a) n a1 · . . . · an + n b1 · . . . · bn ≤ n (a1 + b1 ) · . . . · (an + bn ), where n ≥ 2, ai > 0, bi > 0, i  1, . . . , n, √ √ n (b) √ (n + 1)! − n n! ≥ 1, where n ≥ 2, n ∈ N, (c) n Fn+1 > 1 + √n 1F , where n ≥ 2, n ∈ N, F1  1, F2  2, Fk+2  n Fk+1 + Fk ,  k  1, 2,. . .,  n 1 (d) n C2n+1 , where n  2, 3, . . ., > 2 1 + √n n+1 n

(e) (1 + a1 ) · . . . · (n + an ) ≥ n 2 , where n ≥ 2, n ∈ N, a1 > 0, . . . , an > 0 and a1 · . . . · an  1,  √ √ n a ·...·a + n b ·...·b +b1 )·...·(an +bn ) 1 n n √ √1 ≥ n ∈ N, bi > 0, (f) n (a(a11−c n a ·...·a − n c ·...·c , where n ≥ 2, )·...·(a −c ) 1 n n 1 n 1 n ai > ci > 0, i  1, . . . , n, √ n

√ n

1 −c1 )·...·(an −cn ) 1 +b1 )·...·(a n +bn ) √ √ Hint. Prove that √n a(a·...·a ≥ 1 ≥ √n a(a1 ·...·a . n n n − c1 ·...·cn 1 n + b1 ·...·bn √ √ √ 3 3 3 ab + cd ≤ (a + c + d )(a + c + b), where a > 0, b > 0, c > 0, d > (g) 0. Hint. Note that    √ √ √ √ 3 3 √ √ 3 a b b + c d d ≤ (a + c)( b + d )2

80

6 Using Symmetry and Homogeneity

≤

  √ 3 (a + c)(b + d ) + 2(a + c) bd ≤ 3 (a + c)(b + d ) + (a + c)2 + bd .

7. x2 (x2 − 1)2 + y2 (y2 − 1)2 ≥ (x2 − 1)(y2 − 1)(x2 + y2 − 1). 8. x2 (x − 1)2 + y2 (y − 1)2 + z 2 (z − 1)2 ≥ 2xyz(2 − x − y − z). Hint. Prove that x2 (x − 1)2 + y2 (y − 1)2 + z 2 (z − 1)2 − 2xyz(2 − x − y − z)  x2 (x − 1)2 + (y(y − 1) − z(z − 1))2 + 2xyz(x + y − 1)(x + z − 1). 9. (x1 − x2 )(x1 − x3 )(x1 − x4 )(x1 − x5 ) + (x2 − x1 )(x2 − x3 )(x2 − x4 )(x2 − x5 )+ · · · + (x5 − x1 )(x5 − x2 )(x5 − x3 )(x5 − x4 ) ≥ 0. 10. 0 ≤ ab+bc+ca −abc ≤ 2, where a ≥ 0, b ≥ 0, c ≥ 0 and a2 +b2 +c2 +abc  4. Hint. Note that min(a, b, c) ≤ 1, and without loss of generality one can assume that (a − 1)(b − 1) ≥ 0. 11. xλ (x − y)(x − z) + yλ (y − z)(y − x) + z λ (z − x)(z − y) ≥ 0, where x, y, z > 0. Hint. Let x ≥ y ≥ z. Note that if λ ≥ 0, then we have xλ (x − y)(x − z) + yλ (y − z)(y − x) + z λ (z − x)(z − y)  (x − y)(xλ (x − z) − yλ (y − z)) + z λ (z − x)(z − y) ≥ 0. Otherwise, if λ < 0, then we have xλ (x − y)(x − z) + yλ (y − z)(y − x) + z λ (z − x)(z − y)  xλ (x − y)(x − z) + (y − z)(z λ (x − z) − yλ (x − y)) ≥ 0. 12.

 3

 a 2 b+c 5 2

+

 3

 b 2 + a+c 5 2

 3

 c 2 ≥ √334 , a+b 5 2

where a > 0, b > 0, c > 0.

13. (a − a + 3)(b − b + 3)(c − c + 3) ≥ (a + b + c)3 , where a > 0, b > 0, c > 0. Hint. Note that a5 − a2 + 3  a2 (a3 − 1) + 3 ≥ 1 · (a3 − 1) + 3  a3 + 13 + 13 . 14. abc + abd + bcd + acd − abcd ≤ 3, where a > 0, b > 0, c > 0, d > 0, and a3 + b3 + c3 + d 3 + abcd  5. Hint.  Prove that  3 a3 + b3 + c3 + d 3 (a + b + c + d ) + 12abcd ≥ 5((abc + abd + bcd + acd )(a + b + c + d ) − 4abcd ). 15. 0 ≤ ab+bc+ca−abc ≤ 2, where a ≥ 0, b ≥ 0, c ≥ 0 and a2 +b2 +c2 +abc  4. 16. a2 + b2 + c2 + 2abc  + 1 ≥22(ab + bc 2+ ca), where  a ≥ 0, b ≥ 0, c ≥ 0. x+y+z 1 (x − y) + (y − z) + (z − x)2 , where x > 0, y > 0, z > 0, 17. xy+yz+zx ≤ 1 + 48 and xy + yz + zx + xyz  4.

Chapter 7

The Principle of Mathematical Induction

A large number of inequalities that at first glance appear difficult can be easily proved by a classical mathematical proof technique called the principal of mathematical induction. This chapter is devoted to some applications of the principal of mathematical induction in proving algebraic inequalities. Note that there are different kinds of mathematical induction. In this chapter we mainly consider applications of the classical form of the principle of mathematical induction called the (first) principle of mathematical induction, since this variety is applied most often. Nevertheless, we also formulate two other relatively important variants of the principle of mathematical induction. Let Sn be a statement about an arbitrary positive integer n. The (first) principle of mathematical induction is used to prove that the statement Sn holds for all positive integral values of n. The principle of mathematical induction implies that in order to prove that statement Sn holds for all such values of n, one needs to prove first that statement S1 holds and then that from the assumption that statement Sk holds for an arbitrary positive integer k, it follows that statement Sk+1 holds as well. Putting together these explanations, we deduce the following formulation. (First) principle of mathematical induction. Let Sn be a statement about an arbitrary positive integer n. (i). The basis (first step/base case): S1 is true. (ii). The inductive step: Whenever Sk is true for k ∈ N, then Sk+1 is also true. The assumption that Sk holds for some k ∈ N is called the induction hypothesis (inductive hypothesis). (iii). Conclusion: If (i) and (ii) hold, then Sn is true for all n ∈ N. Remark Note that sometimes instead of n ∈ N, one can consider n ∈ N0  {0, 1, 2, . . .}, in which case one needs to verify (the basis) that S0 is true.

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_7

81

82

7 The Principle of Mathematical Induction

So, the (first) principle of mathematical induction implies that if the basis and the inductive step are proved, then Sn is true for all positive integers n. Let us consider another useful variant of the principle of mathematical induction called the second principle of mathematical induction. (It is also called “strong induction,” because a stronger induction hypothesis is used.) Second principle of mathematical induction. Let Sn be a statement about an arbitrary positive integer n. (i). The basis: S1 is true. (ii). The inductive step: Whenever Si is true for all i ≤ k, k ∈ N, it follows that Sk+1 is true. (iii). Conclusion: If (i) and (ii) hold, then Sn is true for all n ∈ N. Now let us consider the following variant of the principle of mathematical induction that is used to prove that statement Sn holds not for all positive integers n, but only for all positive integers greater than or equal to a certain positive integer k. Different starting point (induction basis other than 1 or 0). Let m be a given positive integer and let Sn be a statement about an arbitrary positive integer n, where n ≥ m. (i). The basis: Sm is true. (ii). The inductive step: Whenever Si is true for all m ≤ i ≤ k, it follows that Sk+1 is true. (iii). Conclusion: If (i) and (ii) hold, then Sn is true for all n ∈ N, n ≥ m. Below we provide examples of inequalities whose proofs make use of the (first) principle of mathematical induction. Example 7.1 Prove that

1 2n+1

+

1 2n+2

+ ··· +

1 2n+n

≥

11 , 30

where n > 1, n ∈ N.

11 11 Proof If n  2, then we have 30 ≥ 30 . Assume that the given inequality is true for n  k, that is,

1 1 11 1 + + ··· + ≥ . 2k + 1 2k + 2 2k + k 30

(7.1)

Let us prove that the given inequality is true for n  k + 1. For n  k + 1 we need to prove that 1 1 1 11 + + ··· + ≥ . 2(k + 1) + 1 2(k + 1) + 2 2(k + 1) + (k + 1) 30 Adding to both sides of inequality (7.1) the expression we obtain

1 , 2k+2

1 3k+1

1 1 1 + 3k+2 + 3k+3 − 2k+1 −

1 1 11 1 1 1 1 1 + ··· + ≥ + + + − − . 2k + 3 3k + 3 30 3k + 1 3k + 2 3k + 3 2k + 1 2k + 2 Since

1 3k+1

+

1 3k+2

+

1 3k+3

−

1 2k+1

−

1 2k+2

> 0 (Problem 1.19), it follows that

7 The Principle of Mathematical Induction

83

1 11 1 + ··· + ≥ . 2k + 3 3k + 3 30 Therefore, the given inequality is true for every positive integer n. This ends the proof. Example 7.2 Prove that 1 1 + · · · + 2 < 1, 22 n

(7.2)

where n  3, 4, . . . . Proof Note that it is impossible to prove this inequality similarly to how we proved Example 7.1. In the proof of this inequality and in the proof of some other inequalities, the method of induction is applied to another inequality (a more general one), from which follows the validity of the considered inequality. In order to prove inequality (7.2), let us prove the following inequality: 1 1 1 + ··· + 2 < 1 − . 2 2 n n

(7.3)

We carry out the proof by mathematical induction on n. For n  3, we have 212 + 312 < 1 − 13 . Assume that (7.3) is true for n  k (k ≥ 3), and let us prove that it is true for n  k + 1. For n  k, we have 212 + · · · + k12 < 1 − k1 . 1 Let us add to both sides of this inequality the expression (k+1) 2 . Then we obtain 1 1 1 1 1 + · · · + k 2 + (k+1)2 < 1 − k + (k+1)2 . 22 1 1 1 1 1 Since 1− k1 + (k+1) 2 < 1− k+1 , we deduce that the inequality 22 +· · ·+ k 2 + (k+1)2 < 1 1 − k+1 coincides with the given inequality for n  k + 1. Therefore, the given inequality holds for every positive integer n ≥ 3. This ends the proof. In some inequalities the principle of mathematical induction is used in the following way: (a) One first proves that the considered inequality holds for values of the positive integer n equal to n 1 , . . . , n k , . . . , where n 1 < · · · < n k < · · ·. (b) From the validity of the considered inequality for arbitrary n  k (k ≥ 2), follows that it holds for n  k − 1. Example 7.3 Ky Fan inequality: Prove that n ≥ 2, 0 < x1 ≤ 21 , . . . , 0 < xn ≤ 21 .

x1 ···xn (x1 +···+xn )n

≤

(1−x1 )···(1−xn ) , ((1−x1 )+···+(1−xn ))n

Proof Let us first prove the given inequality for n equal to 21 , . . . , 2k , . . . . (1−x1 )(1−x2 ) 1 x2 For n  2, we have (x x+x ≤ ((1−x (Problem 1.20). )2 )+(1−x ))2 1

2

1

2

where

84

7 The Principle of Mathematical Induction

Assume that the given inequality holds for n  2k (k ∈ N), and let us prove that it holds for n  2k+1 . We have 

p p x1 + · · · + x p x p+1 + · · · + x2 p  2 p x1 + · · · + x p x p+1 + · · · + x2 p x1 + · · · + x2 p x1 + · · · + x2 p ⎛   ⎞p x p+1 +···+x2 p x1 +···+x p       1 − x p+1 · · · 1 − x2 p (1 − x1 ) · · · 1 − x p p p ⎟ ⎜ ≤   p ·     p · ⎝    ⎠ x +···+x2 p 2 x1 +···+x p 1 − x p+1 + · · · + 1 − x2 p (1 − x1 ) + · · · + 1 − x p + p+1 x1 · · · x2 p

2 p  

x1 · · · x p

p · 

x p+1 · · · x2 p



p ·

p

  (1 − x1 ) · · · 1 − x2 p ≤   p     p  1 − x p+1 + · · · + 1 − x2 p (1 − x1 ) + · · · + 1 − x p ⎛    ⎞p x +···+x x +···+x   1 − 1 p p 1 − p+1 p 2 p (1 − x1 ) · · · 1 − x2 p ⎟ ⎜ × ⎝    2 ⎠   2 p ,  x +···+x2 p x +···+x p (1 − x1 ) + · · · + 1 − x2 p 1− 1 + 1 − p+1 p

p

p

where p  2k . (1−x1 )···(1−x2 p ) x1 ···x2 p We obtain ≤ , whence the given inequality holds (x1 +···+x2 p )2 p ((1−x1 )+···+(1−x2 p ))2 p k+1 for n  2 . Now let us prove that if the given inequality holds for m, then it holds true for m − 1, where m ≥ 3 (m ∈ N). (1−x1 )···(1−xm ) 1 ···x m . We have (x1 x+···+x m ≤ ((1−x1 )+···+(1−xm ))m m) x1 +···+xm−1 Taking xm  m−1 , where by Problem 1.10, 0 < xm ≤ 21 , we obtain 

x 1 +···+x m−1 m−1  x 1 +···+x m−1 m xm−1 + m−1

x1 · · · xm−1 · x1 + · · · +

  m−1 (1 − x1 ) · · · (1 − xm−1 ) · 1 − x1 +···+x m−1 ≤  m .  m−1 (1 − x1 ) + · · · + (1 − xm−1 ) + 1 − x1 +···+x m−1

Therefore, we have x1 · · · xm−1 (1 − x1 ) · · · (1 − xm−1 ) ≤ . m−1 (x1 + · · · + xm−1 ) ((1 − x1 ) + · · · + (1 − xm−1 ))m−1

Problems Prove the following inequalities (7.1–7.4, 7.6, 7.7, 7.10, 7.11, 7.13–7.20, 7.24, 7.25). 7.1. (a)

1 2

· · · 2n−1 ≤ 2n

√1 , 3n+1 1 · · · + 2n

where n ∈ N.

+ < 25 , where n ≥ 2, n ∈ N. 36

√ √ 7.2. (a) a + a + · · · + a ≤ 1+ 24a+1 , where a ≥ 0,    1 (b) n+1

+

1 n+2

n

Problems

85

 √ (b) 2 3 4 . . . n < 3, where n ≥ 2, n ∈ N. 7.3. x12 + · · · + (2n − 1)xn2 ≤ (x1 + · · · + xn )2 , where x1 ≥ · · · ≥ xn ≥ 0. 7.4. (a) |sin(x1 + · · · + xn )| ≤ |sin x1 | + · · · + |sin xn |, (b) sin(x1 + · · · + xn ) ≤ sin x1 + · · · + sin xn , where x1 , . . . , xn ∈ [0, π ], (c) |cos x1 | + |cos x2 | + |cos x3 | + |cos x4 | + |cos x5 | ≥ 1, where x1 + x2 + x3 + x4 + x5  0. 7.5. Prove (a) Bellman’s inequality: if a function f (x) is defined in [0, a) (or [0, +∞)) and for arbitrary numbers x ≥ y ≥ z from that interval we have f (x)− f (y)+ f (x) ≥ f (x − y + z) and f (0) ≤ 0, then for all numbers a > x1 ≥ · · · ≥ n−1 f (xn ) ≥ xn ≥ 0, the following inequality  holds: f (x1 )− f (x2 )+· · ·+(−1) n−1 f x1 − x2 + · · · + (−1) xn ,   (b) tan x1 − tan x2 + · · · + (−1)n−1 tan xn ≥ tan x1 − x2 + · · · + (−1)n−1 xn , where π2 > x1 ≥ · · · ≥ xn ≥ 0, r  (c) a1r − a2r + · · · + (−1)n−1 anr ≥ a1 − a2 + · · · + (−1)n−1 an , where a1 ≥ · · · ≥ an ≥ 0, r ≥ 1. 7.6. (a) (x1 + · · · + x5 )2 ≥ 4(x1 x2 + x2 x3 + x3 x4 + x4 x5 + x5 x1 ),where x1 > 0, . . . , x5 > 0.

2 2 (b) x1 xn2 + x22 + x2 x12 + x32 +· · ·+ xn−1 xn−2 + xn2 + xn xn−1 + x12 ≤ 21 (x1 +

7.7. 7.8. 7.9. 7.10. 7.11.

· · · + xn )2 , where n ≥ 3 and x1 > 0, . . . , xn > 0. 1 + · · · + xn )2 ≤ (x1 + · · · + nxn ) · max(x1 , . . . , xn ), where x1 2 (x 1 0, . . . , xn ≥ 0. Prove Theorem 11.1 (Chapter 11). Prove Theorem 11.2 (Chapter 11). (a) a1 + · · · + ann ≤ na1 · . . . · an , where a1 ≥ · · · ≥ an ≥ 1; (b) a1 + · · · + ann ≥ na1 · . . . · an , where 0 ≤ a1 ≤ · · · ≤ an ≤ 1. a2 an2 ≥ 4(an − a1 ), where a1 > 0, . . . , an > 0; (a) a21 + · · · + an−1 a3

a3

≥

3

(a1 +···+an ) , where ai > 0, bi > 0, ci > 0, i  (b) b1 1c1 + · · · + bn ncn ≥ (b1 +···+b n )(c1 +···+cn ) 1, . . . , n. 7.12. Prove that (a) if f (x) is defined in I 1 and is a convex function,2 then

(x2 + x1 )( f (x2 ) − f (x1 )) + · · · + (xn + xn−1 )( f (xn ) − f (xn−1 )) ≥ (xn + x1 )( f (xn ) − f (x1 )),

where√n ≥ 2, xn , x√ 1, . . . , x √n ∈ I . √ x1 < √· · · <√ (b) a b + b c + c a ≥ a c + b a + c b, where a ≥ b ≥ c ≥ 0; x (c) x1x2 · x2x3 · . . . · xnx1 ≥ x2x1 · x3x2 · . . . · xn n−1 · x1xn , where xn ≥ · · · ≥ x1 > 0, n ≥ 3; a(c−b) b(a−c) c(b−a) + (a+c)(2b+a+c) + (b+a)(2c+b+a) ≤ 0, (d) (c+b)(2a+b+c) where a ≥ b ≥ c > 0. 1I

is defined as the domain of function f , so I = D(f) Chapter 11.

2 See

86

7 The Principle of Mathematical Induction a1 +···+an−1 n−1

n+1 n + a1 +···+a ≥ 2 a1 +···+a , where n ≥ 2, ak +a2 k+2 ≥ ak+1 , k  n+1 n 1, . . . , n − 1. +···+a2n 2n−1 ≥ a0 +a2n+1 , where ak ≥ ak−12+ak+1 , k  1, 2, . . . , 2n − 7.14. (a) a1 +a3 +···+a n 1, n ∈ N, a 0 +a 2 +···+a 2n n+1 , where a > 0, n ∈ N. (b) a+a 3 +···+a 2n−1 ≥ n 1 7.15. 1 + · · · + n > ln(n + 1), where n ∈ N. 7.16. 1 + · · · + n √1 n ≤ 3 − √2n , where n ∈ N.

7.13.

α 2 , where α ≥ 0, n ∈ N. 7.17. (1 + α)n ≥ 1 + nα + n(n−1) 2  k+1 k 7.18. k! ≥ e , where k ∈ N. √ n    sin(2i x) ≤ 1 + 3 n, where n  0, 1, 2, . . . . 7.19. 2 i0

7.20. cos α + · · · + cosnnα ≥ − 21 , where n ∈ N, 0 ≤ α ≤ π2 . 7.21. Let the numbers a1 , . . . , an (n ≥ 2) be greater than 1 and |ak+1 − ak | < 1 for k  1, . . . , n − 1. Prove that the sum aa21 + aa23 + · · · + aan−1 + aan1 is less than n 2n − 1. . Prove that 7.22. Let √1  x1 ≤ x2 ≤√· · · ≤ xn+1√ , where n ∈ N, (a) xx2 2−x1 + · · · + xxn+1n+1−xn ≤ 4n−3 2 √

√

(b) xx2 2−x1 + · · · + xxn+1n+1−xn < 1 + 21 + · · · + n12 , where x2 , . . . , xn+1 ∈ N. 7.23. Prove that, if α1 > 0, . . . , αn > 0, β1 > 0, . . . , βn > 0, and α1 + · · · + αn ≤ β1 βn αn α1 + · · · + cos ≤ cos + · · · + cos . β1 + · · · + βn ≤ π, then cos sin α1 sin αn sin α1 sin αn 2012 2012 2012 2011 2011 7.24. 2(a + 1)(b + 1)(c + 1) ≥ (1 + abc)(a + 1)(b + 1)(c2011 + 1), where a > √ 0, b > 0,c > 0. 7.25. (a) b1 ≥ b2 , (b) Newton’s inequality: bk2 ≥ bk−1 bk+1 , for k  2, . . . , n − 1, where n ≥ 2, n ∈ N, a1 > 0, . . . , an > 0, and bk 

1 (a1 · · · ak−1 ak + a1 · · · ak−1 ak+1 + · · · + a1 · · · ak−1 an + · · · + an−k+1 · · · an−1 an ). Cnk

Proofs 7.1. (a) We proceed by mathematical induction on n. The given inequality holds for n  1, since 21 ≤ √14 . Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. 1 For n  k, we have 21 · · · 2k−1 ≤ √3k+1 . 2k Multiplying both sides of this inequality by

2(k+1)−1 , 2(k+1)

we deduce that

2k − 1 2k + 1 1 1 2k + 1 ··· · ≤√ . · 2 2k 2k + 2 3k + 1 2k + 2

Proofs

87

Taking into consideration that

√1 3k+1

√1 3k+4

(Problem 1.21), we obtain

··· · ≤ Therefore, the given inequality holds for every positive integer n. (b) In order to prove the given inequality, let us prove the following inequality: 1 2

2k−1 2k

2(k+1)−1 2(k+1)

√ 1 . 3(k+1)+1

2k+1 · 2k+2 ≤

1 1 1 25 1 + + ··· + + ≤ , where n ≥ 2, n ∈ N. n+1 n+2 2n 4n + 1 36

(1)

We proceed by mathematical induction on n. Note that inequality (1) holds for n  2, since 13 + 41 + · · · + 19  25 . 36 Assume that inequality (1) holds for n  k, where k ≥ 2, k ∈ N. Let us prove that inequality (1) holds for n  k + 1. We have 1 1 1 1 1 1 + + ··· + +  + k+2 k+3 2(k + 1) 4(k + 1) + 1 k+1 k+2   1 25 1 1 1 1 1 1 + ··· + ≤ + + + − + − 2k 4k + 1 2k + 1 2k + 2 k + 1 4k + 5 4k + 1 36   25 1 1 1 1 1 1 ≤ + + − + − + 2k + 1 2k + 2 k + 1 4k + 5 4k + 1 36 (2k + 1)(2k + 2) 4 25 − ≤ . 36 (4k + 1)(4k + 5)

Then inequality (1) holds for n  k + 1. Therefore, the given inequality holds for every positive integer n ≥ 2. 7.2. (a) We proceed by induction on n. √ √ The given inequality holds for n  1, since a ≤ 1+ 24a+1 .

√ √ Assume that the inequality a + a + · · · + a ≤ 1+ 24a+1 holds. We have    k   √ √

 √ √ 1 + 4a + 1 1 + 4a + 1  a + a + · · · + a  a + a + · · · + a ≤ a +  ,  2 2      



k+1

k

and thus it follows that the given inequality holds for n  k + 1. Therefore, the inequality holds for every positive integer n.  √ (b) We prove the inequality k (k + 1) . . . n < k + 1, where 1 ≤ k ≤ n and k ∈ N , by induction √ on p  n − k (n is assumed to be a constant). For p  0, we have n < n + 1, whose proof is obvious. Assume that the given inequality holds for p  n − m, and let us prove that it holds for p  n − m + 1  n − (m − 1).

 √ For p  n −m, we have m (m + 1) . . . n < m +1, and hence we obtain

 √ √ (m − 1) m (m + 1) . . . n < (m − 1)(m + 1) < m.

88

7 The Principle of Mathematical Induction

 √ Hence, the inequality k (k + 1) . . . n < k+1 holds for 1 ≤ k ≤ n, k ∈ N.



 √ For k  2, we have 2 3 4 . . . n < 3, which ends the proof. 7.3. We proceed by induction on n. For n  1, we have x12 ≤ x12 . Assume that the inequality x12 + · · · + (2k − 1)xk2 ≤ (x1 + · · · + xk )2 holds. 2 to both sides of this inequality, we obtain the following Adding (2k + 1)xk+1 inequality: 2 2 x12 + · · · + (2k − 1)xk2 + (2k + 1)xk+1 ≤ (x1 + · · · + xk )2 + (2k + 1)xk+1 . (7.4)

Let us prove that 2 ≤ (x1 + · · · + xk+1 )2 , (x1 + · · · + xk )2 + (2k + 1)xk+1

(7.5)

2 ≤ (x1 + · · · + xk+1 )2 − (x1 + · · · + xk )2  which is equivalent to (2k + 1)xk+1 (2(x1 + · · · + xk ) + xk+1 )xk+1 ,or to the following inequality:

0 ≤ xk+1 (2(x1 + · · · + xk ) − 2kxk+1 )  2xk+1 ((x1 − xk+1 ) + · · · + (xk − xk+1 )).

This inequality holds because according to the assumption of the problem, we have 0 ≤ xk+1 ≤ xi , i  1, . . . , k. 2 ≤ Hence, from (7.4) and (7.5), it follows that x12 +· · ·+(2k−1)xk2 +(2k + 1)xk+1 2 (x1 + · · · + xk+1 ) , which is the given inequality for n  k + 1. Therefore, the given inequality holds for every positive integer n. 7.4. (a) We proceed by induction on n. For n  2, we have |sin(x1 + x2 )| ≤ |sin x1 | + |sin x2 |. Indeed, since sin(x1 + x2 )  sin x1 cos x2 + sin x2 cos x1 and |cos α| ≤ 1, it follows that |sin(x1 + x2 )| ≤ |sin x1 cos x2 | + |sin x2 cos x1 | ≤ |sin x1 | + |sin x2 |. Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. Indeed, we have |sin((x1 + · · · + xk ) + xk+1 )| ≤ |sin(x1 + · · · + xk )| + |sin xk+1 | ≤ (|sin x1 | + · · · + |sin xk |) + |sin xk+1 |,

which is the given inequality for n  k + 1. Therefore, the given inequality holds for every positive integer n. (b) As for x ∈ [0, π ], we have that sin x ≥ 0, and then from part (a), it follows that sin x1 + · · · + sin xn ≥ |sin(x1 + · · · + xn )| ≥ sin(x1 + · · · + xn ). (c) We have  π  π       |cos x1 | + |cos x2 | + |cos x3 | + |cos x4 | + |cos x5 |  sin − x1  + · · · + sin − x5 , 2 2

Proofs

89

and then from part (a) it follows that   π  π     π  π   5π         1. − x1  + · · · + sin − x5  ≥ sin − x1 + · · · + − x5   sin sin 2 2 2 2 2 

Therefore, |cos x1 | + |cos x2 | + |cos x3 | + |cos x4 | + |cos x5 | ≥ 1. 7.5. (a) Let us first prove the given inequality for n  1, 3, 5, 7, . . . , 2k + 1, . . .. For n  1, we have f (x1 ) ≥ f (x1 ). Assume that the given inequality holds for n  2k − 1 (k ∈ N), and let us prove that it holds for n  2k + 1. We have f (x1 ) − f (x2 ) + · · · + f (x2k−1 ) ≥ f (x1 − x2 + · · · + x2k−1 ), and therefore,                 f x1 − f x2 + · · · + f x2k−1 − f x2k + f x2k+1 ≥ f x1 − x2 + x3 − · · · + x2k−1 − f x2k + f x2k+1 .

Note that x1 ≥ x  x1 − x2 + x3 − · · · + x2k−1 ≥ y  x2k ≥ z  x2k+1 ≥ 0, since x1 − x  (x2 − x3 ) + · · · + (x2k−2 − x2k−1 ) ≥ 0 and x − y  (x1 − x2 ) + · · · + (x2k−1 − x2k ) ≥ 0.

Therefore, f (x) − f (y) + f (x) ≥ f (x − y + z). Thus f (x1 ) − f (x2 ) + · · · + f (x2k−1 ) − f (x2k ) + f (x2k+1 ) ≥ f (x1 − x2 + · · · + x2k−1 ) − f (x2k ) + f (x2k+1 ) ≥ f (x1 − x2 + x3 − · · · + x2k−1 − x2k + x2k+1 ), from which it follows that the given inequality holds for n  2k + 1. Hence, the given inequality holds for an arbitrary odd number n. Now let us prove that the given inequality holds for n  2k. Let x1 ≥ x2 ≥ · · · ≥ x2k . Let us take x2k+1  0, from which we obtain f (x1 ) − f (x2 ) + · · · − f (x2k ) + f (0) ≥ f (x1 − x2 + · · · − x2k + 0), or f (x1 ) − f (x2 ) + · · · − f (x2k ) ≥ f (x1 − x2 + · · · − x2k ), as f (0) ≤ 0. (b) Let us prove that tan x − tan y + tan z ≥ tan(x − y + z), where π2 > x ≥ y ≥ z ≥ 0. The last inequality can be rewritten as tan x − tan y ≥ tan(x − y + z) − tan z, sin(x−y) ≥ cos zsin(x−y) . or cos x cos y cos(x−y+z) If x  y, then the proof is obvious. If x  y, then the given inequality is equivalent to the following inequality: cos z cos(x − y + z) ≥ cos x cos y, which holds because 0 ≤ z ≤ y < π2 and 0 ≤ x − y + z ≤ x < π2 . Let us consider the function f (x)  tan x. Since f (0)  0 and the assumption of Problem 7.5(a) holds, it follows that the given inequality holds.

90

7 The Principle of Mathematical Induction

(c) The proof follows from Problem 7.5(a) and Problem 9.34. 7.6. (a) Let us begin by proving the more general inequality (x1 + · · · + xn )2 ≥ 4(x1 x2 + x2 x3 + · · · + xn−1 xn + xn x1 ),

(7.6)

where n ≥ 4, x1 > 0, . . . , xn > 0. We proceed by induction on n. If n  4, then the inequality (x1 + x2 + x3 + x4 )2 ≥ 4(x1 x2 + x2 x3 + x3 x4 + x4 x1 ) holds, since (x1 + x2 + x3 + x4 )2 − 4(x1 x2 + x2 x3 + x3 x4 + x4 x1 )  (x1 − x2 + x3 − x4 )2 ≥ 0.

Assume that (7.6) holds for n  k, and let us prove that (7.6) holds for n  k + 1. Let max(x1 , . . . , xk+1 )  xi . It follows that (x1 + · · · + (xi−2 + xi−1 ) + xi + · · · + xk+1 )2 ≥ 4(x1 x2 + x2 x3 + · · · + xi−3 (xi−2 + xi−1 ) + (xi−2 + xi−1 )xi + xi xi+1 + · · · + xk xk+1 + xk+1 x1 ) > 4(x1 x2 + x2 x3 + · · · + xi−3 xi−2 + xi−2 xi + xi−1 xi + xi xi+1 + · · · + xk xk+1 + xk+1 x1 ) ≥ 4(x1 x2 + x2 x3 + · · · + xi−3 xi−2 + xi−2 xi−1 + xi−1 xi + xi xi+1 + · · · + xk xk+1 + xk+1 x1 )

(here we assume that x0  xk+1 , x−1  xk ). (b) If n ≥ 4, then we have



2 2 + xn2 + xn xn−1 + x12 x1 xn2 + x22 + x2 x12 + x32 + · · · + xn−1 xn−2 ≤ x1 (xn + x2 ) + x2 (x1 + x3 ) + · · · + xn−1 (xn−2 + xn ) + xn (xn−1 + x1 ) 1  2(x1 x2 + x2 x3 + xn−1 xn + xn x1 ) ≤ (x1 + · · · + xn )2 . 2 The last inequality holds by to Problem 7.6(a). If n  3, then without loss of generality one can assume that max(x1 , x2 , x3 )  x3 . Thus, it follows that

  x2  x1  + x2 x3 + x1 x32 + x22 + x2 x12 + x32 + x3 x12 + x22 ≤ x1 x3 + 2 2 1 x32 + x12 + x22 2  (x1 + x2 + x3 ) . + 2 2 7.7. We proceed by induction on n. If n  1, then 21 x12 ≤ x12 . Obviously, the last inequality holds. Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. If n  k, we have 21 (x1 + · · · + xk )2 ≤ (x1 + · · · + kxk )max(x1 , . . . , xk ). Adding to both sides of this inequality the expression

Proofs

91 x2

xk+1 (x1 + · · · + xk ) + k+1 , we obtain on the left-hand side the expres2 2 1 sion 2 (x1 + · · · + xk+1 ) , and on the right-hand side the expression (x1 + · · · + kxk )max(x1 , . . . , xk ) + xk+1 (x1 + · · · + xk ) + not greater than (x1 + · · · + (k + 1)xk+1 )max(x1 , . . . , xk+1 ). Indeed, since max(x1 , . . . , xk ) ≤ max(x1 , . . . , xk+1 )  c, cxk+1 , we have (x1 + · · · + kxk )max(x1 , . . . , xk ) + xk+1 (x1 + · · · + xk ) +

2 xk+1 , 2

which is

2 xk+1 2

2 ≤ xk+1 ≤

2 xk+1 ≤ (x1 + · · · + kxk )c 2

+ kcxk+1 + cxk+1  (x1 + · · · + (k + 1)xk+1 )c.

Therefore, the given inequality holds for every positive integer n. 7.8. Let us first prove that for arbitrary numbers x1 , . . . , xn belonging to the domain I, we have x + ··· + x  f (x1 ) + · · · + f (xn ) 1 n . ≥ f n n

(7.7)

We proceed by induction on n. If n  2, then inequality (7.7) coincides with the condition of Theorem 11.1. Assume that for n  k, inequality (7.7) holds, and let us prove that it holds for n  k + 1. Let us consider the following expression: x + ··· + x  1 k+1 f (x1 ) + · · · + f (xk ) + f (xk+1 ) + (k − 1) f  A. k+1 Note that A  ( f (x1 ) + · · · + f (xk )) + f (xk+1 ) + f 

x + ··· + x  x + ··· + x  1 k+1 1 k+1 + ··· + f k+1 k+1   k−1

  k+1 x + ··· + x  xk+1 + (k − 1) x1 +···+x 1 k k+1 + kf ≥ kf k k ⎛ x1 +···+xk+1 ⎞ x +(k−1) x1 +···+xk k+1   + k+1 k k ⎠  2k f x1 + · · · + xk+1 . ≥ 2k f ⎝ 2 k+1

Hence, we deduce that f (x1 ) + · · · + f (xk+1 ) + (k − 1) f   k+1 2k f x1 +···+x , k+1 and therefore,

 x1 +···+xk+1  k+1

≥

x + ··· + x  f (x1 ) + · · · + f (xk+1 ) 1 k+1 ≥ f . k+1 k+1 Thus, it follows that inequality (7.7) holds. Now let us prove the inequality (11.3). Let αi  pi , qi ∈ N, i  1, . . . , n.

pi qi

, i  1, . . . , n, where

92

7 The Principle of Mathematical Induction

Let us denote the least common multiple of the numbers q1 , . . . , qn by q Therefore, qpii  lqi , where li ∈ N, i  1, . . . , n. Let us estimate the left-hand side of the given inequality, α1 f (x1 ) + · · · + n f (x n )  B. αn f (xn )  l1 f (x1 )+···+l q Applying inequality (7.7) to the numbers x1 , . . . , x1 , . . . , xn , . . . , xn , note       l1

ln

that their  total numberis equal to q and since α1 + · · · + αn  1, we obtain B ≥ f l1 x1 +l2 x2q+···+ln xn  f (α1 x1 + · · · + αn xn ). 7.9. We proceed by induction on n. If n  2, then inequality (11.7) coincides with the condition (11.5) of Theorem 11.2. Assuming that (11.7) holds for n  k, let us prove that it holds for n  k + 1, that is, that β1 f (a1 ) + · · · + βk+1 f (ak+1 ) ≥ f (β1 α1 + · · · + βk+1 αk+1 ), where β1 ≥ 0, . . . , βk+1 ≥ 0 and β1 + · · · + βk+1  1, a1 , . . . , an ∈ I. We have β1 f (a1 ) + · · · + βk+1 f (ak+1 )   β1 βk  (β1 + · · · + βk ) f (a1 ) + . . . + f (ak ) + βk+1 f (ak+1 ) β1 + · · · + βk β1 + · · · + βk   β1 a1 + · · · + βk ak + βk+1 f (ak+1 ) ≥ (β1 + · · · + βk ) f β1 + · · · + βk   β1 a1 + · · · + βk ak ≥ f (β1 + · · · + βk ) + βk+1 ak+1  f (β1 a1 + · · · + βk ak + βk+1 ak+1 ), β1 + · · · + βk

therefore β1 f (a1 ) + · · · + βk+1 f (ak+1 ) ≥ f (β1 α1 + · · · + βk+1 αk+1 ). βi , i We have used (11.7) for n  k, substituting xi by ai , and αi by β1 +···+β k 1, . . . , k. Hence inequality (11.7) holds for every positive integer n. 7.10. (a) We proceed by induction on n. For n  1, the given inequality holds because a1 ≤ a1 . Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. For n  k, we have a1 + · · · + akk ≤ ka1 · · · ak . On the other hand, by to the assumption of the problem, we have 1 ≤ ak+1 . Multiplying these inequalities, we obtain a1 + · · · + akk ≤ ka1 · · · ak ak+1 . By to the assumption of the problem, k+1 ≤ a1 · · · ak+1 . Summing these inequalities, we deduce that we have ak+1 k+1 ≤ ka1 · · · ak+1 + a1 · · · ak+1  (k + 1)a1 · · · ak+1 . a1 + · · · + akk + ak+1

Therefore, the given inequality holds for every positive integer n. (b) We proceed by induction on n. For n  1, the given inequality holds because a1 ≥ a1 .

Proofs

93

Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. For n  k, we have a1 +· · ·+akk ≥ ka1 · · · ak . On the other hand, by assumption, we have 1 ≥ ak+1 . Multiplying these two inequalities, we deduce that a1 + · · · + akk ≥ ka1 · · · ak ak+1 . k+1 ≥ a1 · · · ak+1 . By to the assumption of the problem, we have ak+1 k+1 ≥ (k + 1)a1 · · · ak+1 . Summing these inequalities, we obtain a1 + · · · +akk +ak+1 Therefore, the given inequality holds for every positive integer n. 7.11. (a) We proceed by induction on n.   a2 For n  2, we have a21 ≥ 4(a2 − a1 ), a22 − 4a1 a2 + 4a12  (a2 − 2a1 )2 ≥ 0 . Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. 2 ak2 a ≥ 4(ak − a1 ). For n  k, we have a21 + · · · + ak−1 Let us add to both sides of this inequality the expression 4(ak+1 − ak ). Then ak2 a2 + 4(ak+1 − ak ) ≥ 4(ak+1 − a1 ). we obtain a21 + · · · + ak−1 It is left to prove that a2 a2 a2 a2 a22 + · · · + k + k+1 ≥ 2 + · · · + k + 4(ak+1 − ak ). a1 ak−1 ak a1 ak−1 Indeed, this inequality is equivalent to the following inequality: 2 ak+1 ≥ 4(ak+1 − ak ), or (ak+1 − 2ak )2 ≥ 0. ak

Therefore, the given inequality holds for every positive integer n. (b) We proceed by induction on n. For n  2, we have 

 a13 a23 + (b1 + b2 )(c1 + c2 )  a13 + a23 b1 c1 b2 c2     c2 b2 b1 c1 c1 b1 b2 c2 + a23 + a23 + a13 + a13 + a23 + a13 c1 b1 b2 c2 c2 b2 b1 c1   c2 b2 3 b1 c1 c1 b1 3 b2 c2 ≥ a13 + a23 + 3 3 a13 · a13 ·a + 3 3 a23 · a23 ·a  (a1 + a2 )3 , c1 b1 2 b2 c2 c2 b2 1 b1 c1

and therefore, a3 (a1 + a2 )3 a13 . + 2 ≥ b1 c1 b2 c2 (b1 + b2 )(c1 + c2 ) Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. 3 a3 a (a1 +···+ak )3 For n  k, we have b1 1c1 + · · · + bk kck ≥ (b1 +···+b . k )(c1 +···+ck )

94

7 The Principle of Mathematical Induction

Let us add to both sides of this inequality the expression obtain

3 ak+1 . bk+1 ck+1

Then we

3 3 ak+1 a3 ak+1 (a1 + · · · + ak )3 a13 + + ··· + k + ≥ b1 c1 bk ck bk+1 ck+1 (b1 + · · · + bk )(c1 + · · · + ck ) bk+1 ck+1 ((a1 + · · · + ak ) + ak+1 )3 , ≥ ((b1 + · · · + bk ) + bk+1 )((c1 + · · · + ck ) + ck+1 )

and hence the given inequality holds for n  k + 1. Therefore, the given inequality holds for every positive integer n. 7.12. (a) We proceed by induction on n. For n  2, we have (x2 + x1 )( f (x2 ) − f (x1 )) ≥ (x2 + x1 )( f (x2 ) − f (x1 )). For n  k, we have 

x2 + x1

                 f x2 − f x1 + · · · + xk + xk−1 f xk − f xk−1 ≥ xk + x1 f xk − f x1 .

Let us add to both sides of this inequality (xk+1 + xk )( f (xk+1 ) − f (xk )), from which we obtain

the

expression

(x2 + x1 )( f (x2 ) − f (x1 )) + · · · + (xk + xk−1 )( f (xk ) − f (xk−1 )) + (xk+1 + xk )( f (xk+1 ) − f (xk )) ≥ (xk + x1 )( f (xk ) − f (x1 )) + (xk+1 + xk )( f (xk+1 ) − f (xk )). In order to prove that the given inequality holds for n  k + 1, it is sufficient to prove the following inequality: (xk + x1 )( f (xk ) − f (x1 )) + (xk+1 + xk )( f (xk+1 ) − f (xk )) ≥ (xk+1 + x1 )( f (xk+1 ) − f (x1 )), or (xk+1 − xk ) f (x1 ) + (xk − x1 ) f (xk+1 ) ≥ (xk+1 − x1 ) f (xk ).

Since f (x) is a convex function, it follows that xk − x1 xk+1 − xk f (x1 ) + f (xk+1 ) ≥ f xk+1 − x1 xk+1 − x1



xk − x1 xk+1 − xk x1 + xk+1 xk+1 − x1 xk+1 − x1

(7.8)   f (xk ),

which implies inequality (7.8). This ends the proof of the given inequality for n  k + 1. Therefore, the given inequality holds for every positive integer n (n > 1). (b) Note that one can rewrite inequality (a) of this problem in the following way:             xn f x1 + x1 f x2 + x2 f x3 + · · · + xn−1 f (xn ) ≥ x2 f x1 + x3 f x2 + · · · + xn f xn−1 + x1 f (xn ),

√ where on taking f (x)  − x, we obtain

Proofs

95

 √   √   √   √   √   √  a − c +c − b +b − a ≥b − c +a − b +c − a . √ √ √ √ √ √ Thus, it follows that a c + c b + b a ≤ b c + a b + c a. (c) Consider the function f (x)  − ln x. By to inequality (a), we have

        xn − ln x1 + x1 − ln x2 + · · · + xn−1 (− ln xn ) ≥ x2 − ln x1 + · · · + xn − ln xn−1 + x1 (− ln xn ), or xn−1 x1 x2 x3 x1 xn xn x1 · x2 · · · xn ≤ x1 · x2 · · · xn−1 · xn .

(d) Consider the function f (x) 

  1 x 1 1 .  − 2 a+b+c−x a+b+c+x (a + b + c − x)(a + b + c + x)

We have f (x) 

1 2





1 1 + (a+b+c+x) 2 . (a+b+c−x)2 1 1 f (x)  (a+b+c−x)3 − (a+b+c+x) 3

If 0 ≤ x ≤ a, then ≥ 0, and thus it follows that a f (c) + c f (b) + b f (a) ≥ b f (c) + a f (b) + c f (a), or a·

c b a +c· +b· (a + b)(a + b + 2c) (a + c)(a + c + 2b) (b + c)(b + c + 2a) b a c +a· +c· . ≥b· (a + b)(a + b + 2c) (a + c)(a + c + 2b) (b + c)(b + c + 2a)

7.13. Note that the given inequality is equivalent to the following inequality: (a1 + · · · + an−1 )n(n + 1) + (a1 + · · · + an−1 )n(n − 1) + (an + an+1 )n(n − 1)     ≥ 2 n 2 − 1 (a1 + · · · + an−1 ) + 2 n 2 − 1 an ,

or   2(a1 + · · · + an−1 ) + n(n − 1)an+1 ≥ n 2 + n − 2 an .

(7.9)

We prove this inequality by induction. For n  2, we obtain the following obvious inequality: 2a1 + 2a3 ≥ 4a2 . Assume that the given inequality holds for n  k, that is,   2(a1 + · · · + ak−1 ) + k(k − 1)ak+1 ≥ k 2 + k − 2 ak ,

(7.10)

and let us prove that it holds for n  k + 1. Let us add to both sides of inequality (7.10) the expression 2ak +(k + 1)kak+2 − k(k − 1)ak+1 , from which we deduce that       2 a1 + · · · + ak−1 + ak + (k + 1)kak+2 ≥ k 2 + k ak + k 2 + k ak+2 − k(k − 1)ak+1         k 2 + k ak + ak+2 − k(k − 1)ak+1 ≥ k 2 + k · 2ak+1 − k(k − 1)ak+1  k 2 + 3k ak+1 .

  Hence, we obtain 2(a1 + · · · + ak ) + (k + 1)kak+2 ≥ k 2 + 3k ak+1 , and thus it follows that the given inequality holds for n  k + 1. Therefore, inequality (7.9) holds for every positive integer n.

96

7 The Principle of Mathematical Induction

7.14. (a) Let us rewrite the given inequality as a1 + a3 + · · · + a2n−1 ≥ n + a2 + · · · + a2n ) and prove it by induction. n+1 (a0 2 . For n  1, we obtain the following obvious inequality: a1 ≥ a0 +a 2 Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1, that is, k (a0 + a2 + · · · + a2k ), or k+1 k a1 + a2 + · · · + a2k−1 + a2k+1 ≥ (a0 + a2 + · · · + a2k ) + a2k+1 . k+1 a1 + a3 + · · · + a2k−1 ≥

Let us prove that k k+1 + a2 + · · · + a2k ) + a2k+1 ≥ k+2 (a0 + a2 + · · · + a2k + a2k+2 ), k+1 (a0 1 k+1 a2k+2 . or a2k+1 ≥ (k+1)(k+2) (a0 + a2 + · · · + a2k ) + k+2 a2k +a2k+2 Since a2k+1 ≥ , it follows that in order to complete the proof of the last 2 a2k +a2k+2 1 inequality, it is sufficient to prove that ≥ (k+1)(k+2) (a0 + · · · + a2k ) + 2 k+1 a , or k+2 2k+2 k(k + 3) k(k + 1) a2k − a2k+2 ≥ a0 + a2 + · · · + a2k−2 . 2 2

We prove this inequality by induction. For k  1, we obtainthe obvious inequality 2a2 − a4 ≥ a0 , since a2 ≥ a1 +a3 2 4 . ≥ 21 a0 +a + a2 +a 2 2 2 Let k  m, then the following inequality holds: m(m + 3) m(m + 1) a2m − a2m+2 ≥ a0 + · · · + a2m−2 . 2 2 Therefore, m(m + 3) m(m + 1) a2m − a2m+2 + a2m ≥ a0 + a2 + · · · + a2m−2 + a2m . 2 2 It is sufficient to prove that m(m + 3) m(m + 1) (m + 1)(m + 4) (m + 1)(m + 2) a2m − a2m+2 + a2m ≤ a2m+2 − a2m+4 , or 2 2 2 2 (m + 1)(m + 2) (m + 1)(m + 2) (m + 1)(m + 2)a2m+2 ≥ a2m + a2m+4 . 2 2

Note that this inequality holds. k−1 k+1 (b) Let ak  −a k . As −a k ≥ −a 2−a . Then by to inequality (a) of this 3 2n−1 2 −···−a 2n 1+a 2 +···+a 2n problem, we have −a−a −···−a ≥ −1−an+1 , and therefore a+a 3 +···+a 2n−1 ≥ n n+1 . n 7.15. We proceed by induction on n. For n  1, we have 1 > ln 2. Assume that the given inequality holds for n  k, that is, 1 + 21 + · · · + k1 > ln(k + 1).

Proofs

97 1 Let us add to both sides of this inequality the expression k+1 . Then 1 + 21 + 1 1 > ln(k + 1) + k+1 . · · · + k1 + k+1 1 > ln(k + 2), which is equivalent to Now let us prove that ln(k + 1) + k+1

  k+2 k + 2 k+1 1 > ln(k + 2) − ln(k + 1)  ln , or 1 > ln . k+1 k+1 k+1  n n  Since lim 1 + n1  e and the sequence 1 + n1 is monotonically increasn→∞ ing (Problem 3.16(a)), it follows that  e > 1+

1 k+1

k+1

 , or 1 > ln

k+2 k+1

k+1 .

7.16. We proceed by induction on n. For n  1, we have 1 ≤ 1. Assume that the given inequality holds for n  k, that is, 1+ 2√1 2 +· · ·+ k √1 k ≤ √2 , and let us add to k 1√ . Then we obtain (k+1) k+1

3−

both sides of the last inequality the expression

1 2 1 1 1 ≤3− √ + . 1 + √ + ··· + √ + √ √ k k (k + 1) k + 1 k (k + 1) k + 1 2 2 In order to complete the proof, it is sufficient to prove that 3− √2k + (k+1)1√k+1 ≤ 3−

√2 . k+1

Indeed, since

1 k+1

≤

√ 2 , k(k+1)+k

we have

√ √  2 k+1− k 1 2 2 2 √   √ − √ ≤ √ .  √ √ √ k(k + 1) (k + 1) k + 1 k k+1 k(k + 1) k + 1 + k

Therefore, the given inequality holds for every positive integer n. 7.17. We proceed by induction on n. For n  1, we have 1 + α ≥ 1 + α. Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1. α2 . For n  k, we have (1 + α)k ≥ 1 + kα + k(k−1) 2 Hence, we deduce that   k(k − 1) 2 k(k − 1) 2 α  1 + kα + α + α + kα 2 (1 + α)k+1 ≥ (1 + α) 1 + kα + 2 2 +

k(k − 1) 3 k(k + 1) 2 k(k − 1) 3 k(k + 1) 2 α  1 + (k + 1)α + α + α ≥ 1 + (k + 1)α + α . 2 2 2 2

It follows that (1 + α)k+1 ≥ 1 + (k + 1)α + k(k+1) α 2 , and therefore, the given 2 inequality holds for every positive integer n. 7.18. We proceed by induction on k. For k  1, we have 1 > 2e .

98

7 The Principle of Mathematical Induction

Assume that the given inequality holds for k  m, that is, m! >

 m+1 m

, and e  m+2 m+1 let us prove that it holds true for k  m + 1, that is, (m + 1)! > e . m  , and therefore We have m! > m+1 e 

m+1 (m + 1)! > e e

m+1 .

(7.11)

 n n  Since lim 1 + n1  e and the sequence 1 + n1 is monotonically increasn→∞ ing (Problem 3.16(a)), it follows that  e > 1+

1 m+1

m+1 .

(7.12)

m+1  . Multiplying inequalities (7.11) and (7.12), we obtain (m + 1)! > m+2 e Therefore, the given inequality holds for every positive integer k. 7.19. Let us first prove that √ 3 3 . (7.13) 2|sin x| + |sin 2x| ≤ 2

   4 3 2 2 √ √ We have 2|sin x|+|sin 2x| ≤ 3 (3 − 3|cos x|)(1 + |cos x|) ≤ 3 46  √ 3 3 . 2

We proceed by induction on n. For n  0, we obtain the following obvious inequality: |sin x| ≤ 1. Assume that the given√ inequality holds for n ≤ k, that is, |sin x| + |sin 2x| + · · · + |sin 2n x| ≤ 1 + 23 n, for all x Let us prove that it holds for n  k + 1. Let us consider √ the following two cases.   √  (a) If |sin x| ≤ 23 , then |sin x| + |sin 2x| + · · · + sin(2k · 2x) ≤ 23 + 1 + √

√

 1 + 23 (k + 1). √ √ (b) If |sin x| > 23 , then from (7.13), it follows that |sin x| + |sin 2x| < 3, whence 3 k 2

  √ √  √   3 3   |sin x| + |sin 2x| + |sin 4x| + · · · + sin(2k−1 · 4x) < 3 + 1 + (k − 1)  1 + (k + 1). 2 2

Therefore, the given inequality holds for every positive integer  n. 7.20. Consider the function f n (α)  cos α + cos22α +· · ·+ cosnnα in 0, π2 and suppose it attains its minimum value in this interval at the point αn . Let us consider the following three cases. (a) If αn  0, then f n (α) ≥ 1 + 21 + · · · + n1 > − 21 .         (b) If αn  π2 , then f n (α) ≥ f n π2 ≥ − 21 , as f 1 π2  0, f 2 π2  f 3 π2            − 21 , f 4 π2  f 5 π2  − 21 + 41 , f 6 π2  f 7 π2  − 21 + 41 − 16 , and so on.

Proofs

99

(c) If 0 < αn < π2 , then f n (αn )  0, that is, − sin αn −sin 2αn −· · ·−sin nαn  0. We obtain that 2 sin α2n sin αn + · · · + 2 sin α2n sin nαn  0, whence cos α2n    cos nαn + α2n .   Thus, sin α2n  ± sin nαn + α2n , and it follows that   αn αn αn αn αn  αn  cos sin cos nαn  cos nαn + + sin nαn +  cos2 ± sin2 > 0. 2 2 2 2 2 2

n We deduce that f n (α) ≥ f n−1 (αn )  cos αn + · · · + cos(n−1)α ≥ − 21 , since n−1  π αn ∈ 0, 2 , and for (n − 1), the statement holds (for n  1, we have f 1 (α)  cos α ≥ 0 > − 21 ). 7.21. We proceed by induction on n. 2 2) For n  2, we have aa21 + aa21  2 + (a1a−a < 2 + a11a2 < 3. 1 a2 Let n ≥ 3. Consider the following two cases. (a) There exists i ∈ {2, . . . , n − 1} such that

(ai − ai−1 )(ai − ai+1 ) ≥ 0.

(7.14)

Then aai−1 + aai+1i ≤ aai−1 + 2, since (ai − ai−1 )(ai − ai+1 ) i i+1 |ai − ai−1 ||ai − ai+1 | < 1 < ai ai+1 . Taking this into consideration and the induction hypothesis, we obtain

≤

ai−1 ai an−1 an a1 ai−2 ai−1 ai+1 a1 a2 + + ··· + + + ··· + + ≤ + ··· + + + a2 a3 ai ai+1 an a1 a2 ai−1 ai+1 ai+2 an + ··· + + 2 < 2(n − 1) − 1 + 2  2n − 1, a1

|ai−1 − ai+1 | ≤ since from (7.14) it follows that max(|ai − ai−1 |, |ai − ai+1 |) < 1. (b) If (ai − ai−1 )(ai − ai+1 ) < 0, i  2, . . . , n − 1, then either a1 < a2 < · · · < an , or a1 > a2 > · · · > an . In the first case, we have aa21 + aa23 + · · · + an−1 + aan1 ≤ 1 + · · · + 1 + aan1 < 2n − 1, an n−1

since an  (an − an−1 ) + (an−1 − an−2 ) + · · · + (a2 − a1 ) + a1 < na1 . + aan1 ≤ 2 + · · · + 2 +1  2n − 1, In the second case, we have aa21 + aa23 + · · · + aan−1 n n−1

since ai  ai+1 + |ai+1 − ai | < ai+1 + 1 < 2ai+1 , i  1, . . . , n − 1. 7.22. (a) We proceed by induction on n.√ For n  1, we need to prove that xx22−1 ≤ 21 , or (x2 − 2)2 ≥ 0. Assume that the inequality holds for n  k, and let us prove that it holds for n  k + 1, where√k ∈ N. We need to prove that if 1 √ x1 ≤ x2 ≤ · · · ≤ √ √ xk+2 −xk+1 x2 −x1 x3 −x2 xk+1 ≤ xk+2 , then x2 + x3 + · · · + ≤ 4k+1 . xk+2 2

For the k +1 numbers 1 ≤ √

4k−3 , 2

and therefore,

x3 x2

≤ ··· ≤

xk+2 , we have x2

x3 x2 −1 x3 x2

+· · ·+

xk+2 xk+1 x2 − x2 xk+1 x2

≤

100

7 The Principle of Mathematical Induction ⎛



⎞ xk+1 xk+2 x2 − x2 ⎟ ⎟ ≤ xk+2 ⎠ xk+2 x2  ⎛ ⎞ √ 2  √ √   2  √  x2 − x1 x2 − x1 2 1 4k − 3 4k − 3 ⎠ 1 ⎝12 + ≤ + √ ≤ ·1+ √ x2 x2 2 x2 x2 2  √ 2x2 − 1 4k + 1 4k + 1  · ≤ 4 2 x22

√

x2 − x1 + x2

√

x3 − x2 + ··· + x3



xk+2 − xk+1

√

x2 − x1 1 ⎜ + √ ⎜ x2 x2 ⎝

x3 x2 − 1 + ... + x3 x2

(see Chapter 4, inequality (4.1)). Hence, the given inequality holds for n  k + 1. Therefore the given inequality holds for every positive integer n. (b) We proceed by induction on n. √ x2 −x1 For n  1, we have x2 ≤ 21 < 1. Assume that the given inequality holds for n  k, and let us prove that it holds for n  k + 1, where k ∈ N. Let 1  x1 ≤ x2 ≤ · · · ≤ xk+1 ≤ xk+2 be positive integers. Then we have  √ xk+2 − xk+1 x − xk+1 x2 − x1 x3 − x2 x − x1 x − x2 ≤ 2 + + ··· + + 3 + · · · + k+2 x2 x3 xk+2 x2 x3 xk+2       1 1 1 1 1 1 + ··· + + + ··· + + ... + + ··· + ≤ x1 + 1 x2 x2 + 1 x3 xk+1 + 1 xk+2

√

≤

1 1 1 + + ··· + . 2 3 xk+2

1 1 1 1 If xk+2 ≤ (k + 1)2 , then 21 + 13 + · · · + xk+2 ≤ 21 + · · · + (k+1) , 2 < 1+ 2 +· · ·+ (k+1)2 and therefore, the given inequality holds for n  k + 1. If xk+2 > (k + 1)2 , then 

xk+2 − xk+1 xk+2

 ≤

xk+2 − 1  xk+2



1 xk+2

 −

1 xk+2

2



<

1 (k + 1)2

−

1 (k + 1)4



 k 2 + 2k . (k + 1)2

Hence, we obtain

  xk+1 − xk xk+2 − xk+1 x2 − x1 1 1 + ≤ 1 + + ··· + + ··· + x2 xk+1 xk+2 2 k2   xk+2 − xk+1 1 1 1 k 2 + 2k 1 2k + 1 ≤ 1 + + ··· + + + < 1 + + ··· + + 2 xk+2 2 2 k (k + 1)2 k 2 (k + 1)2

√

1+

1 1 1 1 1 1 1 < 1 + + + ··· + + + ··· + . + ··· + 2 2 3 k 2 (k + 1)2 (k + 1)2 (k + 1)2    2k+1

Hence, the given inequality holds for n  k + 1. Therefore, the given inequality holds for every positive integer n.

Proofs

101

Remark If 1  x1 ≤ x2 ≤ · · · ≤ xn ≤ xn+1 are positive integers, then ⎛ 2 ⎞ √ √ n  x2 − x1 xn+1 − xn 1⎠ 1 ⎝ + ··· + ≤ − . x2 xn+1 i 2 i1 7.23. We proceed by induction on n. β1 α1 ≤ cos . For n  1, we have 0 < α1 ≤ β1 ≤ π, and therefore cos sin α1 sin α1 For n  2, we have α1 > 0, α2 > 0, β1 ≥ 0, β2 ≥ 0 and α1 + α2 ≤ β1 β2 α1 α2 + cos ≤ cos + cos . β1 + β2 ≤ π , and we need to prove that cos sin α1 sin α2 sin α1 sin α2 cos x1 cos x2 Let α1 and α2 be constants. Consider the expression sin α1 + sin α2 , where α1 + α2 ≤ x1 + x2 ≤ π, x1 ≥ 0, x2 ≥ 0. Assume that this expression attains its maximum value at x1  β1 , x2  β2 . Let α1 ≤ α2 . Then one can assume that β1 ≤ β2 . Otherwise, we have cos β1 < cos β2 , and therefore (cos β1 − cos β2 )(sin α2 − sin α1 ) ≤ 0, or cos β1 cos β2 cos β2 cos β1 + ≤ + . sin α1 sin α2 sin α1 sin α2 If β1  0, then cos β1 cos β2 cos α1 cos α2 1 cos(α1 + α2 ) cos α1 cos α2 + − − ≤ + − − sin α1 sin α2 sin α1 sin α2 sin α1 sin α2 sin α1 sin α2 α1 α1 α1 2 2 sin 2 sin( 2 + α2 ) sin 2 sin(α1 + α2 ) α1 −  tan − ≤ 0. 2 sin α2 cos α21 sin α2

If 0 < β1 ≤ β2 , then on decreasing the value of β1 , the value of the expression cos β1 β2 + cos increases, and therefore β1 + β2  α1 + α2 . Let us prove that sin α1 sin α2 sin β1 sin β2  sin α2 . sin α1 1 +x) 2 −x) Note that the function f (x)  cos(β + cos(β in [−β1 , β2 ] attains its sin α1 sin α2 maximum value at the point x  0, and hence by Fermat’s theorem on sin β1 sin β2 + sin  0. stationary points, f (0)  − sin α1 α2 Lemma. If αi > 0, βi > 0, i  1, . . . , n, α1 +· · ·+αn  β1 +· · ·+βn ≤ π , and sin β1 sin βn  ···   λ, sin α1 sin αn

(7.15)

then λ  1. Proof . Let λ  1. Then without loss of generality one can assume that λ < 1. Let α1 ≤ · · · ≤ αn . Then from (7.15) it follows that β1 ≤ · · · ≤ βn and β1 < α1 , . . . , βn−1 < αn−1 ≤ π2 . Note that sin(β1 + β2 )  λ(sin α1 cos β2 + sin α2 cos β1 ) > λ(sin α1 cos α2 + sin α2 cos α1 )  λ sin(α1 + α2 ).

102

7 The Principle of Mathematical Induction

In a similar way, we obtain sin((β1 + β2 ) + β3 )  λ(sin(β1 + β2 ) cos β3 + sin β3 cos(β1 + β2 )) > λ(sin(α1 + α2 ) cos β3 + sin α3 cos(β1 + β2 )) > λ(sin(α1 + α2 ) cos α3 + sin α3 cos(α1 + α2 ))  λ sin(α1 + α2 + α3 ), and so on. In general, we have sin(β1 + · · · + βn−1 ) > λ sin(α1 + · · · + αn−1 ). sin βn sin(ϕ − αn ), where ϕ  α1 + · · · + αn  Therefore, sin(ϕ − βn ) > sin αn β1 + · · · + βn . Hence, sin αn sin(ϕ − βn ) > sin βn sin(ϕ − αn ), or sin αn sin ϕ cos βn − sin αn sin βn cos ϕ > sin βn sin ϕ cos αn − sin βn sin αn cos ϕ, sin ϕ sin(αn − βn ) > 0. Since 0 < ϕ ≤ π, we must have ϕ  π and sin ϕ > 0, whence αn − βn > 0. It follows that αn > βn . We obtain β1 < α1 , . . . , βn−1 < αn−1 , βn < αn , and therefore β1 +· · ·+βn < α1 + · · · + αn , which leads to a contradiction. Thus λ  1. This ends the proof of the lemma. Since sin α1  sin β1 and sin α2  sin β2 , it follows that α1  β1 and α2  β2 , or α1  β1 and π − α2  β2 (α1 + α2  α1 + π − α2 , α2  π2 , whence α2  β2 ). β1 β2 α1 α2 + cos  cos + cos . Therefore, cos sin α1 sin α2 sin α1 sin α2 Let n ≥ 3 and suppose that the given inequality holds for n − 1. Let us prove that it holds for n. xn x1 +· · ·+ cos , Let 0 < α1 ≤ · · · ≤ αn be constants. Consider the expression cos sin α1 sin αn where α1 + · · · + αn ≤ x1 + · · · + xn ≤ π, x1 ≥ 0, . . . , xn ≥ 0, and suppose that this expression attains its maximum value at x1  β1 , . . . , xn  βn . Then β1 ≤ β2 ≤ · · · ≤ βn (see the case n  2). Without loss of generality one can assume that β1 > 0, for otherwise, for n − 1 we would have cos βn cos(α1 + α2 ) cos α3 cos αn cos β3 cos β2 + ··· + ≤ + ··· + . (7.16) + + sin(α1 + α2 ) sin α3 sin αn sin(α1 + α2 ) sin α3 sin αn

Note that 1 cos(α1 + α2 ) cos β2 cos α1 cos α2 cos β2 − . + − − ≤ sin α1 sin α2 sin α1 sin α2 sin(α1 + α2 ) sin(α1 + α2 ) (7.17) Indeed, since

Proofs

103

 cos β2

1 1 − sin α2 sin(α1 + α2 )

 ≤

1 1 − sin α1 sin(α1 + α2 )

(α2 + (α1 + α2 ) ≤ α1 + α2 + α3 ≤ π, sin(α1 + α2 ) ≥ sin α2 ), it is sufficient α1 α2 1 +α2 ) + 1−cos − 1−cos(α ≤ 0, or to prove that 1−cos sin α1 sin α2 sin(α1 +α2 )  α2 α1 α2  α1 + tan − tan + ≤ 0, tan 2  2 2  2 α1 α2 α1 α2 − tan 2 + tan 2 tan 2 tan 2 ≤ 0. 1 − tan α21 tan α22 The last inequality holds because 0 < α21 ≤ α22 < π4 . Summing (7.16) and (7.17), we obtain the inequality for n. Then we have 0 < β1 ≤ β2 ≤ · · · ≤ βn , β1 + and if the value of β1 decreases, then the value of the expression cos sin α1 cos βn · · · + sin αn increases, and therefore α1 + · · · + αn  β1 + · · · + βn . Then sin β1 sin βn  · · ·  sin (see the case n  2), and therefore by the lemma, we sin α1 αn have α1  β1 , . . . , αn  βn . 7.24. We first prove by induction that if a > 0 and n ∈ N, then 2(1 + a n+1 )3 ≥ (1 + a 3 )(1 + a n )3 .

(1)

For n  1, we need to prove that 2(1 + a 2 )3 ≥ (1 + a 3 )(1 + a)3 , which is equivalent to the following obvious inequality: (a − 1)4 (a 2 + a + 1) ≥ 0. Assume that the inequality holds for n  k, where k ∈ N, and let us prove that it holds for n  k + 1. We have 2(1 + a k+1 )3 ≥ (1 + a 3 )(1 + a k )3 , as (1 + a k+2 )(1 + a k ) ≥ (1 + a k+1 )2 , whence  2(1 + a k+2 )3  2(1 + a k+1 )3 ·

1 + a k+2 1 + a k+1

3

 ≥ (1 + a k )3 (1 + a 3 ) ·

1 + a k+1 1 + ak

3

 (1 + a 3 )(1 + a k+1 )3 ,

and therefore 2(1 + a k+2 )3 ≥ (1 + a 3 )(1 + a k+1 )3 . Thus, inequality (1) holds. Note that (13 + a 3 )(13 + b3 )(13 + c3 ) ≥ (1 + abc)3 (Problem 6.2) and (2(a 2012 + 1)(b2012 + 1)(c2012 + 1))3  2(a 2012 + 1)3 · 2(b2012 + 1)3 · 2(c2012 + 1)3 ≥ (1 + a 3 )(1 + a 2011 )3 · (1 + b3 )(1 + b2011 )3 · (1 + c3 )(1 + c2011 )3 ≥ ((1 + abc)(a 2011 + 1)(b2011 + 1)(c2011 + 1))3 ,

and therefore 2(a 2012 + 1)(b2012 + 1)(c2012 + 1) ≥ (1 + abc)(a 2011 + 1)(b2011 + 1)(c2011 + 1). 7.25. (a) The inequality b12 ≥ b2 follows from the inequality of Problem 2.2. (b) Let b0  1, bn+1  0. We then need to prove that bk2 ≥ bk−1 bk+1 , for k  1, 2, . . . , n.

104

7 The Principle of Mathematical Induction

For the numbers a1 , . . . , an−1 , the numbers b1 , . . . , bn−1 are defined in a similar way as we defined the numbers b1 , . . . , bn for the numbers a1 , . . . , an , and b0  1, bn  0. 2 For n  2, we have b0  1, b1  a1 +a , b2  a1 a2 , b3  0, and hence 2 2 2 b1 ≥ b0 b2 ,b2 ≥ b1 b3 . Assume that this statement holds for n − 1 numbers, and let us prove that it holds for n (n ≥ 3) numbers. k−1 b C k +b Cn−1 an  n−k bk + nk bk−1 · an , for k  1, 2, . . . , n. Note that bk  k n−1 Ck−1 k n n For k  2, . . . , n − 1, we have 2  n 2 (bk2 − bk−1 bk+1 )  (n − k)bk + kbk−1 · an    − (n − k + 1)bk−1 + (k − 1)bk−2 · an (n − k − 1)bk+1 + (k + 1)bk · an   2 2  (n − k)2 − 1 (bk − bk−1 bk+1 ) + (k 2 − 1)an2 (bk−1 − bk−2 bk ) + (k − 1)(n − k − 1)(bk bk−1 − bk−2 bk+1 )an ≥ 0, 2

2

as bk · bk−1 ≥ bk−1 bk+1 · bk−2 bk , and therefore bk2 − bk−1 bk+1 ≥ 0. This ends the proof.

Problems for Independent Study Prove the following inequalities (1–20).     n 1. aa21 − aa21 + · · · + aan−1 ≤ aan1 − aan1 , where n ≥ 2, 0 < a1 ≤ · · · ≤ an . − aan−1 n 2. (1 + a1 ) · · · (1 + an ) ≥ 1 + a1 + · · · + an , where a1 > −1, . . . , an > −1 and the numbers a1 , . . . , an have the same sign. 3. C ≤ D ≤ 2C, where C  (a1 − b1 )2 + · · · + (an − bn )2 , D  (a1 − bn )2 + (a2 − bn )2 + · · · + (an − bn )2 , bk 

a1 + · · · + ak , k  1, . . . , n. k

√ √ 4. n n > n+1 n + 1, where n ≥ 3, n ∈ N.  k 2 5. (a) 1 + n1 < 1 + nk + nk 2 , where k ≤ n, n, k ∈ N.  n (b) 1 + mn m < 3, where m ∈ N, n ∈ N. Hint. See Problems 5(a) and 3.6(c). 2

3

2k

x > 0, where k ∈ N. 6. 1 − x + x2! − x3! + · · · + (2k)!   n n  1  1 7. ≤ a1 an n(a1 + an ) − ai , where 0 < a1 ≤ · · · ≤ an . ai

8.

i1 a1 +···+ak k

<

a1 +···+an n

<

i1 ak+1 +···+an , n−k

where a1 < · · · < an , n > k, n, k ∈ N.

Problems for Independent Study

105

2 2 9. a12 − a22 + · · · − a2n + a2n+1 ≥ (a1 − a2 + · · · − a2n + a2n+1 )2 , where a1 ≥ a2 ≥ · · · ≥ a2n+1 ≥ 0. (n−1)xn2 +2xn +n−1 , where n ≥ 2. 10. x1 + x1 (x2 − x1 ) + · · · + xn−1 (xn − xn−1 ) ≤ 2n 11. 11 · . . . · n n > (2n)! , where n ≥ 5, n ∈ N. 1 )! n )! · · · (2m ≥ 2 S , where m 1 , . . . , m n ∈ Z0 and m 1 + · · · + m n  S. 12. (2m m1! mn ! 1 1 n , where a > 0, b > 0 and n ∈ N. 13. a+b + · · · + a+nb < √a(a+nb) n−1  n · 1 < 4, where n ≥ 2, n ∈ N. 14. n−k 2k−1 k1 k  2 k2 15. (a) 1 + n1 < 1 + nk + 2n 2 , where k, n ∈ N and (k − 1) < n;         n 1 1 n+1 1 (b) 1 + n1 1 + 4n 1 + 4(n+1) , where n ∈ N. < 1 + n+1 n    cos 2i x  ≥ n , where n ∈ N. 16. 2 i0

17. (a) sin α + sin22α + · · · + sinnnα ≥ 0, where n ∈ N and 0 ≤ α ≤ π , (b) cos α + cos22α + · · · + cosnnα ≥ −1, where n ∈ N. 18. (a) xx21 + xx23 + · · · + xxn1 ≥ xx21 + xx23 + · · · + xxn1 , where 0 < x1 ≤ · · · ≤ xn , n ≥ 2; (b) x13 + · · · + xn3 ≥ x1 + · · · + xn , where x1 > 0, . . . , xn > 0 and x1 · . . . · xn  1. x13 xn3 (x1 +···+xn )3 + · · · + 1·(x1 +···+x ≥ n·n(x . Hint. Note that 1·(x1 +...+x n) n) 1 +···+x n ) 19. (1 + a1 )(2 + a2 ) · . . . · (n + an ) ≤ 2 · n! , where n ≥ 2, a1 > 0, . . . , an > 0 and a1 + · · · + an  1. Hint. Let ai  iαi , i  1, . . . , n, and we need to prove that (1 + α1 )(1 + α2 ) · · · (1 + αn ) ≤ 2, having that α1 + 2α2 + · · · + nαn  1. Note that if (n − 1)αn−1 +nαn  (n −1)βn−1 , then (1+αn−1 )(1+αn ) ≤ 1+βn−1 . 20. (1 − a)(a k1 + a k2 + · · · + a kn )2 < (1 + a)(a 2k1 + a 2k2 + · · · + a 2kn ), where n ≥ 2, n ∈ N, 0 < a < 1, 0 ≤ k1 ≤ · · · ≤ kn and k1 , k2 , . . . , kn ∈ N. 2k1 +1 Hint. Note that a k1 (a k2 + · · · + a kn ) < a 2k1 +1 + a 2k1 +2 + · · ·  a1−a . 21. For arbitrary positive integer n > 1 find the smallest possible value of C n 1 + · · · + aann−b < C holds for all positive numbers if the inequality aa11−b +b1 +bn a1 , . . . , an , b1 , . . . , bn satisfying the equality a1 + · · · + an  b1 + · · · + bn . 1 ···x n 1 < 1−x + 22. Prove that for an arbitrary positive integer n > 1, one has 1−x 1−y1 ···yn 1−y1 1−xn · · · + 1−yn , where 0 < yi ≤ xi < 1, i  1, . . . , n. 1−a 1−b 1−b < 1−cd + 1−cd < 1−a + 1−d , where 0 < c ≤ a < 1, 0 < Hint. Note that 1−ab 1−cd 1−c d ≤ b < 1.

Chapter 8

A Useful Inequality

In 1997 Nairi Sedrakyan has published an article (in Russian) [17] called “On the applications of a useful inequality,” in which the author proves a very useful inequality and provides some applications. In 1998, this inequality was re-published by the author, this time in Armenian, in the book [16], in which the author devotes an entire chapter to its applications. In 2002 this book was published in Moscow in Russian [15]. Russian-speaking reader sometimes calls this inequality Sedrakyan’s inequality in reference to Sedrakyan’s 1997 article [17] on its numerous applications, while the English-speaking reader sometimes calls this inequality Engel’s form or Titu’s lemma, in reference to the book [4] published in 1998, and the book [1], published in 2003. Nevertheless, even though Sedrakyan in his article [17] stated and proved this inequality without using the Cauchy–Bunyakovsky–Schwarz inequality, it turns out that this inequality is nothing but another form of that inequality. Probably this form was known even before this article, with the difference that in his article [17], Sedrakyan noticed that written in this form, the inequality has very useful new applications, and he provided numerous ways in which this inequality can be used as a mathematical proof technique to prove inequalities of various types. In this chapter we consider that inequality, along with its generalizations and applications. Lemma 1 (A useful inequality) Let a1 , a2 , . . . , an be real numbers and b1 , b2 , . . . , bn positive real numbers. Then a12 a2 (a1 + · · · + an )2 + ··· + n ≥ . b1 bn b1 + · · · + bn Moreover, equality holds if and only if

a1 b1

 ··· 

(8.1)

an bn

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_8

107

108

8 A Useful Inequality

Generalizations of inequality (8.1). Lemma 2 (Generalization 1) a13 a3 a3 (a1 + a2 + · · · + an )3 , + 2 + ··· + n ≥ b1 c1 b2 c2 bn cn (b1 + b2 + · · · + bn )(c1 + c2 + · · · + cn )

(8.2)

where ai > 0, bi > 0, > ci > 0, i  1, 2, . . . , n. Lemma 3 (Generalization 2) a1n b1n−1

+ ··· +

ann bnn−1

≥

(a1 + · · · + an )n , (b1 + · · · + bn )n−1

(8.3)

where ai > 0, bi > 0, i  1, 2, . . . , n. Lemma 4 (Generalization 3) n a1,1

n a2,1

n am,1 + ··· + ≥ a1,2 · a1,3 · · · a1,n a2,2 · a2,3 · · · a2,n am,2 · am,3 · · · am,n n  a1,1 + a2,1 + · · · + am,1     , ≥ a1,2 + a2,2 + · · · + am,2 · a1,3 + a2,3 + · · · + am,3 · · · a1,n + a2,n + · · · + am,n (8.4)

+

where ai, j > 0, i  1, 2, . . . , m,

j  1, 2, . . . , n.

Proof of Lemma 1 Let us at first prove that for all real numbers a, b and positive real numbers x, y, one has a 2 b2 (a + b)2 + ≥ . x y x+y

(8.5)

Note that this inequality is equivalent to the inequality (ay − bx)2 ≥ 0. Obviously, here the equality holds here if and only if ay − bx  0, that is Now, let us proceed to the proof of this useful inequality. Applying inequality (8.5) several times, we obtain a2 (a1 + a2 )2 a32 a2 a12 a22 a32 + + + ··· + n ≥ + + ··· + n b1 b2 b3 bn b1 + b2 b3 bn (a1 + a2 + a3 )2 an2 ≥ + ··· + b1 + b2 + b3 bn ≥ ··· (a1 + · · · + an )2 . ≥ b1 + · · · + bn Obviously, equality holds if and only if

a1 b1



a2 b2

 ··· 

an . bn

a x

 by .

8 A Useful Inequality

109

This ends the proof. We provide proofs of Lemmas 2–4 after the remark following Lemma 6. Let us consider the following examples and applications of this useful inequality (Lemma 1). Cauchy–Bunyakovsky–Schwarz inequality. Let  x1 , x2 , . . . , xn and  y1 , y2 , . . . , yn be real numbers, then x12 + · · · + xn2 y12 + · · · + yn2 ≥ (x1 y1 + · · · + xn yn )2 . Proof For real numbers a1 , a2 , . . . , an and positive real numbers b1 , b2 , . . . , bn , we 2 a2 a2 n) have b11 + · · · + bnn ≥ (ab11+···+a . +···+bn 2  a2i  xi yi 2and  bi  yi . Then for2 all 1 ≤ i ≤ n, we obtain  2Let us take 2 x1 + · · · + xn y1 + · · · + yn ≥ (x1 y1 + · · · + xn yn ) . This ends the proof. Remark We have obtained a very beautiful inequality (8.1). However, it turns out that this is one of the forms of the Cauchy–Bunyakovsky–Schwarz √ inequality. Indeed, making a change of variables xi √abi and yi  bi we i  2  2  2 obtain Cauchy–Bunyakovsky–Schwarz inequality x1 + · · · + xn y1 + · · · + yn2 ≥ (x1 y1 + · · · + xn yn )2 . Note that in order to prove certain inequalities the form (8.1) is more convenient than the classical form of Cauchy–Bunyakovsky–Schwarz inequality. Example 8.1 Prove that

1 a+b

+

1 b+c

+

1 a+c

≥

9 , 2(a+b+c)

where a > 0, b > 0, c > 0.

Proof Note that according to Lemma 1 (for n  3), we have 1 1 1 12 12 12 + +  + + a+b b+c a+c a+b b+c a+c 9 (1 + 1 + 1)2  . ≥ (a + b + b + c + a + c) 2(a + b + c) This ends the proof. Example 8.2 Prove that 12 1 1 ≤ a+b + a+c + a+b+c+d 0, b > 0, c > 0, d > 0.

1 a+d

+

1 b+c

+

1 b+d

+

1 c+d

≤



3 1 4 a

+

1 b

+

1 c

+

1 d

 , where a >

Proof Note that according to Lemma 1 (for n  6), we have 1 1 1 1 1 1 + + + + + a+b a+c a+d b+c b+d c+d 12 (1 + 1 + 1 + 1 + 1 + 1)2  . ≥ 3(a + b + c + d) a+b+c+d On the other hand, on applying Lemma 1 (for n  2) multiple times for each of the following sets of parentheses, we obtain

110  3

8 A Useful Inequality 1 1 1 1 + + + a b c d

≥



 

1 1 + a b



 +

1 1 + a c



 +

1 1 + a d



 +

1 1 + b c



 +

1 1 + b d



 +

1 1 + c d

 ≥

(1 + 1)2 (1 + 1)2 (1 + 1)2 (1 + 1)2 (1 + 1)2 (1 + 1)2 + + + + + . a+b a+c a+d b+c b+d c+d

Thus, it follows that 12 1 1 1 1 1 1 ≤ + + + + + a+b+c+d a+  b a + c a +d b + c b + d c + d 3 1 1 1 1 + + + , ≤ 4 a b c d which completes the proof. A more general inequality. Let us try to obtain a more general inequality. To this end, we write ai2 as the product of two factors. On proving this inequality for n  2, we obtain new conditions. It turns out that the following lemma holds. Lemma 5 (Generalization 4) If ac11 ≥ · · · ≥ acnn , bc11 ≥ · · · ≥ bbnn (that is acii and bi have the same order) and ci > 0, i  1, . . . , n, then the following inequality ci holds: n  ai bi i1

ci

n

≥

i1

ai · n

n i1

bi .

(8.6.1)

ci

i1

If ai  bi , then inequality (8.6.1) becomes inequality (8.1). Proof In order to prove inequality (8.6.1), we use the following fact: if ac11 ≥ · · · ≥ acnn k and ci > 0, i  1, . . . , n, then ac11 +···+a ≥ acnn for all k  1, . . . , n (see Problem 1. +···+ck 11). Let us prove inequality (8.6.1) by induction on n 1 +b2 ) , or (a1 c2 −a2 c1 )(b1 c2 −b2 c1 ) ≥ For n  2, we have that ac1 b1 1 + ac2 b2 2 ≥ (a1 +ac21)(b +c2 a1 a2 b1 b1 0, which follows from c1 ≥ c2 and c1 ≥ c2 . Assume that inequality (8.6.1) holds for n = k Then we have a1cbi 1 + · · · + ack bk k ≥ (a1 +···+ak )(b1 +···+bk ) . c1 +···+ck From the above-mentioned fact and the inequality for n  2 it follows that ac1 b1 1 + )(b1 +···+bk+1 ) bk+1 ·bk+1 k )(b1 +···+bk ) · · · + ack bk k + ak+1 ≥ (a1 +···+a + ak+1ck+1 ≥ (a1 +···+ac1k+1 or ac1 b1 1 + · · · + ck+1 c1 +···+ck +···+ck+1 )(b1 +···+bk+1 ) ak+1 bk+1 ≥ (a1 +···+ac1k+1 . ck+1 +···+ck1 In a similar way, one can prove the following lemma.

8 A Useful Inequality

111

Lemma 6 (Generalization 5) If acii and bcii have opposite orders and ci > 0, i  1, . . . , n, then the following inequality holds: n  i1

n

ai bi ≤ ci

i1

ai · n

n

bi

i1

.

(8.6.2)

ci

i1

Remark (Chebyshev’s inequality) If the conditions a1 ≥ · · · ≥ an , b1 ≥ · · · ≥ bn , and 0 < c1 ≥ · · · ≥ cn hold, then the conditions of inequality (8.6.1) hold, and therefore this inequality holds as well. From inequalities (8.1) and (8.6.1) we obtain the following well-known inequalities: n  1 n2 ≥ n , where ci > 0, i  1, . . . , n, c i1 i ci i1

⎛ n

ai2

n 1 2 ⎜ i1 a ≥⎜ ⎝ n n i1 i n  i1

ai bi ≥

(8.7)

⎞2 ⎟ ⎟ , ⎠

(8.8)

n n  1 ai · bi , n i1 i1

(8.9)

where a1 ≥ · · · ≥ an and b1 ≥ · · · ≥ bn . Inequality (8.9) is called Chebyshev’s inequality. Note that inequality (8.7) follows from Chebyshev’s inequality (taking ai  ci , bi  − c1i ). Note also that inequality (8.8) follows from Chebyshev’s inequality (taking ai  bi ). ci , we obtain the following form Now, taking ai  ci xi , bi  ci yi , Pi  n ci

i1

of Chebyshev’s inequality: n i1

Chebyshev’s inequality (alternative form). If xi and yi have the same order, n Pi  1, Pi > 0, i  1, . . . , n, then for the means M z  z i Pi , one has

M x · M y ≤ M(x y).

i1

Remark Therefore, the Cauchy–Bunyakovsky–Schwarz inequality is a particular case of Chebyshev’s inequality.

112

8 A Useful Inequality

Proof of Lemma 2 Indeed, inequality (8.2) is one of the forms of the Cauchy–Bunyakovsky–Schwarz inequality  3      a1 + a23 + · · · + an3 · b13 + b23 + · · · + bn3 · c13 + c23 + · · · + cn3 ≥ (a1 b1 c1 + a2 b2 c2 + · · · + an bn cn )3 ,

(8.10)

where ai > 0, bi > 0, ci > 0, i  1, 2, . . . , n. Note that in inequality (8.10), by making the change of variables ai  √3 xyii zi , √ √ bi  3 yi , ci  3 z i , we obtain for the numbers xi , yi , z i inequality (8.2). In order to prove inequality (8.10), note that if the variables a1 , a2 , . . . , an are replaced by variables λa1 , λa2 , . . . , λan , where λ is any positive number, then we obtain an equivalent inequality. Choose a number λ such that (λa1 )3 + (λa2 )3 + · · · + (λan )3  1. Therefore, without loss of generality one can assume that a13 + a23 + · · · + an3  1, b13 + b23 + · · · + bn3  1, c13 + c23 + · · · + cn3  1, and hence it is sufficient to prove that 1 ≥ a1 b1 c1 + a2 b2 c2 + · · · + an bn cn Indeed, if x > 0, y > 0, z > 0, then x 3 + y 3 + z 3 − 3x yz  21 (x + y + z) (x − y)2 + (y − z)2 + (z − x)2 ≥ 0, and a 3 +b3 +c3

therefore ai bi ci ≤ i 3i i , i  1, 2, . . . , n. Summing all these inequalities, we deduce that 1 ≥ a1 b1 c1 +a2 b2 c2 +· · ·+an bn cn . This ends the proof. Proof of Lemma 3 Lemma 3 is a direct consequence of Lemma 4. Proof of Lemma 4 See Problem 45 in the end of this chapter. Let us consider the following examples (applications of Lemma 2). Example 8.3 Prove that

a3 bc

+

b3 ca

+

c3 ab

≥ a + b + c, where a > 0, b > 0, c > 0 .

Proof According to inequality (8.2) we have that, a + b + c. This ends the proof. Example 8.4 Prove that y > 0, z > 0.

a3 x

3

3

+ by + cz ≥

(a+b+c)3 3(x+y+z)

+

b3 ca

+

c3 ab

≥

(a+b+c)3 (b+c+a)+(c+a+b)



where a > 0, b > 0, c > 0, x > 0,

Proof According to inequality (8.2), we have (a+b+c)3 (1+1+1)(x+y+z)

a3 bc

a3 x

+

b3 y

+

c3 z



a3 1·x

+

b3 1.y

+

c3 1.z

≥

(a+b+c)3 . 3(x+y+z)

 This ends the proof.

Example 8.5 Prove that a > 0, b > 0, c > 0.

a b2 c2

+

b c2 a 2

+

c a 2 b2

≥

9 , a+b+c

where a 2 + b2 + c2  3abc and

 2 3  2 3  2 3 a b c b c a + 2 2 + 2 2  2 3 2 2 + 2 3 2 2 + 2 3 2 2, 2 2 b c c a a b a ·a b c b ·a b c c ·a b c then according to inequality (8.2), we have b2ac2 + c2ba 2 + a 2cb2 ≥ (a 2 +b2 +c2 )3 9  a+b+c . (a 2 +b2 +c2 )(a 3 b2 c2 +a 2 b2 c3 +a 2 b2 c3 ) This ends the proof. Proof Since

Problems

113

Problems Prove the following inequalities (8.1)–(8.8). 1 2 3 + x3x+x + x1x+x ≥ 23 , where x1 > 0, x2 > 0, x3 > 0. 8.1. Nesbitt’s inequality: x2x+x 3 1 2 a b c d + c+d + d+a + a+b ≥ 2, where a > 0, b > 0, c > 0, d > 0. 8.2. b+c xn x1 x2 n−1 + x1 +x ≥ 2, where n ≥ 4 and x1 > 0, . . . , xn > 0. 8.3. x2 +xn + x3 +x1 +· · · + xnx+x n−2 n−1 3

3

3

a b c a+b+c 8.4. (a) a 2 +ab+b , where a > 0, b > 0, c > 0. 2 + b2 +bc+c2 + c2 +ca+a 2 ≥ 3 a b c a+b b+c (b) b + c + a ≥ b+c + a+b + 1, where a > 0, b > 0, c > 0. (c) ab + bc + ac ≥ a+c + b+a + c+b , where a > 0, b > 0, c > 0.  a b c  b+c a+bc+ab+c a+b (d) 2 b + c + a ≥ b+c + a+b + a+b + a+c + a+c + b+c where a > 0, b > 0, c > 0. a+c a+b b+c a+c an a1 n 8.5. p−2a1 + · · · + p−2an ≥ n−2 , where p is the perimeter of a polygon with side lengths a1 , . . . , an . 1 1 1 + b3 (a+c) + c3 (a+b) ≥ 23 , where abc = 1 and a,b,c > 0. 8.6. (a) a 3 (b+c)  1 1 1 (b) 1+x + · · · + < + · · · + x1n , where xi > 0, i  1, . . . , n 1+x1 +···+xn x1 1 y (c) y 2x−z + z 2 −x + x 2z−y > 1, where 2 < x < 4, 2 < y < 4, 2 < z < 4.  3      2 8.7.  8 x + y 3 + z 3 ≥ 9x 2 + yz y 2 +x z z 2 +x y , where x > 0, y > 0 z > 0. n n n n n 1 1 1 1 1 · · · ≤ 8.8. a b c d ai · bi · ci · · · di , i i i i n n n n n i1

8.9.

8.10.

8.11.

8.12. 8.13. 8.14. 8.15. 8.16.

i1

i1

i1

i1

where 0 ≤ a1 ≤ · · · ≤ an , 0 ≤ b1 ≤ · · · ≤ bn , 0 ≤ c1 ≤ · · · ≤ cn , · · · , 0 ≤ d1 ≤ · · · ≤ dn . Consider a sequence with positive terms (x k ) where 1  x0 ≥ x1 ≥ · · · ≥ xn ≥ · · · . Prove that there exists a positive integer n such that for every such x2 x2 x2 ≥ 3.999. sequence (x k ), one has x01 + x12 + · · · + xn−1 n Let M be an arbitrary point in the interior of a given triangle ABC. Consider perpendiculars M A1 , M B1 , MC1 drawn from the point M to lines BC, C A, AB, respectively. For which point M in the interior of triangle AB attain the smallest possible value? ABC does the quantity MBCA1 + MC BA1 + MC 1 Let G be the intersection point of the medians of triangle A1 A2 A3 and let C be the circumcircle of triangle A1 A2 A3 Let the lines G A1 , G A2 , G A3 intersect the circle C a second time at points B1 , B2 , B3 , respectively. Prove that G A1 + G A2 + G A3 ≤ G B1 + G B2 + G B3 . 2 √ 2 +···+a2k−1 −a2k ) , where Prove that a1 +a2 +· · ·+a2k −2k 2k a1 · a2 · · · a2k ≥ (a1 −a 2(a1 +a2 +···+a2k ) k ∈ N and a1 , a2 , . . . , a2k > 0. √ √ √ + a2 + · · · + an ), Prove that a1 a2 + a1 a3 + · · · + an−1 an ≤ n2 ≤ n−1 2 (a1 1 1 where n ∈ N, n ≥ 2, a1 , . . . , an ≥ 0 and 1+a + · · · +  n − 1. 1+an 1 1 1 1 Prove that 1+a1 +···+an−1 + 1+a1 +···+an−2 +an + · · · + 1+a2 +···+an ≤ 1, where n ∈ N, n ≥ an  1.  2, a1 , . . . , an ≥ 0, and  a1 . . .  Prove that a 2 + bc b2 + cd c2 + da d 2 + ab ≥ (a +1)(b+1)(c+1)(d +1), where abcd = 1 and a > 0, b > 0, c > 0, d > 0. 1 1 1 1 + c5 (a+2b) Prove that a 5 (b+2c) 2 + 5 2 ≥ 3 , where abd = 1 and a > 0, b > b (c+2a)2 0, c > 0.

114

8 A Useful Inequality

8.17. Prove that a 3 +ba3 +c2 + a 2 +bb3 +c3 + a 3 +bc2 +c3 ≤ 1, where ab + bc + ac ≥ 3 and a > 0, b > 0, c > 0. 8.18. Prove that y3 x3 z3 + (1+x)(1+z) + (1+x)(1+y) ≥ 34 , where x > 0, y > 0, z > 0 and (a) (1+y)(1+z) x yz  1. √ 3 3 x yz y x z + 1−y + 1−z ≥ 1− √3 x yz , where 0 < x < 1, 0 < y < 1, 0 < z < 1. (b) 1−x y x z 9 + (x+z)(x+z−y) + (x+y)(x+y−z) ≥ 2(x+y+z) , where x, y, z are the (c) (y+z)(y+z−x) side lengths of some triangle. a3 b3 c3 d3 + a+c+d + a+b+d + a+b+c ≥ 13 , where a > 0, b > 0, c > 0, d > 0 and (d) b+c+d ab +bc + cd + da  1.   a3 a3 a3 ≥ (1 + a1 )(1 + a2 ) · · · (1 + an ), where (e) 1 + a12 1 + a22 · · · 1 + an2 2 3 1 a1 , . . . , an > 0.     2 ≥ n2 , where n ∈ N, n ≥ (f) x12 + · · · + xn2 x 2 +x1 x + · · · + x 2 +x1 n x1 1 2 n 1 2, x1 , . . . , xn > 0.   n n 1 2 ≥ n+1 + a 2a+a + · · · + a 2a+a , where (g) (a1 a2 + a2 a3 + · · · + an a1 ) a 2a+a 2 3 1 2 3 1 n ∈ N, n ≥ 3, a1 , . . . , an > 0 and a1 + · · · + an  1. x15 x25 + x1 +x3 +···+x + ··· + 8.19. Find the smallest value of the expression x2 +x3 +···+x n n xn5 , where x1 +x2 +···+xn−1 · · · + xn2  1.

n ≥ 3, x1 > 0, x2 > 0, . . . , xn > 0, and x12 + x22 +

8.20. Prove that for all positive numbers x, y, z such that x yz  1, the following 3 3 3 inequality holds: x 2x+y + y 2y+z + z 2z+x ≥ 23 .

Proofs 8.1. In similar problems one often needs to write x12

x22

a2 ab

instead of ab . It follows that

x2

2 +x 3 ) 2 3 + x3x+x + x1x+x  x1 (x2 +x3 ) + x2 (x3 +x1 ) + x3 (x13+x2 ) ≥ 2(x1(xx12 +x ≥ 23 +x2 x3 +x3 x1 ) 1 2 (here we have used that x1 x2 + x2 x3 + x3 x1 ≤ x12 + x22 + x32 ). (a+b+c+d)2 a2 b2 c2 d2 + b(c+d) + c(d+a) + d(a+b) ≥ ab+2ac+ad+bc+2bd+cd ≥ 2. 8.2. We have that, a(b+c) In order to prove the last inequality, it is sufficient to expand (a + b + c + d)2 and use that a 2 + c2 ≥ 2ac, b2 + d 2 ≥ 2bd. 2 x2 x2 xn2 2 +···+x n ) 8.3. We have x1 (x21+xn ) + x2 (x32+x1 ) + · · · + xn (x1 +x ≥ 2(x1 x2 +x(x21x+x  An . n−1 ) 3 +···+x n−1 x n +x n x 1 ) If n ≥ 4, then An ≥ 2 (see the proof Problem 7.6). 4 4 4 8.4. (a) Using inequality (8.1), we obtain a 3 +aa2 b+ab2 + b3 +bb2 c+bc2 + c3 +c2ca+ca 2 ≥ 2 (a 2 +b2 +c2 ) ≥ a+b+c . 3 (a+b+c)(a 2 +b2 +c2 ) Here we have used that a 2 + b2 + c2 ≥ 13 (a + b + c)2 , or inequality (8.6.1) for n = 3.

x1 x2 +x3

2

Proofs

115

b c (b) By inequality (8.1), it follows that ab + bc + ac  ab + bc + ac ≥ ab + (b+c)  bc+ac a b b+c b+c + c · a+b + a+b . b b+c Hence, in order to complete the proof, one needs to prove that ab + bc · a+b ≥   2 a+b 2 +1, which is true because it is equivalent to the inequality ac − b ≥ 0. b+c (c) Without loss of generality one can assume that (b − a)(b − c) ≤ 0. b+c + a+b + 1. On the other hand, by Problem 8.4 (b), we have ab + bc + ac ≥ a+b b+c a+b a+c a+b Therefore, it is sufficient to prove that b+c + 1 ≥ b+c + a+c , which holds because it is equivalent to the inequality (b − a)(b − c) ≤ 0. a+c b+c + a+b + a+c . (d) In a similar way, one can prove that ab + bc + ac ≥ a+b b+c 8.5. Without loss of generality one can assume that a1 ≥ · · · ≥ an , whence 0 < p − 2a1 ≤ · · · ≤ p − 2an . a1 ·1 + ··· + According to the remark of inequality (8.6.1), we have that p−2a 1 2

an ·1 p−2an

(a1 +···+an )n np−2(a1 +···+an )

≥



2

2

n . n−2

√

( ) + ( b1 ) + ( 1c ) ≥ ( a1 + b1 + 1c )  ab+bc+ac ≥ 3 3 a 2 b2 c2  8.6. (a) We have that, a(b+c) b(a+c) c(a+b) 2(ab+bc+ac) 2 2 3 . 2 1 (b) Set 1+x1 +···+x  yi , i  1, . . . , n, in which case 1 > y1 > · · · > yn > 0 i yi−1 yi 1 and xi  yi−1 −yi , i  1, . . . , n, y0  1. From inequality (8.1), we deduce that     n  n      yi2 1 yi−1 yi 1   yi + + ··· +   x1 xn y − yi yi−1 − yi i1 i−1 i1   n  yi2 y1 + · · · + yn > ≥ √ > y1 + · · · + yn , y − y 1 − yn i i1 i−1 1 a



2

2

2

2

1 1 1 > 1+x + · · · + 1+x1 +···+x . xn 1 n 2 2 y y2 x z x (c) We have that, y 2 −z + z 2 −x + x 2 −y  x y 2 −z + y z 2 −x + z xz2 −y ≥ ( ) ( ) ( ) (x+y+z)2 (x+y+z)2 ≥ x y 2 +yz 2 +zx 2 −x z−yx−zy > 4x y+4yz+4zx−x z−yx−zy ≥ 1, and therefore y 2x−z y + x 2z−y > 1. z 2 −x

and therefore

1 x1

+ ··· +

+

8.7 We have 

2

x + yz



2

y + xy







x 2 + y 2 + z 2 + x y + yz + x z z + xy ≤ 3    3 2 2 x + y2 + z2  . 3 2

On the other hand, from inequality (8.1) we obtain x 3 +y 3 +z 3 x 2 +y 2 +z 2



x4 x (x 2 +y 2 +z 2 )

+

y4 x (x 2 +y 2 +z 2 )

+

z4 x (x 2 +y 2 +z 2 )

≥

x 2 +y 2 +z 2 x+y+z

≥

3

(8.11) 

x 2 +y 2 +z 2 , 3

116

8 A Useful Inequality

whence 2  3  3 x 3 + y3 + z3 ≥ x 2 + y2 + z2 .

(8.12)

From (8.11) and (8.12) it follows that (x 2 + yz)(y 2 + x z)(z 2 + x y) ≤ 89 (x 3 + y 3 + z 3 )2 . 8.8. Using Chebyshev’s inequality, we obtain ⎛

⎞⎛ ⎞⎛ ⎞ ⎛ n n n  1  ⎠⎝ 1  ⎠ 1 ⎝1 ⎠ ⎝ ai bi ci · · · ⎝ n n n n i1 i1 i1 ⎛ ⎞ ⎛ ⎞ n n   1 1 ⎝ ⎠ ⎝ ai bi ci · · · di ⎠ ≤ · · · ≤ ≤ n n i1

i1

n  i1

⎞

⎛

1 di ⎠ ≤ ⎝ n

n  i1

⎞⎛ 1 ai bi ⎠⎝ n

n  i1

⎞

⎛

1 ci ⎠ · · · ⎝ n

n 

⎞ di ⎠ ≤

i1

n 1 ai bi ci · · · di . n i1

x2

x2

x2

1 +···+x n−1 ) 8.9. From inequality (8.1), we obtain x01 + x12 + · · · + xn−1 ≥ (x0x+x  Kn . n 1 +x 2 +···+x n Let us prove that there exists a positive integer n 0 such that for n > n 0 , K n ≥ 3.999. We have



x0 + x1 + · · · + xn−1

2

2

2   − 3.999(x1 + x2 + · · · + xn )  1 − x1 + x2 + · · · + xn−1 +

  3.999 + 0.001 x1 + x2 + · · · + xn−1 − 3.999xn ≥ 0.001(n − 1)xn − 3.999xn ≥ 0, if n ≥ + 1. 0.001

Therefore, one can take n0 equal to 4000. 8.10. By transforming the given expression and using inequality (8.1), we obtain that BC 2 C A2 AB 2 + + BC · M A1 C A · M B1 AB · MC1 p (BC + C A + AB)2 4 p2 2 . ≥  BC · M A1 + C A · M B1 + AB · MC1 2S r AB Therefore, the smallest possible value of the expression MBCA1 + MC BA1 + MC is 1 2p BC CA AB equal to r , when BC·M A1  C A·M B1  AB·MC1 that is, when M A1  M B1  MC1 Therefore, the point M is the incenter of triangle ABC. 8.11. Let us denote by A1 , A2 , A3 and a1 , a2 , a3 the midpoints and the lenghts of sides A2 A3 , A1 A3 , A1 A2 , respectively. Letting a1 ≤ a2 ≤ a3 , one can easily prove that G A3 ≤ G A2 ≤ G A1 . a2 We have 23 G A1 · B1 A1  41 a12 , and hence G B1  G2A1 + 6G1A1 .

Hence, it is sufficient to prove that

a12 3G A1

a2

a2

+ 3G2A2 + 3G3A3 ≥ G A1 + G A2 + G A3 .

Proofs

117 a2

a2

a2

a12 +a22 +a32 G A1 +G A2 +G A3

From inequality (8.6.1), it follows that 3G1A1 + 3G2A2 + 3G3A3 ≥ 3(G A21 +G A22 +G A23 ) ≥ GA + GA + GA , G A1 +G A2 +G A3

a12 3 G A1

1

2

a22 3 G A2

a32 3 G A3



3

and therefore, + + ≥ G A1 + G A2 + G A3 . 8.12. From Problem 2.1, it follows that √ √ √ √ √ √ a1 a2 + a1 a2 + a3 a4 + a3 a4 + · · · + a2k−1 a2k + a2k−1 a2k √ ≥ 2k 2k a1 · a2 · · · a2k , and hence we obtain that a1 + a2 + · · · + a2k − 2k

√ √ √ 2 √ √ 2 a1 · a2 · · · a2k ≥ a1 − a2 + · · · + a2k−1 − a2k 

2k

(a1 − a2 )2 (a2k−1 − a2k )2 (a1 − a2 )2 (a2k−1 − a2k )2 ≥ √ . √ 2 + · · · + √ √ 2 ≥ 2(a + a ) + · · · + 2(a 1 2 2k−1 + a2k ) a1 + a2 a2k−1 + a2k (a1 −a2 )2 2(a1 +a2 )

From inequality (8.1), it follows that (a1 −a2 +···+a2k−1 −a2k )2 , 2(a1 +a2 +···+a2k )

It follows that a1 + a2 + · · · + a2k − 2k

+ ··· +

√ 2k a1 · a2 · · · a2k ≥ 2

(a2k−1 −a2k )2 2(a2k−1 +a2k )

≥

(a1 −a2 +···+a2k−1 −a2k )2 , 2(a1 +a2 +···+a2k )

1 1 n + · · · + 1+a ≥ (1+a1 )+···+(1+a , and hence 8.13. We have n − 1  1+a 1 n n) n−1 n + a2 + · · · + an ) ≥ 2 . 2 (a1 √ √ ( a1 +···+ an )2 , and therefore n ≥ √a a + an a1 Note that 1  1+a + · · · + ≥ 1 2 1+a 2 (1+a )+···+(1+a n 1 n) √ √1 a1 a3 + · · · + an−1 an . S−an S−a1 8.14. Letting S  a1 + · · · + an we need to prove that 1+S−a + · · · + 1+S−a ≥ n − 1. 1 n From inequalities (8.1) and (3.5), it follows that √ √ √ 2 2 2 √ S − a1 S − an S − a1 + · · · + S − an S − a1 S − an  + ··· +  + ··· + ≥ 1 + S − a1 1 + S − an 1 + S − a1 1 + S − an n + (n − 1)S    √ √ S − ai − a j + ai a j S − ai S − a j (n − 1)S + 2 (n − 1)S + 2 



≥

i  j

i  j

≥

 n + (n − 1)S   √ √ 2 ai + a j + 2 ai a j ai a j (n − 1) S + 2 (n − 1)S + n(n − 1)S − 2 n + (n − 1)S

i  j

i  j

n + (n − 1)S  √ ai a j (n − 1)2 S + 2Cn2 Cn2 i  j

n + (n − 1)S





i  j

n + (n − 1)S

(n − 1)2 S + 2Cn2  n − 1, n + (n − 1)S

S−an S−a1 + · · · + 1+S−a ≥ n − 1. 1+S−a1 n (a+1)3 a 1 2 . Since a +bc  a·1 + a·d , it follows from inequality (8.2) that a 2 +bc ≥ 2a(d+1) (b+1)3 (c+1)3 2 2 2 In a similar way, we obtain b + cd ≥ 2b(a+1) , c + da ≥ 2c(b+1) , d + ab ≥ (d+1)3 . 2d(c+1)

whence we obtain 8.15.

≥

3

Multiplying these inequalities, we obtain that  2     2 2 2 2 a + bc b2 + cd c2 + da d 2 + ab ≥ (a+1) (b+1) 8(c+1) (d+1) .

118

8 A Useful Inequality

√ √ √ √ Since (a + 1)(b + 1)(c + 1)(d + 1) ≥ 2 a · 2 b · 2 c · 2 d  8,      2 2 2 2 we have a 2 + bc b2 + cd c2 + da d 2 + ab ≥ (a+1) (b+1) 8(c+1) (d+1) ≥ (a + 1)(b + 1)(c + 1)(d + 1), and therefore, (a 2 + bc)(b2 + cd)(c2 + da)(d 2 + ab) ≥ (a + 1)(b + 1)(c + 1)(d + 1). 8.16. Since 1 a 5 (b + 2c)2

1

+

+

b5 (c + 2a)2  1 3 a



(ab + 2ac)(ab + 2ac)

+

1  c5 (a + 2b)2  1 3 (bc + 2ab)(bc + 2ab)

then from inequality (8.2), it follows that ab+bc+ac . 9

Note that, 1 c5 (a+2b)2

≥

ab+bc+ac 9 1 . 3

≥

 1 3

b

√ 3 3 ab·bc·ac 9



1 , 3

+

1 a 5 (b+2c)2

6a+3ac 6b+3ab + (a+b+c) 3 (a+b+c)3 6 1 2 1 + ≤ 3+3 3(ab+bc+ac) a+b+c

Note that

+

a3 1·1

1 b5 (c+2a)2 1 a 5 (b+2c)2 3

+ + 3

,

1 c5 (a+2b)2 1 b5 (c+2a)2

b c + 1·1 + 1·c ≥

+

≤

6c+3bc (a+b+c)3



3(ab+bc+ac) ≤ (a+b+c)2 (a+b+c) a b + + c a 3 +b3 +c2 a 2 +b3 +c3 a 3 +b2 +c3

6 (a+b+c)2

≤  1, and therefore, 8.18 (a) From inequality (8.2), it follows that,

3

≤

+

≤ 1.

x3 y3 z3 (x + y + z)3 + + ≥  (1 + y) · (1 + z) (1 + z) · (1 + x) (1 + x) · (1 + y) (1 + y + 1 + z + 1 + x)(1 + z + 1 + x + 1 + y)  2 x+y+z x+y+z 1 3 √  ≥ . (x + y + z) ≥ , as x + y + z ≥ 3 3 x yz  3 and 3+x +y+z 4 3+x +y+z 2 3

≥

(a+b+c)3 , 3(2+c)

6b+3ab c 6c+3bc , ≤ (a+b+c) 3. (a+b+c)3 a 3 +b2 +c3 a b deduce that a 3 +b3 +c2 + a 2 +b3 +c3 + a 3 +bc2 +c3

b a 2 +b3 +c3

Summing these inequalities, we 6a+3ac 6b+3ab 6c+3bc + (a+b+c) . 3 + (a+b+c)3 (a+b+c)3

+

and therefore,

8.17. From inequality (8.2), it follows that, a 3 + b3 + c2  6a+3ac and therefore, a 3 +ba3 +c2 ≤ (a+b+c) 3. In a similar way, we obtain

c

(ac + 2bc)(ac + 2bc)

3

y x z Therefore, (1+y)(1+z) + (1+x)(1+z) + (1+x)(1+y) ≥ 34 . (b) From inequality (8.2), it follows that  √ 3 √   √ 3  √ 3 3x+ √ 3x 3z 3z 3 √ 3 y+ √ 3y 27 3 x yz x y z + + ≥ ≥  + +    √ 1−x 1−y 1−z 1 · (1 − x) 1 · (1 − y) 1 · (1 − z) 3(3 − x − y − z) 3 3 − 3 3 x yz √  √ √ √ 3 3 x yz √ √ √ 3 √    , as 3 x + 3 y + 3 z ≥ 3 3 x · 3 y · 3 z and x + y + z ≥ 3 3 x yz. √ 1 − 3 x yz

√ 3 3 x yz

y x z Therefore, 1−x + 1−y + 1−z ≥ 1− √3 x yz . (c) From inequality (8.2), it follows that x y z + +  (y + z)(y + z − x) (x + z)(x + z − y) (x + y)(x + y − z) 

x3 y3 z3    + +  ≥ (x y + x z) x y + x z − x 2 (x y + yz) x y + yz − y 2 (x z + yz) x z + yz − z 2

≥

9 (x + y + z)3   ≥ , 2(x + y + z) 2(x y + yz + x z) 2x y + 2yz + 2x z − x 2 − y 2 − z 2

Proofs

119   as x 2 + y 2 + z 2 − x y − yz − x z  0, 5 (x − y)2 + (y − z)2 + (z − x)2 ≥ 0 and (x + y + z)2 (x + y + z)2 ≥ 3. ≥ 3, x y + yz + x z 2x y + 2yz + 2x z − x 2 − y 2 − z 2

y x z Therefore, (y+z)(y+z−x) + (x+z)(x+z−y) + (x+y)(x+y−z) ≥ (d) From inequality (8.2), it follows that

9 . 2(x+y+z)

a3 b3 c3 d3 + + + ≥ 1 · (b + c + d) 1 · (a + c + d) 1 · (a + b + d) 1 · (a + b + c) ≥

1 (a + b + c + d)3 (a − b + c − d)2 + 4(ab + bc + cd + da)  ≥ , 4(3a + 3b + 3c + 3d) 12 3 3

3

3

3

a b c d and therefore, b+c+d + a+c+d + a+b+d + a+b+c ≥ 13 . (e) From inequality (8.2), it follows that 3 (1+x)3 13 x3 1 + xy 2  1·1 + y·y ≥ (1+y)(1+y) , whence we obtain

     (1 + a1 )3 (1 + a2 )3 (1 + an )3 a13 a23 an3 1+ 2 1 + 2 ··· 1 + 2 ≥ · · · · a2 a3 a1 (1 + a2 )2 (1 + a3 )2 (1 + a1 )2  (1 + a1 ) (1 + a2 ) · · · (1 + an ) . (f) From inequality (8.2), it follows that,  3 13 n3 1 2 2 + · · · + xn ·(x1n +x1 ) ≥ 2(x +···+x + 2 , and therefore (x 1 + · · · + x n ) x1 ·(x1 +x2 ) x12 +x1 x2 1 n)    2 2 n 3 x 2 +···+x 2 x 2 +···+x 2 n · · · + x 2 +x1 n x1 ≥ 2(x( 1+···+x )n2) ≥ n2 , since 1 n n ≥ x1 +···+x . n 1 n n     2 1 1 n2 2 Therefore, x1 + · · · + xn x 2 +x x + · · · + x 2 +xn x1 ≥ 2 . 1

1 2

n

(g) For A  a1 a2 + a2 a3 + · · · + an−1 an + an a1 < n1 , from inequality (8.2), it follows that a3 a3 a a1 a2 an a3  1  +  2  + ··· +  n  ≥ + + · · · + n−1 +  a1 a2 a1 a2 + a1 a2 a3 a2 a3 + a2 an a1 an a1 + an a22 + a2 a32 + a3 an2 + an a12 + a1   a + a2 + · · · + an 3 1 n 1   > · , ≥  1 A(A + 1) A n+1 A A + a1 + a2 + · · · + an ⎞ ⎛   a a1 a2 an ⎠ n . + + · · · + n−1 + ≥ a1 a2 + a2 a3 + · · · + an−1 an + an a1 ⎝ n+1 a22 + a2 a32 + a3 an2 + an a12 + a1

and therefore,

For A ≥ n1 , from inequality (3), we have  a1

a2

an−1

an

 a1 2 a2



 a2 2 a3



 an 2 a1

+ + ··· + 2 +  + + ··· + ≥ a2 + 1 a3 + 1 a1 + 1 a22 + a2 a32 + a3 an + an a12 + a1 2     a1 a2 an      a2 + a3 + a1 a1 a2 an a1 a2 an n2 + + ··· + ≥nn · ···  n. ≥ , as ≥ n+1 n+1 a2 a3 a1 a2 a3 a1

Hence, we obtain that,     a1 n2 n a2 an−1 an 1 a1 a2 + a2 a3 + · · · + an−1 an + an a1  . + + · · · + + ≥ · n n+1 n+1 a22 + a2 a32 + a3 an2 + an a12 + a1

120

8 A Useful Inequality

8.19. Letting x1 + x2 + · · · + xn  S,

x15 x2 +x3 +···+xn

x5

x5

n 2 + x1 +x3 +···+x + · · · + x1 +x2 +···+x  A, n n−1 3 3 3 x12 ) x22 ) xn2 ) ( ( ( from inequality (8.2), it follows that A  x1 (S−x1 ) + x2 (S−x2 ) + · · · + xn (S−xn ) ≥ x3 x3 x3 (x12 +x22 +···+xn2 )3 1  ≥ 1 , as 1 1 + 2 + · · · + n ≥

S·(S−x1 +S−x2 +···+S−xn ) 2 S3  Sn . nS

(n−1)S 2

n(n−1)

1·x1

1·x2

1·xn

1 and for x1  x2  · · ·  xn  √1n , we have n(n−1) 1 A  n(n−1) , and therefore, the smallest possible value of the given expression 1 is equal to n(n−1) . 3 3 3 (x+y+z)3 From inequality (8.2), it follows that x 2x+y + y 2y+z + z 2z+x ≥ 3 x 2 +y+y 2 +z+z 2 +x . (  2  ) 3 2 2

We have obtained A ≥

8.20.

It is sufficient to prove that, 2(x + y + z) ≥ 9 x + y + z + x + y + z .  √ Letting x + y + z  d, we have√x y + yz + zx  (x y + yz + zx)2 ≥ √ 3x y · yz + 3x y · zx + 3yz · zx  3d. ≥ One needs to prove  that, 2(x + y + z)3 + 18(x y + yz + zx)  9 (x + y + z)2 + x + y + z . √ Let us prove that, 2d 5 + 18 3√≥ 9d 3 + 9d. √ √ √ Note that, 2d 5 −9d 3 −9d +18 3  (d − 3)2 (2d 3 +4 3d 2 +9d +6 3) ≥ 0.

Problems for Independent Study Prove the following inequalities (1–22, 24–45, 47, 48). n ≥ ai , where an+1  a1 , ai > 0 i  1, . . . , n. i1 i1 2 2 n  n 2 +1 xi + x1i ≥ ( n ) , where x1 + · · · + xn  1, xi > 0, i  1, . . . , n. 2.

1. 2

n

ai2 ai +ai+1

i1

2

2

2

2

2

2

3

3

3

+b +c +c a b c + b 2a + a 2b ≤ bc + ac + ab , where a > 0, b > 0, c > 0. 3. a + b + c ≤ a 2c (a1 +···+an )(b1 +···+bn ) an bn a1 b1 4. (a) a1 +b1 + · · · + an +bn ≤ (a1 +···+an )+(b1 +···+bn ) , where ai > 0, bi > 0, i  1, . . . , n, 2

n n 1 + · · · + an +bann +b ≥ a(a1 +···+a + (b) a1 +ba11 +b +aa1 b1 +aan bn n )+2n where a > 0, ai > 0, bi > 0, i  1, . . . , n,

(c)

n

√ai 2bi 2 ai +bi i1

n

≤



ai

i1 n

ai

n2 , a(b1 +···+bn )+2n

n

bi i1 2  n

2 , bi

where ai > 0, bi > 0, i  1, . . . , n,

  b1 +···+bn 1 +···+an ≥ n, where ai > + ··· + + n2 a1a+···+a + (d) b1 +···+bn +2 n +2 0, bi > 0, i  1, . . . , n, n 2 (c1 +···+cn ) , where ai > 0, bi > 0, ci > 0, i  (e) a1c·b1 1 + · · · + anc·bn n ≥ (a1 +···+a n )(b1 +···+bn ) 1, . . . , n,    aj bj ai bi ≤ 0, i, j ∈ {1, . . . , n}. − − ci cj ci cj i1

a1 +b1 a1 +b1 +na1 b1

i1

an +bn an +bn +nan bn

Problems for Independent Study

(f)

1 √ a1 b1 −c1 d1

+ ··· +

121

1 √ an bn −cn dn

≥

n2  , √ √ 2 (a1 +···+an )(b1 +···+bn )−( c1 d1 +···+ cn dn )

where ai > 0, bi > 0, ci > 0, di > 0, ai bi ≥ ci di , i  1, . . . , n. √ Remark Letting ai bi − ci di  xi , we have 

  2  1 1 + ··· + √ c1 d1 + · · · + cn dn √ (a1 + · · · + an )(b1 + · · · + bn ) − an bn − cn dn a1 b1 − c1 d1         2 2 2  1 1  (a + · · · + a ) x1 + · · · + xn + (a + · · · + a ) c1 d1 + · · · + cn dn −  + ··· + c1 d1 + · · · + cn dn . 1 n 1 n x1 xn a1 an a1 an

x1 2 + 3x+x +· · ·+ xxnn−1 + xn ≥ n , where n  5 or n  6, xi > 0, i  1, . . . , n. 4  +x1 x1 +x22  2 x2 +x3 2x 2 a a a 6. 1 + a12 1 + a23 · · · 1 + an1 ≥ (1 + a1 ) · · · (1 + an ), where n ≥ 2, a1 > 0, . . . , an > 0.

5.

Remark Note that 1 + 7.

a b+c

a2 b



12 1

+

a2 b

(1+a)2 , 1+b

≥

where b > 0.

b c + c+a + a+b + ab+bc+ca ≤ 25 , where a, b, c are the side lengths of some triangle. a 2 +b2 +c2

Remark Note that b+c−a c+a−b a+b−c + +  b+c c+a a+b (b + c − a)2 (c + a − b)2 (a + b − c)2 (a + b + c)2 .  + + ≥  2 (b + c)(b + c − a) (c + a)(c + a − b) (a + b)(a + b − c) 2 a + b2 + c2

8. 9. 10. 11.

b c d + d+2a+3b + a+2b+3c ≥ 23 , where a > 0, b > 0, c > 0 d > 0. c+2d+3a n √ xi n n i1 √ xi √ ≥ , where n ≥ 2, x > 0, i  1, . . . , n, and xi  1. i 1+xi n−1 i1 i1 2 2 x1 x xn2 4 + 1+x1 +x32+···+xn + · · · + 1+x1 +x2 +···+x ≥ 3n−2 , where n ≥ 2, 1+x2 +x3 +···+xn n−1 xi > 0, i  1, . . . , n and x1 + · · · + xn  2. x1k + · · · + xnk ≥ x1 + · · · + xn , where k ∈ N and n ≥ 2, xi > 0, i  a b+2c+3d

+

1, . . . , n, x1 · · · xn  1. √ √ a1 + · · · + an ≤ a1 + · · · + an , where n ≥ 2, ai > 0, i  1, . . . , n and a1 · · · an  1. (x+y)z 4z ≤ 4z+3x+3y , where x > 0, y > 0, z > 0; 13. (a) (x+y) 2 +z 2 12.

(b)

(b+c−a)2 (b+c)2 +a 2 √ x1√ x1 + x2

+

(c+a−b)2 (a+b−c)2 + (a+b) ≥ 35 , where a > 0, b > 0, c 2 +c2 (c+a)2 +b2 √ √  x x √ 2√ + · · · + √ n√ ≥ 1 x1 + · · · + xn , x2 + x3 xn + x1 2

> 0.

14. (a) + > 0. where n√≥ 3, x1 > √ 0, . . . , x n √ a c b√ √ √ ≥ 1, where a, b, c ∈ (1, 2). (b) 4b√bc−ca √a + 4c√ca−a + b 4a b−b c Remark Note that 15. 16.

√ b a √ √ 4b c−c a

√

ab)  4b(√ac−ac . 2

(a1 +···+an )2 n 1 2 ≤ a2a+a + a3a+a + · · · + a1a+a , where n ≥ 3, a1 > 0, . . . , an > 0. 2(a12 +···+an2 ) 3 4 2 9 9 9 9 9 9 x +y y +z z +x + y 6 +y 3 z 3 +z 6 + z 6 +z 3 x 3 +x 6 ≥ 2, where x > 0, y > 0, z > 0 and x 6 +x 3 y 3 +y 6

x yz  1.

122

17. 18. 19. 20. 21.

8 A Useful Inequality √ y + 1+zz 2 ≤ 3 4 3 , where x 2 + y 2 + z 2  1. 1+y 2 2 2 2 b3 c3 + c+2a + a+2b ≥ a +b3 +c , where a > 0, b > 0, c > 0. 1 1 1 3n+1 ≤ n+1 + n+2 + · · · + 2n ≤ 4(n+1) , where n ∈ N  √ √ √ 2 an a2 4 + a 2 +1 + · · · + a 2 +1 ≥ 5 a1 a1 + a2 a2 + · · · + an an , 3 1 4, a1 > 0, . . . , an > 0 and a12 + · · · + an2  1. a b c + 2b+c + 2c+a ≤ 1, where a > 0, b > 0, c > 0. 2a+b x 1+x 2 a3 b+2c 2n 3n+1 a1 a22 +1

+

Remark We have

1 2

−

a 2a+b

+

1 2

+

b 2b+c



b2 2b(2a+b)

+

c2 2c(2b+c)

≥

where n ≥

(b+c)2 2(2ab+(b+c)2 )

≥

c . 2c+a

9 , where a > 0, b > 0, c > 0 and a 2 + b2 + c2  3abc. 22. b2ac2 + c2ba 2 + a 2cb2 ≥ a+b+c 23. Solve the following system of equations:  x1 + x2 + · · · + xn  n, . x14 + x24 + · · · + xn4  x13 + x23 + · · · + xn3 1 1 1 27 + b(b+c) + c(c+a) ≥ 2(a+b+c) 24. a(b+a) 2 , where a > 0, b > 0, c > 0. 4 4 4 4 25. a b+b c+c d+d a ≥ abcd(a + b + c + d), where a > 0, b > 0, c > 0, d > 0 a b c 9 + a(a+b) 26. b(b+c) 2 + 2 ≥ 4(ab+bc+ac) where a > 0, b > 0, c > 0. c(a+c)2

Remark We have

a b(b+c)2



a3 . ab·a(b+c)2

    27. 3 x 2 − x + 1 y 2 − y + 1 z 2 − z + 1 ≥ (x yz)2 − x yz + 1.  3 Remark Prove that 3 t 2 − t + 1 ≥ t 6 + t 3 + 1.         28. 2 a 2012 + 1 b2012 + 1 c2012 + 1 ≥ (1 + abc) a 2011 + 1 b2011 + 1 c2011 + 1 , where a > 0, b > 0, c > 0. a b c 9 + (a+b) 29. (b+c) 2 + 2 ≥ 4(a+b+c) , where a > 0, b > 0, c > 0. (a+c)2 Remark Note that

a (b+c)2



a3 . (ab+ac)·(ab+ac)

30. x 8 + y 8 + z 8 ≥ x 2 y 2 z 2 (x y + yz + zx). Remark Note that x yz  0, whence 31. 32.

x3 x+yz 

+

3 y3 + z y+zx z+x y

abc a+b−c

abc + b+c−a some triangle.

x 8 +y 8 +z 8 x 2 y2 z3

3 3 3 x2 y2 z2  (y 2 ·z)2 + (z 2 ·x)2 + (x 2 ·y)2 .

≥ 14 , where x + y + z  1 and x > 0, y > 0, > z > 0.  abc + a+c−b ≥ a + b + c, where a, b, c are the side lengths of

Remark Note that abc(a + b + c) ≥ a 3 (b + c − a) + b3 (a + c − b) + c3 (a + b − c). √ √ √ 33. 3 ab + 3 cd ≤ 3 (a + c + d)(a + c + b), where a > 0, b > 0, c > 0, d > 0.

Problems for Independent Study

Remark Note that

ab (a+c)b

+

123

cd d(a+c)

≥

√ √ ( 3 ab+ 3 cd)3 . (a+c+d)(b+a+c)

34. (a 5 − a 2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3 , where a > 0, b > 0, c > 0. Remark Note that x 5 − x 2 + 3 ≥ x 3 + 2, where x > 0. 35. 3(a 3 + b3 + c3 ) ≥ (a 2 + b2 + c2 )3 , where a > 0, b > 0, c > 0. Remark Note that 3(a 3 + b3 + c3 ) ≥ (a + b + c)(a 2 + b2 + c2 ).       a3 a3 a3 36. 1 + a12 · 1 + a22 · · · 1 + an2 ≥ (1 + a1 ) · (1 + a2 ) · · · (1 + an ), where a1 > 2 3 1   2      0, 3a2 > 0,. .3. , an > 0. 3 37. a1 + 1 · a2 + 1 · · · an + 1 ≥ a1 a2 + 1 · a22 a3 + 1 · · · an2 a1 + 1 , where a1 > 0, a2 > 0, . . . , an > 0.      3 Remark We have x 3 + 1 x 3 + 1 y 3 + 1 ≥ x 2 y + 1 , where x > 0, y > 0. 38.

x1 1−x12

+

x2 1−x22

+···+

xn 1−xn2

≥

(x1 +x2 +···+xn )3 2, (x1 +x2 +···+xn )2 −(x12 +x22 +···+xn2 )

where 0 < x1 < 1, 0 <

x2 < 1, . . . , 0 < xn < 1.  39. n n (x1n + 1) · (x2n + 1) · · · (xnn + 1) ≥ x1 + x2 + · · · + xn + where x1 · x2 · · · xn  1 and x1 < 0, x2 < 0, . . . , xn < 0.

1 x1

+

1 x2

+ ··· +

1 xn

n

,

Remark Note that (x1n + 1) · (1 + x2n ) · · · (1 + xnn ) ≥ (x1 · 1 · · · 1 + 1 · x2 · · · xn )n   n x1 + x11 . √ √ √ 40. x1 3 1 + xn − x2 + x2 · 3 1 + x1 − x3 +· · ·+ xn · 3 1 + xn−1 − x1 ≤ x1 + x2 +· · · xn , where x1 > 0, x2 > 0, . . . , xn > 0. √ √ √ √ Remark Note that x1 · 3 1 + xn − x2  3 x1 · 3 x1 · 3 x1 + xn x1 − x1 x2 .   2 41. (x12 + x22 + · · · + xn2 ) x 2 +x1 x + x 2 +x1 x + · · · + x 2 +x1 n x1 ≥ n2 , 1 2 2 3 n 1 2 where x1 > 0, x2 > 0, . . . , xn > 0. Remark Note that 42. 43.

13 x12 +x1 x2

+

13 xn2 +x2 x3

+ ··· +

13 xn2 +xn x1

≥

n3 . 2(x1 +x2 +···+xn )2

x13 xn3 +···+xn + · · · + (axn +bx1 )(ax ≥ x1(a+b) 2 , where a (ax1 +bx2 )(ax2 +bx1 ) 1 +bx n ) 0, x2 > 0, . . . , xn > 0. a13 a3 + · · · + bnn ≥ 1, where ai > 0, bi > 0 (i  1, . . . , n) and b1 b12 + · · · + bn2 .

Remark Note that

a13 b1

+ ··· +

an3 bn

> 0, b > 0, x1 > 3  2 a1 + · · · + an2 

3 3 a2 a2  (a3 1 ) . + · · · + (an3 n )2 . 1 b1

·b12

bn

·bn

3 2   44. 4 x 3 + y 3 + z 3 + x yz ≥ x 2 + y 2 + z 2 + t 2 , where x > 0, y > 0, z > 0, yz and t 2  maxx (x,y,z) .

124

8 A Useful Inequality

2 2 2 x2 y2 z2 Remark Note that x + y + z + x yz  ( x ) + ( y ) + ( z ) +

3

45.

3

√ 3

3

n a1,1 a1,2 ·a1,3 ···a1,n

an

x 2 y2 z2 √ 3 x yz

2

.

an

2,1 m,1 + + ... + ≥ a2,2 ·a2,3 ···a2,n am,2 ·am,3 ···am,n n (a1,1 +a2,1 +···+am,1 ) , where ai, j > 0, i  1, 2, (a1,2 +a2,2 +···+am,2 )·(a1,3 +a2,3 +···+am,3 )···(a1,n +a2,n +···+am,n ) . . . , m, j  1, 2, . . . n.

Remark Without loss of generality one can assume that a1,1 + a2,1 + · · · + am,1  a1,2 + a2,2 + · · · + am,2  · · ·  a1,n + a2,n + · · · + am,n  1. an Note that ak,2 ·ak,3k,1···ak,n + ak,2 + · · · + ak,n ≥ nak,1 . 46. Find the smallest possible value of the expression a4 +2b4 +3c4 , if a +b +c  1. Remark Note that a 4 + 2b4 + 3c4 

a4 b4 c4 (|a| + |b| + |c|)4 + + ≥   ≥        3 3  3 13 3 1 3 1 3 1 1 2 3 1+ 2 + 3 3 6 (|a + b + c|)4   √ √ √ 3 .   3 3 3 2+ 3+ 3 6 1 + 3 21 + 3 13

≥

47.

(b+c)5 a

+ (c+a) + (a+b) ≥ 32 + bc + ca), where b c 9 (ab a > 0, b > 0, c > 0 and a + b + c  1. 5

5

(b+c) (c+a)  1·1·1·a + 1·1·1·b + (a+b) ≥ 32 . 1·1·1·c 27   √  k−1 48. 2 − 1 (a1 + a2 + · · · + an ) < k 2a1k + 22 a2k + · · · + 2n ank , where a1 > 0, . . . , an > 0 and k ≥ 3, k ∈ N.

Remark Note that

(b+c)5 a

+

(c+a)5 b

+

(a+b)5 c

5

5

5

Remark Note that a1k a2k ank 2a1k + 22 a2k + · · · + 2n ank   √ k−1 + · · · +  k−1 √ k−1 ≥ √ k−1 +  k−1 k−1 2 2 2n 1/ 1/ 2 1/ ≥ 

(a1 + a2 + · · · + an )k (a1 + a2 + · · · an )k k−1 . √ k−1 >  k−1 √ √ √ k−1 k−1 k−1 2 + 1/ 22 2 + 1/ 22 + · · · + 1/ 2n + · · · 1/ 1/ √ k−1

Problems for Independent Study

125

49. Let G be the intersection point of the medians of triangle A1 A2 A3 , and let C be the circumcircle of triangle A1 A2 A3 . Let the lines G A1 , G A2 , G A3 intersect the circle C a second time at points B1 , B2 , B3 , respectively. Prove that G B1 +  G B2 + G B3 ≥

A1 A22 + A2 A23 + A3 A21 .

Remark See the proof of Problem 8.11.

Chapter 9

Using Derivatives and Integrals

Historical origins. Derivatives and integrals are foundational proof technique tools in mathematics. The derivative of a function y  f (x) of a variable x is a measure of the rate at which the value y of the function changes with respect to the change of the variable x. In general, in the literature the following two distinct notations are commonly used for derivatives: 1. Leibniz’s notation, introduced by German mathematician and philosopher Gottfried Wilhelm von Leibniz, born 1 July 1646 in Leipzig, Holy Roman Empire (now Leipzig, Germany), died 14 November 1716 in Hanover, Holy Roman Empire (now Hanover, Germany). According to Leibniz’s notation, the derivative of y with respect to x is denoted by ddyx , where dy represents the change in y and d x represents the change in x. 2. Lagrange’s notation, introduced by the Italian–French mathematician Joseph Louis Lagrange, born 25 January 1736 in Turin, Kingdom of Sardinia (now Turin, Italy), died 10 April 1813 in Paris, France. According to Lagrange’s notation, the derivative of a function f (x) with respect to x is denoted by f  (x), or sometimes it is denoted by f x (x). In this chapter, for simplicity and brevity we use Lagrange’s notation for the derivative f  (x). Suppose we need to prove the inequality f (x) ≥ g(x)

(9.1)

on an interval [a, b]  I or [a, +∞)  I , where the functions f (x) and g(x) are defined and continuous on I . Theorem 9.1 If functions f (x), g(x) are differentiable on the domain I , f (a) ≥ g(a), and on I we have that h  (x) ≥ 0, where h(x)  f (x) − g(x), then f (x) ≥ g(x) holds for all x ∈ I .

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_9

127

128

9 Using Derivatives and Integrals

Proof If on the domain I we have h  (x) ≥ 0, then the function h(x) on this domain does not decrease, and consequently, h(x) ≥ h(a) at each point x of the domain I , that is, f (x) − g(x) ≥ f (a) − g(a) ≥ 0. Therefore, f (x) ≥ g(x). This ends the proof. Remark If on the domain I we have h  (x) > 0(x  a), then for x ∈ I and x > a, one has f (x) > g(x). Now let us consider the following examples in order to see how Theorem 9.1 can be applied. Example 9.1 Prove that 2x+1 > x + 2 if x ≥ 1. Proof Let us consider the function h(x)  2x+1 − x − 2 on [1 , + ∞). We have h(1)  1 and h  (x)  2x+1 ln 2 − 1. Note that the function y  2x is increasing on [1 , +∞), whence h  (x) ≥ 4 ln 2 − 1 > 0. Therefore, if x ≥ 1, then h(x) ≥ h(1) or 2x+1 ≥ x + 3, and hence 2x+1 > x + 2. This ends the proof. Example 9.2 Prove that

k k   ( n1 m1

am an ) m+n

≥ 0.

Proof Let us consider the function h(x) 

k k   ( n1 m1

We have xh  (x)

x m am .x n an ) m+n

on [0 , +∞).

k k   ( x m am · x n a n )  (xa1 + · · · + x k ak )2 ≥ 0, and therefore, n1 m1

for x > 0, we have h  (x) ≥ 0. Hence h(1) ≥ h(0)  0, that is,

k k   ( n1 m1

am an ) m+n

≥ 0.

This ends the proof. Theorem 9.2 If f (b) ≥ g(b) and on the domain I one has h  (x) ≤ 0,(h(x)  f (x) − g(x)), then on I one has f (x) ≥ g(x), where I  [a, b] or I  (−∞, b]. Proof If h  (x) ≤ 0 on I , then the function h(x) in this domain is not increasing, and it attains its minimum value at the point x  b. On the other hand, h(b) ≥ 0, and therefore for every x in I , one has h(x) ≥ 0. Therefore, on I , one has f (x)−g(x) ≥ 0, or equivalently, f (x) ≥ g(x). This ends the proof.  Integrals are another crucial proof technique tool in mathematics. The notation was also introduced by Gottfried Wilhelm von Leibniz in 1675. It represents the first letter, s, of the word sum (summa in Latin). Let us provide the following useful theorem as an application of integrals to prove inequalities. Theorem 9.3 If for every x in the domain I one has the inequality f (x) ≥ g(x), then

9 Using Derivatives and Integrals

129

x

x f (t)dt ≥

a

g(t)dt,

(9.2)

a

where I  [a, b] or I  [a, +∞). Proof If inequality (9.1) holds, then F  (x) ≥ G  (x), where F  (x)  f (x), G  (x)  g(x), and F(a)  G(a)  0. x It follows that F(x) ≥ G(x). Hence, using that F(x)  f (t)dt, G(x)  x

a

g(t)dt, we obtain inequality (9.2).

a

This ends the proof. Now let us consider the following example in order to see how Theorem 9.2 can be applied. Example 9.3 Prove that ln(2sin x) > 21 x(π − x) −

5 π 2, 72

if x ∈ (π/6, π/2).

Proof Let us consider the inequality cot x > π2 − x, which is seen to be valid if we use the fact that tan α > α for 0 < α < π2 and substitute α by π2 − x. Integrating the inequality under consideration, we obtain x x  π  cot tdt > − t dt, and hence 2 π π 6 6     2  2 , or ln(sin x) − ln 21 > π2 · x − x2 − π2 · π6 − 21 · π6 ln(2sin x) > 21 x(π − x) − This ends the proof.

5 2 π . 72

Problems Prove the following inequalities (9.1–9.22, 9.27–9.32) 9.1. 3a 3 + 7b3 ≥ 9ab2 , where a ≥ 0, b ≥ 0. 9.2. 2n−1 (x n + y n ) ≥ (x + y)n , where x > 0, y > 0, n ∈ N. 2 9.3. (a) cos x ≥ 1 − x2 , 3 (b) sin x ≥ x − x3! , where x ≥ 0, 2 4 (c) cos x ≥ 1 − x2! + x4! ; 3 5 (d) sin x ≥ x − x3! + x5! , where x ≥ 0 (use that sin x ≤ x). √ 9.4. x − sin x ≤ 1 − cos x ≤ x 2 − sin x, where 0 ≤ x ≤ π2 . 9.5. tan x + sin x ≥ 2x, where 0 ≤ x < π2 . 2 3 2 3 x 2n+1 x 2n+2 − 2n+2 ≤ ln(1 + x) ≤ x − x2 + x3 − . . . + 9.6. x − x2 + x3 − . . . + 2n+1 x ≥ 0,n ∈ N.

x 2n+1 , 2n+1

where

130

9.7. 9.8. 9.9. 9.10. 9.11. 9.12. 9.13. 9.14. 9.15. 9.16. 9.17.

9 Using Derivatives and Integrals

ln(cos x) ≤ − x2 , where 0 ≤ x < π2 . , where 0 ≤ x ≤ π2 . sin x ≤ x(π−x) 2 tan3 x tan x − 3 ≤ x, where 0 ≤ x < π2 . (x + x1 )arccot x > 1, where x > 0. 1 1 1 + n+2 + · · · + 3n < ln 3, where n ∈ N. n+1 3 cos x sin x 3 < < , where 0 < x ≤ π2 . 1+2 cos x x 4−cos x p q Young’s inequality: ab ≤ ap + bq , where a > 0, b > 0, p > 0, q > 0 and 1 + q1  1. p p n n a −b p > aa n−b , where a > b > 0, p > n. a p +b p +bn 1 t 1t (1 + x ) − (1 + x t )− t ≤ x, where x ≥ 0, t ≥ 2. a where b ≥ 1. ab  ≤ e +2 b(ln b − 1), 2 (ln x) 2+ 3 ln x ≤ x x−1 , where x ≥ 1. 2

9.18. 2sin x + 2cos x ≥ 3, where 0 ≤ x ≤ π2 .  k+1 k+1 · 9.19. nk+1 < 1k + 2k + · · · + n k < 1 + n1  n n 9.20. (a) e < n! , where n ∈ N,  n (b) n! < n ne , where n ≥ 7, n ∈ N.

n k+1 , k+1

where n, k ∈ N.

1

9.21. (a α + bα ) α > (a β + bβ ) β , where a > 0, b > 0, 0 < α < β.  a+b  a a  b b 9.22. a+b ≤ c · d , where a > 0, b > 0, c > 0, d > 0. c+d 9.23. Find the integer part of the expression √314 + √31 + · · · + √3 1 6 . 1

9.24. Prove that 9.25. Prove that

1 2

+

1 √ 3 2

+

1 √ 4 3

+ ··· +

5

1√ (n+1) n

10

< 2, where n ≥ 2, n ∈ N.

(a) Bernoulli’s inequality: (1 + x)α > 1 + α x, if α > 1, x > −1, x  0, (b) Bernoulli’s inequality: (1 + x)α < 1 + α x, if 0 < α < 1, x > −1, x  0. (c) (S − x1 )x1 + . . . + (S − xn )xn > n − 1, where n ≥ 2, x1 > 0, . . . , xn > 0, and x1 + . . . + xn  S. 9.26. Let f (x)  a1 sin x + a2 sin 2x + . . . + an sin nx, where a1 , · · · , an are real numbers and n is a positive integer. We have that | f (x)| ≤ |sinx| for all real numbers x. Prove that |a1 + 2a2 + · · · + nan | ≤ 1. 2 2 9.27. x √ ≥ (1 + x)ln (1 + x), where x > −1. √ n+1 n n + 1 < n, where n ≥ 3, n ∈ N. 9.28. 9.29. a1 b1x +a2 b2x +· · ·+an bnx ≥ a1 +· · ·+an , where ai > 0, bi > 0, i  1, . . . , n, x > 0, and b1a1 · . . . · bnan  1.  x+1 x+1 , where a > 0, x > a1 . 9.30. x x > a a+1 1 1  α   β β  a1 +···+anα α a1 +···+an β 9.31. ≥ , where a1 > 0, · · · , an > 0, α ≥ β, α  0, β  n n 0.α ≥√β, α  0, β  0. ab < ln a−b < a+b , where a > 0, b > 0, a  b, 9.32. (a) a−ln b 2 x(x+2) 2x (b) x+2 < ln(x + 1) < 2(x+1) , where x > 0, ln x (c) x−1 < √1x , where x > 0, x  1,

Problems

131

2 2x+1 (d) 2x+1 < ln(1 + x1 ) < 2x(x+1) , where x > 0, (e) |x − y| ≤ |ln x − ln y| , where 0 < x ≤ 1, 0 < y ≤ 1, x−ln y , where 0 < y < x ≤ 1, (f) ln 1y < ln x−y   ln x−ln y (g) ln 1 + 1y < x−y , where x > 0, y > 0 and x + y ≤ 1. 2

3

2k

x 9.33. (a) Prove that 1 − x + x2! − x3! + · · · + (2k!) > 0, where k ∈ N. (b) Prove that if a polynomial P(x) of degree n is nonnegative for x, then P(x) + P  (x) + P  (x) + · · · + P (n) (x) ≥ 0 for all values of x. 9.34. Prove that a r − br + cr ≥ (a − b + c)r , where a ≥ b ≥ c ≥ 0 and r ≥ 1. 9.35. Prove that (1 − x1k )m + . . . + (1 − xnk )m ≥ n − 1, where x1 ≥ 0, . . . , xn ≥ 0, k, m ∈ N, k ≥ m, and x1 + . . . + xn ≤ 1. n 2 n   ij 9.36. Prove that ai ≤ · ai a j . i+ j−1 i1

i, j1

Proofs 9.1. Let us consider the function h(a)  3a 3 + 7b3 − 9ab2 on [0, +∞). If b > 0, then h(0)  7b3 > 0, h  (a)  9a 2 − 9b2  9(a − b)(a + b). Note that the function h(a) is decreasing on [0, b], and is increasing on [b, +∞), and since h(b)  3b3 + 7b3 − 9b3  b3 > 0, we have h(a) > 0 on [0, +∞). Therefore, 3a 3 + 7b3 > 9ab2 . If b  0, then obviously 3a 3 + 7b3 ≥ 9ab2 . 9.2. Consider the function h(y)  2n−1 (x n + y n ) − (x + y)n on [0, +∞). If x > 0, then h  (y)  n2n−1 y n−1 − n(x + y)n−1  n((2y)n−1 − (x + y)n−1 ), h  (y)  0 if y  x, h  (y) < 0 for 0 ≤ y < x, and h  (y) > 0 for y > x. It follows that the function h(y) is decreasing on [0, x], and increasing on [x, +∞). On the other hand, h(x)  0, and hence on [0, +∞) we have that h(y) ≥ 0, that is, 2n−1 (x n + y n ) ≥ (x + y)n . 9.3. (a) If x > 0, then integrating the inequality sin x ≤ x, we obtain x x 2 2 sin t dt ≤ t dt, or −cos x + 1 ≤ x2 − 0, that is, cos x ≥ 1 − x2 . 0

0

If x  0, then cos x  1  1 − Since cos x  cos |x| ≥ 1 −

|x|2 2

x2 . 2

1−

x2 , 2

cos x ≥ 1 −

it follows that

x2 . 2

(b) By Theorem 9.3, for inequality (9.3) we have

(9.3) x 0

and hence the given inequality holds.

cos tdt ≥

x  0

1−

t2 2

 dt,

132

9 Using Derivatives and Integrals

(c) If x > 0, then by part (b) and using to Theorem 9.3, we obtain  x x  t3 t − 3! dt, and thus the given inequality holds. If x  0, sin tdt ≥ 0

0

then cos x  1  1 −

x2 2!

+

x4 . 4!

Since cos x  cos |x| ≥ 1 − |x| + |x|  1 − x2! + x4! , it follows that 2! 4! x2 x4 cos x ≥ 1 − 2! + 4! . (d) The proof follows from part (c) and Theorem 9.3. √   9.4. Let us evaluate sin x −cos x. We have sin x +cos x  2 sin x + π4 , and note √ that for 0 ≤ x ≤ π2 , on integrating the inequality 1 ≤ sin x + cos x ≤ 2, we x x x √ obtain 1 dt ≤ (cos t + sin t) dt ≤ 2dt, therefore x ≤ sin x −cos x +1 ≤ 0 0 0 √ x 2. ≥ 9.5. According to inequality (3.2), we have cos x + cos12 x   1 1 2 cos x · cos2 x  2 cos x ≥ 2, and from Theorem 9.3, it follows that x  x  cos t + cos12 t dt ≥ 2dt, and hence the given inequality holds. 0

2

4

2

4

0

9.6. Let us first us prove that if x ≥ 0, then 1 − x + x 2 − · · · + x 2n − x 2n+1 ≤

1 ≤ 1 − x + x 2 − · · · + x 2n . 1+x

(9.4) 2n+2

≤ Using the formula for the sum of a geometric progression, we obtain 1−x 1+x 1 1+x 2n+1 2n+2 2n+1 ≤ 1+x , or 1 − x ≤1≤1+x (x ≥ 0). 1+x From Theorem 9.3 and inequality (9.4), it follows that the given inequality holds.  2  9.7. Since (ln(cos x))  − tan x, − x2  −x, from the inequality − tan x ≤   −x 0 ≤ x < π2 and Theorem 9.3 it follows that the given inequality holds.

9.8. Consider the function F(x)  sin x − x(π−x) on 0, π2 and the function G(x)  2 F  (x)  cos x + x − π2 on [0, π/2]. Since G  (x) ≥ 0, it follows that for x ≤ π2 , we have G(x) ≤ G(π/2), or F  (x)  cos x + x − π2 ≤ 0, and since x ≥ 0, we have F(x) ≥ F(0)  0. ≥ 0. It follows that sin x − x(π−x) 2  3 9.9. Consider the function f (x)  x −tan x + tan3 x on 0, π2 . We have that f  (x)  1 − (tan x) + tan2 x · (tan x)  tan4 x ≥ 0, and therefore, f (x) ≥ f (0)  0, 3 that is, x ≥ tan x − tan3 x . x 9.10. Let us prove that for x > 0, we have arccot x > 1+x 2. x Consider the function f (x)  arccot x − 1+x 2 on (0, +∞) . 1 1−x 2 2 We have f  (x)  − 1+x 2 − (1+x 2 )2  − (1+x 2 )2 < 0, and hence the function f (x) is decreasing on (0, +∞). It follows that f (x) > 0 as lim f (x)  0.

9.11. We have

n+k  n+k−1

1 dx n+k

<

n+k  n+k−1

x→+∞

1 d x, x

k  1, 2, . . . , 2n, and hence

Proofs

133

1 1 1 + + ... + < n+1 n+2 3n

n+1

1 dx + x

n

n+2

n+1

3n 

3n

1 dx + . . . + x

1 dx x

3n−1

 1 d x  ln x 3n n  ln 3. x

n 3 cos x 9.12. Let us prove first that if 0 < x ≤ π2 , then sinx x > 1+2 , or for 0 < x < π2 , cos x tan x + 2 sin x > 3x.   Indeed, consider the function f (x)  tan x + 2 sin x − 3x on 0, π2 . Since  π f (0)  0 and the derivative of the function f (x) on 0, 2 is positive, we have  x·cos x f  (x)  cos12 x +2 cos x −3  cos12 x +cos x +cos x −3 > 3 3 coscos −3  0, 2x 1 since cos2 x  cos x.  Therefore, the function f (x) on 0, π2 is increasing, and hence f (x) > f (0).   3 The inequality sinx x < 4−cos on 0, π2 is equivalent to the inequality 4sin x − x sin x cos x < 3x In order to prove this, let us consider the function F(x)  4sin x − sin x cos x − 3x in the given domain. As for 0 < x < π2 , we have that

F  (x)  4 cos x − cos 2x − 3  4 cos x − 2 cos2 x − 2  −2(cos x − 1)2 < 0.  Then on 0, π2 , the function F (x) is decreasing. It follows that F (x) < F (0)  0. x p−1 pb

9.13. Consider the function f (x)  p−1 p

p−2

x −b qbx 2

q−1

p

· x b − bq x 2  q

q

+

bq−1 qx 

in (0, +∞) . We have f  (x)  q

, and therefore, f (x) > 0 for x > b p and f  (x) < 0

for 0 < x <  q   b p .q  Hence on 0, b p , the function f (x) is decreasing, and on b p , +∞ it is increasing. q Therefore, the function f (x) attains its smallest value for x  b p , that is, 

x p−1 pb

+

bq−1 qx

≥

q

bp

 p−1

+

pb

bq−1 q

qb p



1 p

+

1 q

 1, or

xp p

+

bq q

≥ xb. Thus for x  a

we obtain the given inequality. p p p n n  cc p−1 , a −b  9.14. Let us denote ab by c. In this case, we have c > 1 and aa p−b +b p +1 a n +bn n x c −1 . Consider the function f (x)  ccx−1 on (−∞, +∞). We have f  (x)  cn +1 +1 2c x ln c > 0, and therefore, the function f (x) is increasing on the given domain. (c x +1)2 p n > ccn−1 . Hence, if p > n, then f ( p) > f (n), that is, cc p−1 +1 +1 9.15. One can easily prove that the given inequality is equivalent to the following √ 1 2 inequality: (1 + x t ) t ≤ x+ 2x +4 . The proof follows from Problem 9.21 and 1

inequality (1 + x 2 ) 2 ≤

√ x+ x 2 +4 . 2

134

9 Using Derivatives and Integrals

9.16. Consider the function f (x)  e x + b(ln b − 1) − xb in (−∞, +∞). We have f  (x)  e x − b, and therefore, the function f (x) is increasing on [ln b, +∞) and decreasing on (−∞, ln b]. Hence for x  ln b, the function attains its smallest value, that is, f (x) ≥ f (ln b). Therefore, f (a) ≥ eln b + b(ln b − 1) − b ln b  0. 3 2 9.17. Consider the function f (x)  x x−1 − 2 ln x − ln3 x in [1, +∞). 2 2 2 − lnx x . We have f  (x)  1 + x12 − 2x − lnx x  (x−1) x2 In order to find the sign of the function f  (x), let us consider the sign of 2 the function g(x)  (x−1) − ln2 x in the given domain. We have g  (x)  x x 2 −1

x 2 −1 x2

−2ln x

− 2lnx x  x 2x . 2 Now let us consider the function F(x)  x x−1 − 2ln x on [1, +∞). 2 2 − 2x  (x−1) ≥ 0, we have x ≥ 1, F(x) ≥ F(1)  0, Since F  (x)  x x+1 2 x2  and therefore, g (x) ≥ 0, whence g(x) ≥ g(1)  0, and f  (x) ≥ 0, that is, f (x) ≥ f (1)  0.

9.18. Consider the function f (x)  2sin x + 2cos x − 3 on 0, π4 . We have f  (x)  2cos x · cos x ln 2(2sin x−cos x − tan x) ≥ 0, since 2sin x−cos x − tan x ≥ 0.

let us consider the function F(x)  sin x − cos x − log2 tan x in Indeed, 0, π4 . We obtain F  (x)  cos x + sin x − ≤

1 1  sin x cos x ln 2 sin x cos x

√ 2 1 1 ( − ) < 0, sin xcos x 2 ln 2

√  2 1 sin(x + π/4) sin 2x − 2 ln 2

and therefore, F(x) ≥ F(π/4)  0 or 2sin x−cos x − tan x ≥ 0. It follows that f  (x) ≥ 0, and hence f (x) ≥ f (0)  0. If π4 ≤ x ≤ π2 , then f ( π2 − x) ≥ 0, and using that f (x)  f ( π2 − x), we deduce that f (x) ≥ 0. 9.19. We have 2 1 + 2 + ··· + n < k

k

k

3 k

x dx + 1

n+1 x dx + · · · + xk dx k

n

2

n+1 1 (n + 1)k+1 1 (n + 1)k+1 n k+1  − <  (1 + )k+1 · . xkdx  k+1 k+1 k+1 n k+1 1

In a similar way, we obtain 1 2 n k n 1k +2k +· · ·+n k > x k d x + x k d x +· · ·+ x dx  xkdx  0

1

n−1

0

x k+1 n | k+1 0



n k+1 . k+1

Proofs

135

9.20. (a) We have 2

3

ln 2 + ln 3 + · · · + ln n >

ln x d x + · · · +

ln x d x + 1

n

2

n ln x d x 

n−1

ln x d x 1

 (x ln x − x) |n1  n ln n − n + 1 > n ln n − n,

and therefore, n! >

 n n e

.

(b) One can easily verify that 7! < 7

 7 7 e

9 ln 2 + ln 3 + · · · + ln n < ln 7! +

ln x d x + · · · + 8

. Letting n ≥ 8, we have

n+1 n+1   ln x d x  ln 7! + ln x d x n

8

< (n + 1) (ln (n + 1) − 1) − 8 (ln 8 − 1) + 8 ln 7 − 7 < (n + 1) ln n − n,  ln 7! + (x ln x − x) |n+1 8

 n or n! < n ne (see Problem 3.16(b)) 1 9.21. Consider the function f (x)  (a x + b x ) x on (0, +∞). We have a x ln a x + b x ln b x − ln(a x + b x )a f (x)  (a + b ) · x 2 (a x + b x ) 

x

x

1 x

x

+b x

.

Consider the function F(t)  t ln t + c ln c − (t + c)ln(t + c) on (0, c], where c > 0. t We have F  (t)  1 + ln t − 1 − ln(t + c)  ln t+c < ln 1  0, and therefore, F(t) ≤ F(c)  2c ln c − 2c ln 2c < 0. Let a x ≥ b x . Taking c  a x , t  b x , we obtain a x ln a x + b x ln b x − (a x + b x )ln(a x + b x ) < 0, hence f  (x) < 0, and thus f (x) is a decreasing function on (0, +∞). 1 1 Therefore, if β > α > 0, then (a α + bα ) α > (a β + bβ ) β . a x 9.22. Consider the function f (x)  (x+d) a+b on (0, +∞). x ad , then We have f  (x)  (x+d) a+b+1 · (ad − bx). Therefore, if 0 < x < b ad   f (x) > 0, for otherwise, if x > b , then f (x) < 0. Hence, the function 

 , +∞ . , and decreasing on ad f (x) is increasing on 0, ad b b Thus, it follows that at the point x  ad , function f (x) attains its greatest b value.     ca , and therefore f (c) ≤ f ad , or (c+d) We deduce that f (x) ≤ f ad a+b ≤ b b a ( adb ) , and we obtain  a+b a+b ≤  a a ·  b b . a+b c+d c d (d+ adb ) a−1

136

9 Using Derivatives and Integrals

9.23. We have 1 1 1 + ··· + √ < +√ √ 3 3 3 4 5 106 106 

4

1 dx + √ 3 x

3

5 4

1 dx + · · · + √ 3 x

106

106 −1

1 dx √ 3 x

  √ 3 1 9 3√ 3√ 3 3 12 10 − 9  14997 + 3 1 − dx  < 14997. √ 3 2 2 2 x

3

In a similar way, we obtain 1 1 1 + ··· + √ > +√ √ 3 3 3 4 5 106

5

1 dx + √ 3 x

4

6 5

1 dx + · · · + √ 3 x

10 6 +1

1 dx √ 3 x

106

10  +1 6

1 3√ 3√ 3√ 3 3 3 3 3 6 + 1)2 − 12 − (10 16 > 10 16 d x  √ 3 2 2 2 2 x 4 √ 3 4 3√ 3√ 8 − 3 3 16 3 3  10 − 16  1500 − 16  14996 + > 14996. 2 2 2 2



It follows that 14996 < √314 + · · · + √3 1 6 < 14997, and therefore, the integer 10 part of the given number is equal to 14996. 9.24. We have 1 1 1 1 + √ + √ + ··· + √ 2 3 2 4 3 (n + 1) n 1 1 1 1 1 1 1 < + √ + √ + √ + √ + √ + ··· + √ 2 3 2 4 3 5 4 5 5 6 6 n n <



1 1 1 1 + √ + √ + √ + 2 3 2 4 3 5 4 1 1 1 1 + √ + √ + + 2 3 2 4 3 10

5 4

n 4

1 √ dx + · · · + x x

n

1 √ dx x x

n−1

1 1 1 1 1 2 1 1 + √ + √ + 1 − √ < 1.6 + √ + √ < 2, √ dx  + 2 10 3 2 4 3 x x n 3 2 4 3

since √ √ 1 2 · 1.5 + 1.8 1 2 2+ 3 <  0.4. √ + √  12 12 3 2 4 3 Therefore,

1 2

+

1 √ 3 2

+ ··· +

1√ (n+1) n

< 2.

Proofs

137

9.25. (a) Consider the function f (x)  (1 + x)α − 1 − αx in (−1, +∞) . Since f  (x)  α(1+ x)α−1 −α and α > 1, it follows that for −1 < x < 0 we have f  (x) < 0. It then follows that f (x) > f (0)  0, and for x > 0 we have f  (x) > 0, whence f (x) > f (0)  0. (b) See the proof of (a). (c) Consider the following two cases. (1) max(x1 , . . . , xn ) ≥ 1. Let max(x1 , . . . , xn )  xi , j  i. Then we x have (S − x j )x j ≥ xi j ≥ 1,and therefore, (S − x1 )x1 + . . . + (S − xn xn ) > n − 1. (2) max(x1 , . . . , xn ) < 1. Then from inequality (c) in Problem 9.25, it follows that S − xk S − xk ≥ (1 + S − xk − 1)1−xk 1 + (1 − xk )(S − xk − 1) S − xk S − xk > , where k  1, . . . , n.  S − xk (S − xk ) S

(S − xk )xk 

Therefore, (S − x1 )x1 + . . . + (S − xn )xn >

S−x1 S

+ ... +

S−xn S

 n − 1.

9.26. We have − |sin x| ≤ f (x) ≤ |sin x| . If 0 ≤ x ≤ π2 , then − sin x ≤ f (x) ≤ ≤ sinx x , and therefore, − lim sinx x ≤ lim f (x) ≤ sin x, whence − sinx x ≤ f (x) x x sin x x→0 x

lim

x→0

 1 (x > 0) .

x→0

On the other hand, lim

x→0

f (x) f (x) − f (0)  lim  f  (0)  a1 + 2a2 + . . . + n · an . x→0 x x −0

It follows that −1 ≤ a1 + 2a2 + . . . + n · an ≤ 1. 9.27. If x ≥ 0, then the given inequality is equivalent to the following inequality: √x ≥ ln(1 + x). In order to prove this, let us consider the function f (x)  x+1 √x − ln(1 + x) in (−1, +∞) . 1+x √

2

x+1−1) Since f  (x)  2(√x+1(x+1) ≥ 0, we have for x ≥ 0 that f (x) ≥ f (0)  0. For −1 < x ≤ 0, the given inequality is equivalent to the following inequality: f (x) ≤ f (0)  0. x 9.28. Consider the function f (x)  lnx x in (0, +∞) . Since f  (x)  1−ln , in x2 if n + 1 > n ≥ 3, then (e, +∞) the function f (x) is decreasing, and hence √ √ < lnn n , and therefore n+1 n + 1 < n n. f (n + 1) < f (n), or ln(n+1) n+1 9.29. Let us first prove that for x ≥ 0, one has f (x)  b x − 1 − x ln b ≥ 0, since f  (x)  ln b(b x − 1). If b ≥ 1, then ln b ≥ 0 and b x − 1 ≥ 0, and therefore f  (x) ≥ 0. If 0 < b < 1, then ln b < 0 and b x − 1 ≤ 0, and therefore f  (x) ≥ 0. Hence, we obtain that f  (x) ≥ 0, that is f (x) ≥ f (0). It follows that b x ≥ 1 + x ln b, and therefore

138

9 Using Derivatives and Integrals

a1 b1x + a2 b2x + · · · + an bnx ≥ a1 (1 + x ln b1 ) + a2 (1 + x ln b2 ) + · · · + an (1 + x ln bn )  a1 + a2 + · · · + an + x ln(b1a1 · b2a2 · · · · ·bnan )  a1 + a2 + · · · + an .

9.30. The given inequality is equivalent to the following inequality: x ln x > ln a + x+1 . (x + 1) ln a+1  x+1 in a1 , +∞ . Consider the function f (x)  x ln x − ln a − (x + 1) ln a+1 As for x > a1 , we have f  (x)  ln x(a+1) > 0, and then the function f (x) in x+1   the given domain is increasing. Hence for x > a1 , we have f (x) > f a1  0. 9.31. Consider the function f (x)  f  (x)  

ln

a1x +···+anx n

x

in (−∞, 0) and (0, +∞). We have

a1x ln a1 +···+anx ln an a x +···+anx · x − ln 1 n a1x +···+anx x2

n   x 2 a1x + . . . + anx



a1x ln a1x + · · · + anx ln anx n

a x + · · · + anx a x + · · · + anx − 1 · ln 1 n n

 ≥ 0,

since the function f (t)  t ln t is convex in (0, +∞). Indeed, f  (t)  ln t + 1 and f  (t)  1t > 0. 1/α  β  α β 1/β a +···+a α a +···+a ≥ 1 n n . Therefore, if α ≥ β > 0 or 0 > α ≥ β, then 1 n n  α     α 1/α √ 1/α a +···+a If α > 0 > β, then 1 n n ≥ n a1α · · · anα  n a1 · · · an ,

 1/β  β β 1/β √ a1 +···+an β β n ≤ a · · · a  n a1 · · · an , and therefore n 1 n

a1α + · · · + anα n



1/α ≥

β

β

a1 + · · · + an n

1/β .

9.32. (a) Without loss of generality one can assume that a ≥ b. In this case, the given inequalities are equivalent to the following inequalities:     2 ab − 1 a a b < ln < − . a + 1 b b a b Consider the function f (x)  ln x − 2 · x−1 in [1, +∞) . x+1 (x−1)2  Note that f (x)  x(x+1)2 > 0 (x > 1), and therefore, the function f (x) is increasing on [1, +∞) and for x > 1, we have f (x) > f (1)  0. 2 a −1 Taking x  ab , we obtain (ab +1 ) < ln ab . In order to prove the second  b inequality, let us define ab  x and consider the function g(x)  2 ln x − x + x1 in [1, +∞)

Proofs

139 2

We have g  (x)  − (x−1) < 0 (x > 1), and therefore, the function g(x) x2 is decreasing on [1, +∞), and then for x > 1, we have g(x) < g(1)  0, a b a that is, ln b < b − a . (b) Let us take a  x + 1 and b  1. From Problem 9.32(a), it follows that 2x x x < ln(x + 1) < √x+1 . Note that √x+1 < x(x+2) . x+2 2(x+1) √ √ (c) Taking a  x, b  1 and using that ab < ln a−b , we obtain x < a−ln b x−1 ln x , and hence x−1 < √1x . ln x (d) Taking a  x + 1, b  x and using Problem 9.32(a), we obtain √ 2 1 x(x + 1) < ln 1x+1 < x+(x+1) , and therefore 2x+1 < ln x+1 < √x(x+1) . 2 x x

2x+1 √ 1 < 2x(x+1) . x(x+1) a−b a+b Using that ln a−ln b < 2 for the numbers x, y and ln a−b > 0, we a−ln b    x−y  x+y 2 see that  ln x−ln y  < 2 (x  y), or |ln x − ln y| > x+y · |x − y| ≥

Note that

(e)

|x − y| , since 0 < x, y ≤ 1, and for x  y, we obtain that the given inequality holds. (f) Let I  (0, 1], x1  y, x2  x, x3  1, and f (x)  − ln x. Then from Problem 7.12(a), it follows that x ln y − ln y ≥ y ln x − ln x > y ln y − ln x.   (g) From Problem 9.25(b), it follows that 1 +   x−ln y we obtain ln 1 + 1y < ln x−y . 9.33. (a) Let us define pk (x)  1 − x +

1 y

x−y

< 1 + x−y  xy ; hence y

x 2k x2 − ... + , k  0, 1, 2, . . . . 2! (2k)!

If x ≤ 0, then pk (x) > 0. Now let us prove that if x > 0, then pk (x) − e−x > 0. For k  0, we have p0 (x) − e−x  1 − e−x > 0. Assume that for k  n (if x > 0), we have pn (x) − e−x > 0, and let us prove that it holds for k  n + 1. (That is, the following inequality holds: pn+1 (x) − e−x > 0.) Indeed, let f (x)  pn+1 (x) − e−x . Then for x > 0 we have f  (x)  pn (x) − e−x > 0, and therefore, f  (x) > f  (0)  0, whence for x > 0 it follows that f (x) > f (0)  0. Therefore, we have obtained pn+1 (x) > e−x if x > 0, and hence for x > 0 it follows that pk (x) > 0. (b) Since for every value of x we have p(x) ≥ 0, it follows that n is an even number. On the other hand, the polynomial F(x)  P(x) + P  (x) + · · · + P (n) (x) has degree equal to n. So the polynomial F(x) has the smallest value. Let min F(x)  F(x0 ), in which case F  (x0 )  0. Thus, it follows (−∞,+∞)

that F  (x0 )  P  (x0 ) + P  (x0 ) + · · · + P (n) (x0 ) + P (n+1) (x0 )  P  (x0 ) + P  (x0 ) + · · · + P (n) (x0 )  0, and F(x0 )  P(x0 ) + P  (x0 ) + P  (x0 ) + · · · + P (n) (x0 )  P(x0 ) ≥ 0. Therefore, F(x) ≥ F(x0 ) ≥ 0, and hence F(x) ≥ 0 for all values of x.

140

9 Using Derivatives and Integrals

Taking P(x)  9.33(a)).

x 2n , (2n)!

we obtain

x 2n (2n)!

2n−1

x + (2n−1)! + · · · + 1 ≥ 0 (see Problem

9.34. Consider the function f (x)  a r − br + x r − (a − b + x)r , in [0, b] . )r −1 ) ≤ 0, since Then f  (x)  r x r −1 − r (a − b + x)r −1  r x r −1 (1 − (1 + a−b x a−b ≥ 0 and r − 1 ≥ 0. x Hence, in [0, b] the function f (x) is nonincreasing, and hence f (x) ≥ f (b)  0, that is, f (c) ≥ 0 or a r − br + cr ≥ (a − b + c)r . 9.35. Note that for 0 ≤ x ≤ 1 we have 0 ≤ x k ≤ x m ≤ 1, and then (1 − x k )m ≥ (1−x m )m , and so it is sufficient to prove that (1−x1m )m +. . .+(1−xnm )m ≥ n−1. Let n ≥ 3 and x1 ≤ . . . ≤ xn . Then x1 + 2x2 ≤ 1. Let us prove that (1 − x1m )m + (1 − x2m )m ≥ 1 + (1 − (x1 + x2 )m )m . Consider the function f (x)  (1 − x m )m − (1 − (x + x2 )m )m in [0, h] , where h  min(x2 , 1 − 2x2) (if h  0, then x1  0, and hence (1) holds).  Since f  (x)  m 2 (x + x2 − (x + x2 )m+1 )m−1 − (x − x m+1 )m−1 and x2 ≥ x2 ((x + x2 ) + x2 )m ≥ x2 ((x + x2 )m + . . . + x2m ) ≥ x2 ((x + x2 )m + . . . + x m ), we have x + x2 − (x + x2 )m+1 ≥ x − x m+1 . It follows that f  (x) ≥ 0, and hence f (x1 ) ≥ f (0). Therefore, (1− x1m )m −(1−(x1 + x2 )m )m ≥ 1−(1− x2m )m . From inequality (1) it follows that it is enough to prove that (1 − x1m )m + . . . + (1 − xnm )m ≥ n − 1, for n  2. Let 0 ≤ x1 ≤ x2 and x1 + x2 ≤ 1. Let us prove that (1 − x1m )m + (1 − x2m )m ≥ 1. We have (1−x1m )m +(1−x2m )m ≥ (1−x1m )m +(1−(1−x1 )m )m and 0 ≤ x1 ≤ 21 .

Consider the function g(x)  (1 − x m )m + (1 − (1 − x)m )m in 0, 21 .   Since g  (x)  m 2 (1 − x − (1 − x)m+1 )m−1 − (x − x m+1 )m−1 and 1 − 2x  (1−2x)(1−x +x)m ≥ (1−2x)((1−x)m +. . .+x m ), we have 1−x −(1−x)m+1 ≥ x − x m+1 . Hence g  (x) ≥ 0, and therefore, g(x1 ) ≥ g(0), that is, (1 − x1m )m + (1 − (1 − x1 )m )m ≥ 1. It follows that (1 − x1m )m + (1 − x2m )m ≥ 1. 9.36. If we set ai  bii , i  1, . . . , n, then we need to prove that p(1) ≥ 0, where

n n   bi i−1 2 bi b j i+ j−1 · x − x · x . p(x)  i+ j−1 i i, j1

i1

We have ⎛ p  (x) 

n  i, j1

⎜ ⎜ bi b j x i+ j−2 − ⎜ ⎝

n 

i1

2 ⎞ bi i

x

· xi

⎟ ⎟ ⎟ ⎠

Proofs

141



n 

2

n 

i1

bi b j x i+ j−2 −

⎜ i1 ⎜ ⎝

bi · x i −

n  i1

x

· x

i

n 

bi · x

i−1

i1

x−

n 

i1

2 bi i

· x

i

x2

i, j1

⎛ n

bi i

bi i

· xi

⎞2

 n 2

 ⎟ b i i−1 ⎟  , bi − · x ⎠ i i1

for x  0. Therefore, for x > 0, we have p  (x) ≥ 0, and hence p(x) is a nondecreasing function on [0, +∞) . It follows that p(1) ≥ p(0)  0.

Problems for Independent Study Prove the following inequalities (1–13, 16–20). 1. 2 sin x ≥ π2 + (x − π2 ) cos x, where 0 ≤ x ≤ π2 . p q 2. x p−1 > x q−1 , where p > q > 0. √ √ √ √ 3. 1 + 2 + · · · + n > 23 n n, where n ∈ N. 4. e x ≥ x e , where x > 0. 2 5. sinα α > 1 − α6 , where 0 < α < π2 . 6. 2 sin α + tan α > 3α, where 0 < α < π2 . 7. sin x + sin22x + sin33x > 0, where 0 < x < π .  b+1  a b ≥ b , where a > 0, b > 0. 8. a+1 b+1 b c −ea < e 2c−1 (b + a) + 1, where 0 ≤ a < b ≤ c. 9. eb−a 10. (a) sinα α > sinβ β , where 0 < α < β < π2 ; 11. 12. 13.

14.

15. 16. 17.

(b) tanα α < tanβ β , where 0 < α < β < π2 . 2 x < sin x < x, where 0 < x < π/2. π (sin x)−2 ≤ x −2 + 1 − π42 , where 0 < x < π2 . 3 (a) tan x > x + x3 , where 0 < x < π2 , (b) x cos x < 0, 6, where 0 < x < π2 , 3  (c) sinx x ≥ cos x, where 0 < x < π2 . Let pn and qn be the respective perimeters of regular n-gons inscribed in and circumscribed about a circle with radius 21 . Let us divide the interval ( pn , qn ) into three equal parts. To which part does the number π belong? Find real values of x such that a x ≥ x a , where x ≥ 0 and a ≥ 1. 2(b+1)(c+1) a+b a+c + a+c+2 ≥ 2bc+3b+c+2 , where a > 0, b > 0, c > 0; (a) a+b+2ab a+b c+d a+c b+d (b) a+b+2ab + c+d+2cd + a+c+2 + b+d+2 ≥ 2, where a > 0, b > 0, c > 0, d > 0. x p + x − p + 2 p ≤ (x + x −1 ) p + 2, where x > 0 and p ≥ 2.

142

9 Using Derivatives and Integrals 3(x 2 −1) , where x > 1, x 2 +4x+1  √  a−b 1 , where < 3 2 ab + a+b ln a−ln b 2 2 2 cos x sin x

18. (a) ln x > (b)

19. (cos x)

> (sin x)

a > 0, b > 0, a  b.

, where 0 < x <

π . 4

Remark According to the Bernoulli’s inequality (see Problem 9.25(a)), we have cos2 x

(cos x) sin2 x > 1 + (cos x − 1) · 20.

cos2 x . sin2 x

1 1  p p  q q a1 + . . . + an p b1 + . . . + bn q ≥ a1 b1 + . . . + an bn , where ai > 0, bi > 0, i  1, . . . , n, p > 0, q > 0 and 1p + q1  1.

p

p

Remark Use the inequality of Problem 9.13 with the condition a1 + . . . + an  1 q q and b1 + . . . + bn  1.

Chapter 10

Using Functions

Consider functions f : Rn → R and g : Rn → R. Assume that we need to prove the inequality f (x1 , x2 , . . . , xn ) ≥ g(x1 , x2 , . . . , xn ) for some values of variables x1 , x2 , . . . , xn ∈ R. Rewriting this inequality as f (x1 , x2 , . . . , xn ) − g(x1 , x2 , . . . , xn ) ≥ 0, we study the dependence on xi (1 ≤ i ≤ n) of the function F(x)  f (x1 , x2 , . . . , xn ) − g(x1 , x2 , . . . , xn ), where the variables x1 , x2 , . . . , xi−1 , xi+1 , . . . , xn are considered constants. Let us consider the following example. Example 10.1 Prove that 

 x12 + x22 + · · · + xn2 cos

π ≥ x1 x2 + x2 x3 + · · · + xn−1 xn , n +1

where n ≥ 2. Proof Let us consider the following function: F(x1 )  cos

  π π x12 − x2 x1 + x22 + · · · + xn2 cos − x2 x3 − · · · − xn−1 xn . n +1 n +1

This is a quadratic expression in x1 . Note that the quadratic expression ax 2 +bx +c b . Hence since cos n π+ 1 > 0, it for a > 0 attains its smallest value at the point x  − 2a follows that F(x1 ) attains its smallest value at the point x1  2 cosx2 π , and therefore, n+1   π x2 2 2 2 . − x1 x2 − · · · − xn−1 xn ≥ F F(x1 )  (x1 + x2 + · · · + xn ) cos n +1 2 cos n π+ 1 One can easily prove that 

x2 F 2 cos n π+ 1



 

sin 2 sin

3π n+1

2π n+1

cos

 π n+1

x22

+

x32

+ ··· +

xn2

cos

π − x2 x3 − · · · − xn−1 xn . n +1

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_10

143

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10 Using Functions

Let us consider the following quadratic function: G(x2 )  F 

G(x2 ) ≥ G

sin sin

2π n+1 3π n+1



x3





sin 2 sin

4π n+1

3π n+1

cos



π n+1

x32 + · · · + xn2 cos



x2 2 cos n π+ 1

. We obtain

π − x3 x4 − · · · − xn−1 xn . n +1

In a similar way, for variables x3 , . . . , xn , we obtain F(x1 ) ≥ G(x2 ) ≥ · · · ≥

sin (n+1)π 2 n+1 π xn 2 sin nnπ + 1 cos n + 1 · · · + xn−1 xn .

cos n π+ 1  0, and therefore, (x12 + x22 + · · · + xn2 ) cos n π+ 1 ≥ x1 x2 +

One can use the following properties of functions in proving a large number of inequalities. Property 1 If a function f (x) is defined on [a, b] and is decreasing on [a, c] and increasing on [c, b], then on [d, e], then the function f (x) attains its maximum value at one of the endpoints of [d, e] (a ≤ d < e ≤ b). Property 2 If the function f (x) is defined on [a, b] and is increasing on [a, c], and is decreasing on [c, b], then on [d, e], the function f (x) attains its smallest value at one of the endpoints [d, e] (a ≤ d < e ≤ b). One can easily note that if the function f  (x) is increasing on [a, b], then the function f (x) attains its greatest value on [a, b] at the point a or b, and if f  (x) is decreasing on [a, b], then f (x) attains its greatest value on [a, b] at the point a or b. These properties of functions can be used to prove a large number of inequalities. Example 10.2 Prove that a 2 + b2 + c2 ≤ a 2 b + b2 c + c2 a + 1, where 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, 0 ≤ c ≤ 1. Proof Consider the function f (a)  a 2 (1 − b) − c2 a + b2 + c2 − bc2 − 1 in [0, 1]. If b  1, then f (a) is a quadratic trinomial in a, and the branches of its graph are directed upward. Hence, the function f (a) attains its greatest value at oneof the endpoints in [0, 1]. Since f (0)  b2 + c2 − b2 c − 1  (1 − c) b2 − (1 + c) ≤ 0, f (1)  1 − b − c2 + b2 + c2 − b2 c − 1  b(b − 1) − b2 c ≤ 0, it follows that f (a) ≥ 0 on [0, 1]. This ends the proof. If b  1, then the proof can be done similarly. Example 10.3 Prove that x1 + x2 + x3 − x1 x2 − x2 x3 − x3 x1 ≤ 1, where 0 ≤ xi ≤ 1, i  1, 2, 3. Proof Let us consider the following monotonic function f (x)  x + x2 + x3 − x x2 − x2 x3 − x x3  x(1 − x2 − x3 ) + x2 + x3 − x3 x2 , which attains its greatest value at one of the endpoints [0, 1]. We have f (0)  x2 + x3 − x2 x3  1 + (1 − x3 )(x2 − 1) ≤ 1, f (1)  1 − x3 − x2 + x2 + x3 − x2 x3  1 − x2 x3 ≤ 1. Therefore, in [0, 1] one has f (x) ≤ 1 or f (x1 )  x1 + x2 + x3 − x1 x2 − x2 x3 − x3 x1 ≤ 1. This ends the proof.

Problems

145

Problems Prove the following inequalities (10.2–10.14).



10.1. Prove that, if ax 2 + bx + c ≤ 1 holds for all numbers

x belonging to [−1, 1], then for those x one has the following inequality: cx 2 − bx + a ≤ 2. 10.2. (a + b + c)2 < 4(ab + bc + ac), where a, b, and c are the side lengths of some triangle. 10.3. b + ac + 1 + c + ab + 1 + a + bc + 1 + (1 − a)(1 − b)(1 − c) ≤ 1, where 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, 0 ≤ c ≤ 1.   10.4. Schweitzer’s inequality: (a + b + c + d + e) a1 + b1 + 1c + d1 + 1e ≤ 25 +   2 p − qp , where 0 < p ≤ a, b, c, d, e ≤ q. 6 q  n n n 10.5. (a) Chebyshev’s inequality: m i ai ( m i bi ) ≤ m i ai bi , where a1 ≤ i1

i1

n

i1

a2 ≤ · · · ≤ an , b1 ≤ b2 ≤ · · · ≤ bn , m i  1, m i > 0(i  1, . . . , n). i1 n  n  n (b) m i ai m i bi ≥ m i ai bi , where a1 ≤ a2 ≤ · · · ≤ an , b1 ≥ i1

i1

b2 ≥ · · · ≥ bn ,

n

i1

m i  1, m i > 0(i  1, . . . , n).

i1

10.6. (1 + a1 )(1 + a2 ) · · · (1 + an ) ≥ 1 + a1 + a2 + · · · + an , where the numbers ai > −1, i  1, . . . , n, and the numbers a1 , . . . , an have the same sign. 2 · n 2 , where x1 , . . . , xn ∈ [a, b], 0 < 10.7. (x1 + · · · + xn ) x11 + · · · + x1n ≤ (a+b) 4ab a < b. − b) c)(1 − a − b + c) 10.8. (1 − a)(1 ≥ (1 − c(a , where 0 < a ≤ c ≤ b, a + b < 1. ab + b − c) 10.9. 1 +1 a + 1 +1 b ≥ 1 +1 m + 1 +1 ab , where 1 < a < m < b. m 

10.10. abc ≤ 18 ( pa + qb + r c), where a, b, c > 0, p, q, r ∈ 0; 21 , a + b + c  p + q + r  1. 10.11. ai + a j > ak , where n ≥ 3, i < j < k, 0 < a1 < . . . < an and 2    2 a1 + · · · + an2 > (n − 1) a14 + · · · + an4 . 10.12. x +y+z ≥ x y+yz+zx, where

x ≥ 0, y ≥ 0, z ≥ 0 and x y+yz+zx

+x yz ≤ 4. 10.13. b2 − 4ac ≤ B 2 − 4 AC , if ax 2 + bx + c ≤ Ax 2 + Bx + C , for all real values of x.  10.14. 1 + √23 (x 2 − x + 1)(y 2 − y + 1)(z 2 − z + 1) ≥ (x yz)2 − x yz + 1. 10.15. (a) (1 − x1 · · · xn )λ + (1 − y1λ ) · · · (1 − ynλ ) ≥ 1, where λ > 1, 0 < xi < 1, 0 < yi < 1, xi + yi  1, i  1, . . . , n. (b) (1− p n )m +(1−q m )n ≥ 1, where p+q  1, p > 0, q > 0, m, n ∈ N.

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10 Using Functions

Proofs 10.1. We have |cx 2 − bx + a|≤ |cx 2 − c|+|c − bx + a| |c|(1 − x 2 ) + |c − bx + a|≤ |c|+|c − bx + a|. 2 If x  0, then from

2|ax + bx + c|≤ 1 it follows that |c|≤ 1. We deduce that cx − bx + a ≤ 1 + |c − bx + a|. Note that the function f (x)  |c − bx + a| in [−1, 1] attains its greatest value at one of the endpoints. Therefore, f (x) ≤ max(

f (−1), f (1))  max(|c + b + a|, |c − b + a|) ≤ 1, whence cx 2 − bx + a ≤ 1 + 1  2. 10.2. Since a, b and c are the side lengths of some triangle, we have |b − c| < a < b + c. Without loss of generality, one can assume that b ≥ c. Consider the function f (a)  (a + b + c)2 − 4(ab + bc + ac) in [b − c; b + c]. We have f (a)  a 2 −2a(b + c) +b2 +c2 −2bc. The graph of this function is a parabola with branches pointing upward. Consequently, in the given domain, f (a) attains its largest value at one of the points b−c, b+c, that is, f (b − c)  4c(c − b) ≤ 0, f (b + c)  −4bc < 0. Hence in (b − c, b + c) we have f (a) < 0. 10.3. Consider the function f (a)  b + ac + 1 + c + ab + 1 + a + bc + 1 +(1 − a)(1 − b)(1 − c) in [0, 1]. Obviously, the derivative is increasing: f  (a) 

b c 1 − − − (1 − b)(1 − c) b + c + 1 (c + a + 1)2 (a + b + 1)2

in [0, 1], and therefore, it satisfies Property 1 above, that is, f (a) attains its greatest value at one of the points 0, 1. Substituting a by 0 or 1 and considering the obtained function as a function depending on b, we obtain that it also attains its greatest value at one of the points b  0, b  1. Since the given inequality is symmetric with respect to a, b, and c, we obtain that the expression in the left-hand side attains its greatest value at one of the following triples: (0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1). Note that in all these cases, the left-hand side is equal to 1. Alternative proof According to Problem 1.26, we have a b c b+c + + + (1 − a)(1 − b)(1 − c) ≤ a − a + 1+b+c 1+a+c 1+a+b 2 a + c abc a + b abc abc +b−b + +c−c + + (1 − a)(1 − b)(1 − c)  1. + 3 2 3 2 3

  10.4. Consider the function f (a)  (a + b + c + d + e) a1 + b1 + 1c + d1 + 1e , or f (a)  Aa + Ba + AB + 1, where A  b1 + 1c + d1 + 1e , B  b + c + d + e. Since f  (a)  A − aB2 (B > 0), the function f (a) is increasing in [ p, q], and therefore, it attains its greatest value at one of the points p, q. It follows that f (a) ≤ max( f ( p), f (q)).

Proofs

147

 Consider the function g(b)  ( p + b + c + d + e) 1p + b1 + 1c + d1 + 1e or  h(b)  (q + b + c + d + e) q1 + b1 + 1c + d1 + 1e in [ p, q]. In a similar way we obtain that g(b) ≤ max(g( p), g(q)) and h(b) ≤ max(h( p), h(q)). Continuing these arguments with respect to variables c, d, e, we get that the expression attains its greatest value when some of the variables are equal to p and the others are equal to q. Therefore,    (a + b + c + d + e) a1 + b1 + 1c + d1 + 1e ≤ (mp + nq) m 1p + n q1 , where m ≥ 0, n ≥ 0, m + n  5, m, n ∈ Z.    2 p q Now let us prove that (mp + nq) mp + qn ≤ 25 + 6 − . q p   2 p q mnp mnq , + ≤ 25 − q p q p     p q p q (m + n)2 + mn + − 2 ≤ 25 + 6 + −2 , q p q p     p q p q + − 2 ≤ 25 + 6 + −2 , 25 + mn q p q p m 2 + n2 +

since one can easily note that mn ≤ 6. Remark Problem 10.4 can be generalized in the following way. Prove that if 0 < p < a1 , a2 , . . . , an < q, then    2 p q − (see Problem (a1 + · · · + an ) a11 + · · · + a1n ≤ n 2 + ψ(n) q p 10.7), where ⎧ 2 ⎨ n , n  2k(k  0, 1, 2, . . .), 4 ψ(n)  ⎩ n 2 −1 , n  2k + 1(k  0, 1, 2, . . .) 4 10.5. (a) Consider the function ⎛

⎞

⎜ ⎟ ⎜ ⎟ n  n n ⎜ ⎟    ⎜ ⎟ f (x)  ⎜m 2 + x m i bi − m i ai ⎟ m i ai bi − m 2 xb2 ⎜ ⎟ i1 ⎜ ⎟ i 1 i 1 ⎝ ⎠ i  2 i  2 in [a1 , a3 ]. Since f (x) is a linear function, it attains its greatest value at one of the points a1 , a3 . In the expression M  f (a2 ), substituting a2 by a1 or a3 and in the obtained expression substituting a3 by x, in a similar

148

10 Using Functions

way, one can prove that this expression attains its greatest value at one of the points a1 , a4 . Continuing these assumptions for the numbers a4 , . . . , an−1 , b2 , . . . , bn−1 , we obtain M ≤ (αa1 + βan )(γ b1 + δbn ) − ka1 b1 − (α − k)a1 bn − (δ − α + k)bn an − (β + α − δ − k)an b1 , where α is the sum of the m i whose corresponding ai were substituted by a1 , and γ is the sum of the m i whose corresponding bi were substituted by b1 and α + β  1, γ + δ  1, k  min(α, γ ). We have M ≤ αγ a1 b1 + (αδ − α)a1 bn + (βγ − β − α + δ)an b1 + (βδ − δ + α)an bn −

   − ka1 b1 + ka1 bn + kan b1 − kan bn  αγ a1 b1 − αγ a1 bn − αγ an b1 + αγ an bn − k a1 − an b1 − bn      (αγ − k) a1 − an b1 − bn ≤ 0,

as k  α or k  γ and 0 ≤ α ≤ 1, 0 ≤ γ ≤ 1. (b) Let us rewrite (a) for numbers a1 ≤ a2 ≤ · · · ≤ an and −b1 ≤ −b2 ≤ · · · ≤ −bn .    n n n m i ai m i (−bi ) ≤ m i ai (−bi ), and on mulIt follows that i1

i1

i1

tiplying both sides by −1, we obtain the given inequality. 10.6. If a1 > 0, . . . , an > 0, then in this case it is sufficient to expand the left-hand side of the inequality. If −2 ≤ a1 < 0, . . . , −2 ≤ an < 0, then let us consider the function f (x)  (1 + x)(1 + a2 ) · · · (1 + an ) − 1 − x − a2 − · · · − an in [−2, 0]. Since it is a linear function, it attains its smallest value at one of the points −2, 0. One can easily prove that the expression (1 + a1 ) · · · (1 + an ) − 1 − a1 − · · · − an attains its smallest value if some of the numbers are equal to −2 and the others are equal to 0. Hence, if a1  · · ·  an  0, then we have (1 + a1 ) · · · (1 + an ) − 1 − a1 − · · · − an  1 − 1  0. If some of the numbers are equal to −2, then (1 + a1 ) · · · (1 + an ) − 1 − a1 − · · · − an  (−1)k − 1 + 2k ≥ 0, where k is thenumber of −2’s. 10.7. Consider the function f (x)  (x + x2 + · · · + xn ) x1 + x12 + · · · + x1n in [a, b]. Note that one can rewrite this function in the following way: f (x)  1 + Ax + Bx + AB, where A  x2 + · · · + xn , B  x12 + · · · + x1n . Since f  (x)  B − xA2 (B > 0), it follows that f  (x) is increasing in [a, b], and therefore it attains its greatest value at one of the points a, b. It follows that f (x1 ) ≤ max( f (a), f (b)). In a similar way, let us consider the  following function: g(x)  (b + x + x3 + · · · + xn ) b1 + x1 + x13 + · · · + x1n , or g(x)  (a +  x + x3 + · · · + xn ) a1 + x1 + x13 + · · · + x1n in [a, b]. We obtain g(x2 ) ≤ max(g(a), g(b)).

Proofs

149

Repeating thesame for the variables x3 , . . . , xn , we obtain that the expression 1 1 (x1 + · · · +xn ) x1 + · · · + xn attains its greatest value if some of the variables equal a and the others equal b. Let us take x1  · · ·  xk  a, xk+1  · · ·  xn b. + b)2 · n2. It is sufficient to prove that (ka + (n − k)b) ak + n −b k ≤ (a4ab 2 This inequality holds because ((n − 2k)a − (n − 2k)b) ≥ 0. x)(1 − a − b + x) in [a, b]. 10.8. Consider the function f (x)  (1 − x(a  1+ b − x)  1  One can easily prove that f (x)  a + b − 1 x + a + b1 − x + 1.  1  + b) − 1 , and hence the function f (x) is We have f  (x)  (2xx−2 (a(a++bb))(a a+b − x)2

a+b

a+b  decreasing on a, 2 and increasing in 2 , b . It follows that the function f (x) attains its greatest value on [a, b] at one of the points a, b. − b) − b) , we have f (x) ≤ (1 − a)(1 , where by Since f (a)  f (b)  (1 − a)(1 ab ab taking x = c, we obtain the given inequality. in [a, b]. 10.9. Consider the function f (x)  1 +1 x + 1 +1 ab  1 +1 x + 1 − x ab + ab √

√

x

(ab − 1)(x − ab)(x + ab) Since f  (x)  , we see that the function f (x) is 2 2 √   √ (x+ 1) (x + ab) ab, b . Therefore, f (x) ≤ decreasing on a, ab and increasing on max( f (a), f (b)). Since f (a)  f (b)  1 +1 a + 1 +1 b and a ≤ m ≤ b, we have f (m) ≤ 1 +1 a + 1 +1 b . 10.10. Note that 21 −q ≤ p ≤ 21 , and so for the function f ( p)  pa+qb+(1− p−q)c, we have

          1 1 1 1 1 1 −q , f  min − q a + qb + c, a + qb + −q c f ( p) ≥ min f 2 2 2 2 2 2   1 1 1 1 1 1 ≥ min a + c, b + c, a + b . 2 2 2 2 2 2

  Let min 21 a + 21 c, 21 b + 21 c, 21 a + 21 b  21 (a + c). Therefore, pa + qb + r c ≥

1 (a + c) ≥ (a + c)2(a + c)(1 − (a + c))  2(a + c)2 b ≥ 8abc. 2

10.11. Letting m ∈ {1, . . . , n}, we see that the given inequality is equivalent to the following inequality:    4 − 2a 2 a 2 + · · · + a 2 − a 2 + (n − 1) a 4 + · · · + a 4 − a 4 − a 2 + · · · + a 2 − a 2 2 < 0, (n − 2)am m 1 n m n m n m 1 1

which means that the quadratic function    2 + (n − 1) a 4 + · · · + a 4 − a 4 − a 2 + · · · + a 2 − a 2 2 (n − 2)x 2 − 2x a12 + · · · + an2 − am n m n m 1 1

for x  am2 attains a negative value, whence D > 0, and so we have  2 2   a1 + · · · + an2 − am2 > (n − 2) a14 + · · · + an4 − am4 .

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10 Using Functions

2   Repeating the same process, we obtain ai2 + a 2j + ak2 > 2 ai4 + a 4j + ak4 , or (ai + a j + ak )(ai + a j − ak )(ai − a j + ak )(−ai + a j + ak ) > 0, and therefore, ai + a j > ak . √ √ √ 10.12. If max(x y, yz, zx) ≤ 1, then x + y + z ≥ x y + yz + zx ≥ x y + yz + zx, and therefore, x + y + z ≥ x y + yz + zx. If x y > 1, let us consider the following two cases. (a) If x + y ≥ 4, then x + y + z ≥ 4 ≥ x y + yz + zx. (b) If x + y < 4, then z ≤ x 4+−y +x yx y , and one needs to prove that z ≤

x + y − xy . x + y −1

Prove that x 4+−y +x yx y ≤ xx++yy−−x1y , or f (b)  b2 − (a − 1)b + 4a − 4 − a 2 ≤ 0, where a  x + y, b  x y.   2 2 Since 1 ≤ b ≤ a4 , we have f (b) ≤ max f (1), f a4 . √ Note that f (1)  (a − 1)(2 − a) ≤ 0, since a ≥ 2 x y ≥ 2, and  a2 4



(a − 2)2 (a + 4)(a − 4) 16

≤ 0, and therefore f (b) ≤ 0.





10.13. Note that if x  0, then a + bx + xc2 ≤ A + Bx + xC2 , and hence letting x → +∞, we obtain |a| ≤ |A|. (1) Consider the following cases.



b + c ≤ (a) If A  0, then from (1) it follows that a  0, and therefore, x







B + C , whence |b| ≤ |B|. It follows that b2 − 4ac  b2 ≤ B 2 

2 x

B − 4 AC . (b) If A  0, let B 2 −4 AC > 0,

roots of the polynomial

and let x 1 , x 2 be the

Ax 2 + Bx + C. Then from ax 2 + bx + c ≤ Ax 2 + Bx + C it follows that axi2 + bxi + c  0, i  1, 2, and hence a  b  c  0 or



2 2 (x2 − x1 )2  b −a 24ac  B −A24 AC . Therefore, b2 − 4ac ≤ B 2 − 4 AC . (c) If A  0, let B 2 − 4 AC ≤ 0,

and without loss of generality one can assume that A > 0. Therefore, ax 2 + bx + c ≤ Ax 2 +Bx+C, and hence, ( A − a)x 2 + (B − b)x + C − c ≥ 0 and ( A + a)x 2 + (B + b)x + C + c ≥ 0, for all real values of x. f

We have (B − b)2 − 4( A − a)(C − c) ≤ 0 and (B + b)2 − 4( A + a)(C + c) ≤ 0, and hence(B − b)2 − 4( A − a)(C − c) + (B + b)2 − 4( A + a)(C + c) ≤ 0.



Therefore, b2 − 4ac ≤ 4 AC − B 2  B 2 − 4 AC . In order to complete the proof, it is left to consider the case b2 − 4ac < 0. 2 + bx + c ≤ Without loss of generality one can assume that a > 0, and thus ax

2 2 4ac − b 4 AC − B 2 2

≤ 4 A , and hence b − 4ac  Ax + Bx + C, and it follows 4a

that 2 2 2



4ac − b ≤ 4 AC − B  B − 4 AC . 

10.14. If x yz  0, then 1 +

√2 3

(x 2 −x +1)(y 2 −y+1)(z 2 −z+1) ≥ 1 +

√2 3

· 43 · 43 >

1  (x yz)2 − x yz + 1. If x yz  0, then without loss of generality one can assume that yz > 0.

Proofs

151

Let k  1 + √23 , A  (y 2 − y + 1)(z 2 − z + 1), B  yz. Then one needs to prove that (k A − B 2 )x 2 − (k A − B)x + k A − 1 ≥ 0.

(1)

Note that k A ≥ k · 43 y 2 · 43 z 2 > y 2 z 2  B 2 , and then it suffices to prove that D  (k A − B)2 − 4(k A − B 2 )(k A − 1) ≤ 0, or 3k 2 A2 − 2k(2B 2 − B + 2)A + 3B 2 ≥ 0. Let us prove that  2B 2 − B + 2 + 2 (B − 1)2 (B 2 + B + 1) . (2) A≥ 3k Note that A  (y + z)2 − (B + 1)(y + z) + B 2 − B + 1. Let t 

√

B+

√1 . B

(3)

Consider the following two cases.

(a) If t > 4, then from (3), it follows that A ≥ 43 (B−1)2 , and let us prove that    9k (B − 1)2 ≥ 2B 2 − B + 2 + 2 (B − 1)2 (B 2 + B + 1), 9k4 B + B1 − 2 ≥ 4         2 B + B1 − 1 + 2 B + B1 − 2 B + B1 + 1 , or 9k4 t 2 − 4 ≥ 2(t 2 −  2) − 1 + 2  (t 2 − 4)(t 2 − 1). Then the obtained inequality holds because 9k 2 t − 4 − (4t 2 − 10) > 9k4 − 4 16 − 9k + 10 > 0, and therefore, 4  2 2 2 2 2(t 2 − 2) − 1 + 2 (t 2 −  4)(t − 1) ≤ 2(t − 2) − 1 + (t − 4) + (t − 1)  9k 2 2 4t − 10 < 4 t − 4 . √ (b) If t ≤ 4, we have y + z ≥ 2 yz ≥ B 2+ 1 , and then from (3)  √ 2 √ it follows that A ≥ 2 B − 2(B + 1) B + B 2 − B + 1, and  √ let us prove that 3k B 2 + 3B + 1 − 2(B + 1) B ≥ 2B 2 − B + 2 +   2 (B − 1)2 (B 2 + B + 1), 3k(t − 1)2 ≥ 2t 2 − 5 + 2 (t 2 − 4)(t 2 − 1),   √ √ √ √ 2  9 + 4 3 t 2 − 18 + 6 3 t + 12 + 2 3 ≥ 0. or t − 1 − 3 This inequality holds because   √  √ √ 2 D  18 + 6 3 − 4 9 + 4 3 12 + 2 3 < 0. 10.15. (a) We proceed by induction. For n  1 we have (1 − x1 )λ + 1 − y1λ  1. Let n ≥ 2 and suppose that the inequality holds for n − 1 numbers. Consider the function f (x)  (1 − (1 − x) · x2 · · · xn )λ + (1 − x λ ) · (1 − y2λ ) · · · (1 − ynλ ) on [0, 1]. Since f  (x)  λx2 · · · xn (1 − (1 − x) · x2 · · · xn )λ−1 − λx λ−1 · (1 − y2λ ) · · · (1 − ynλ ), we have 0 ≤ x < x0 , where x0 

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10 Using Functions 

1 − x2 ··· xn  1

(1 − y2λ ) ··· (1 − ynλ ) x2 ··· xn

λ−1

, and then we obtain f  (x) > 0, and if x0 < −x2 ··· xn

x ≤ 1, then we obtain f  (x) < 0. (Note that (1 − y2λ ) · · · (1 − ynλ ) > (1 − y2 ) · · · (1 − yn )  x2 · · · xn , whence 0 < x0 < 1.) It follows that f (y1 ) ≥ min( f (0); f (1)). Since f (0)  (1 − x2 · · · xn )λ + (1 − y2λ ) · · · (1 − ynλ ) ≥ 1 and f (1)  1, we have f (y1 ) ≥ 1. Hence, the given inequality holds for every positive integer n. (b) If m  1, then (1 − p n )m + (1 − q m )n  1. If m > 1, then in Problem 10.15(a), taking λ  m, x1  · · ·  xn  p, we obtain (1 − p n )m + (1 − q m )n ≥ 1. This ends the proof.

Problems for Independent Study Prove the following inequalities (2–12, 14–20). 1. Given that 0 ≤ p ≤ 1, 0 ≤ r ≤ 1, and the identity ( px + (1 − p)y)(r x + (1 − r )y)  ax 2 + bx y + cy 2 , prove that one of the numbers a, b, c is not less than 4/9.

 2. x1 + · · · + xn − x1 x2 − x2 x3 − · · · − xn−1 xn − xn x1 ≤ n2 , where 0 ≤ xi ≤ 1 (i  1, . . . , n) and n ≥ 3, n ∈ N. +  x2 + · ·· + xn + 1)2 ≥ 4(x12 + · · · + xn2 ), where x1 , . . . , xn ∈ [0, 1]. 3. (x  1n n 2  √ √ 2 ai bi2 AB ab i1 i1 ab + AB n 2 4. ≤ , where 0 < a ≤ ai ≤ A, 0 < b ≤ bi ≤ B, i  2 ai bi

i1

1, . . . , n. 5. a 2 b2 + b2 c2 + c2 a 2 > 21 (a 4 + b4 + c4 ), where a, b, c are the side lengths of some triangle. 6. (a) x 2 + y 2 + z 2 + x y + yz + zx + x + y + z + 38 ≥ 0, (b) x y(1 − z) + yz(1 − x) + x z(1 − y) ≤ 1, where 0 ≤ x, y, z ≤ 1. 7. 2n ≥ (1 + a1 ) · · · (1 + an ) + (1 − a1 ) · · · (1 − an ) ≥ 2, where 0 ≤ ai ≤ 1, i  1, . . . , n. 8. 2a+b+ 2b+c + 2c+a < 1 + 2a+b+c+1, where a > 0, b > 0, c > 0. 2 2 2 9. min (a − b)2 , (b − c)2 , (a − c)2 ≤ a + b2 + c . 10. a 4 + ab + b2 ≥ 4a 2 b + a 3 b, where a, b ≥ 0. 11. a B + bC + c A < k 2 , where a > 0, b > 0, c > 0, A > 0, B > 0, C > 0 and a + A  b + B  c + C  k. 12. bca+ 1 + cab+ 1 + abc+ 1 ≤ 2, where a, b, c ∈ [0, 1]. 13. Prove that the perimeter of the quadrilateral section of a regular tetrahedron with edge length a is smaller than 3 a.

Problems for Independent Study n

14.

n ai ai xi x i1 i1 i  n 2 ai

≤

1 4



M m

+

153

 m 2 M

, where a1 , . . . , an > 0 and 0 < m ≤ xi ≤ M,

i1

i  1, . . . , n. 15. 8ac + 8bd ≤ 3a 2 + 3b2 + 3c2 + 3d 2 + ab + bc + cd + da, where 0 ≤ a ≤ b ≤ c ≤ d. 2 +b2 −c2 16. (a) a a+b−c ≤ a + b − 2, where 2 ≤ a, b, c ≤ 3.

17. 18. 19.

20.

a12 + a22 − a32 a1 + a2 − a3

a2 + a2 − a2

a2 + a2 − a2

+ a22 + a33 − a44 + · · · + ann + a11 − a22 ≤ 2a1 + 2a2 + · · · + 2an − 2n, where n ≥ 3, 2 ≤ ai ≤ 3, i  1, . . . , n. a(c − d) + 2d 1 < ≤ 2, where 1 ≤ a ≤ 2, 1 ≤ b ≤ 2, 1 ≤ c ≤ 2, 1 ≤ d ≤ 2. 2 b(d − c) + 2c αγ − β 2 ≤ 0, where aγ − 2bβ + cα  0 and ac − b2 > 0. (a1 + · · · + an − b1 − · · · − bn )2 + 2|a1 b2 −

a2 b1 | + · · · + 2|a1 bn − an b1 | + · · · + 2|an−1 bn − an bn−1 | ≤ a12 − b12 + · · · + an2 − bn2 + 2|a1 a2 − b1 b2 | + · · · + 2|a1 an − b1 bn | + · · · + 2|an−1 an − bn−1 bn |, where n ≥ 2 and ai > 0, bi > 0, i  1, . . . , n. a1 + · · · + 1 + a1 + ···an+ an − an + (1 − a1 ) · · · (1 − an ) ≤ 1, where n ≥ 2, 1 + a1 + ··· + an − a1 0 ≤ ai ≤ 1, i  1, . . . , n. (b)

Chapter 11

Jensen’s Inequality

Historical origins. Jensen’s inequality is named after the Danish mathematician Johan Ludwig William Valdemar Jensen, born 8 May 1859 in Nakskov, Denmark, died 5 March 1925 in Copenhagen, Denmark. This inequality was proved in a paper that Jensen published in 1906. He never received an academic degree in mathematics. In 1876 he entered Copenhagen College of Technology and studied mathematics among various other subjects. Mathematics was his favorite subject, and he published his first paper in mathematics when he was still a student. Jensen studied advanced topics of mathematics later by himself. He never held an academic position. In 1881 he began work as an engineer for a Copenhagen division of the international Bell company (from 1882 Copenhagen telephone company) and did mathematics in his spare time. From 1890 to 1924 he was head of the technical department of the Copenhagen telephone company, and from 1892 to 1903 he was president of the Danish Mathematical Society. Theorem 11.1 (Jensen’s inequality) Suppose the following inequalities hold for all a and b in D( f )  I :   a+b f (a) + f (b) ≥ f , (11.1) 2 2    f (a) + f (b) a+b ≤ f . (11.2) 2 2 Then for all α1 , . . . , αn ∈ Q + 1 , where α1 + · · · + αn  1, and all numbers x1 , . . . , xn ∈ I , one has the following inequalities:

1Q

+

α1 f (x1 ) + · · · + αn f (xn ) ≥ f (α1 x1 + · · · + αn xn ),

(11.3)

[α1 f (x1 ) + · · · + αn f (xn ) ≤ f (α1 x1 + · · · + αn xn )].

(11.4)

is the set of the positive rational numbers

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_11

155

156

11 Jensen’s inequality

Proof See Problem 7.8. Note that if on I the function f (x) satisfies (11.1), then f (x) is said to be Jensen convex or midconvex on I , and if f (x) satisfies (11.2), then f (x) is said to be Jensen concave or midconcave on I .    xn n Example 11.1 Prove that cos x1 +···+cos ≤ cos x1 +···+x , where x1 , . . . , xn ∈ 0, π2 . n n  Proof Prove that for all numbers a, b ∈ 0, π2 inequality (11.2) holds, that is,   cos a+cos b b . Indeed, cos a+cos ≤ cos a+b  cos a+b cos a−b ≤ cos a+b . 2 2 2 2 2 2 1 Assume that α1  · · ·  αn  n . From (11.4), it follows that x1 + · · · + xn cos x1 + · · · + cos xn ≤ cos . n n This ends the proof. Theorem 11.2 If for all numbers a and b from D( f )  I and for all α, β such that α ≥ 0, β ≥ 0, α + β  1, one has α f (a) + β f (b) ≥ f (αa + βb),

(11.5)

[α f (a) + β f (b) ≤ f (αa + βb)],

(11.6)

then for all α1 , . . . , αn , x1 , . . . , xn , one has α1 f (x1 ) + · · · + αn f (xn ) ≥ f (α1 x1 + · · · + αn xn ),

(11.7)

[α1 f (x1 ) + · · · + αn f (xn ) ≤ f (α1 x1 + · · · + αn xn )],

(11.8)

where α1 ≥ 0, . . . , αn ≥ 0, α1 + · · · + αn  1, and x1 , . . . , xn ∈ I . Proof See Problem 7.9. Note that if on I the function f (x) satisfies (11.5), then f (x) is said to be convex in I , and if f (x) satisfies (11.6), then f (x) is said to be concave in I . Theorem 11.3 If for a function f (x) in D( f )  I one has f  (x) ≥ 0,  or f  (x) ≤ 0 , then (11.7) or (11.8) holds. Proof Assume that (11.9) holds. Let us first prove that (11.5) holds.

(11.9) (11.10)

11 Jensen’s inequality

157

Indeed, from the finite increments formula, it follows that α f (x1 ) + β f (x2 ) − f (αx1 + βx2 )  α( f (x1 ) − f (αx1 + βx2 )) + β( f (x2 ) − f (αx1 + βx2 ))  α f  (c1 )(x1 − αx1 − βx2 ) + β f  (c2 )(x2 − αx1 − βx2 )    αβ f  (c2 ) − f  (c1 ) (x2 − x1 )  αβ f  (c)(c2 − c1 )(x2 − x1 ),

where x1 < c1 < αx1 + βx2 < c2 < x2 and c1 < c < c2 . Hence, the sign of the left-hand side is the same as the sign of f  (c), and thus (11.5) holds. According to Theorem 11.2, we have that (11.7) holds. This ends the proof. n  n k−1  n

k

k−1 k k ai ci ≥ ai ci , where ai > Example 11.2 Prove that i1

i1

i1

0, ci > 0 i  1, . . . , n, and k ∈ / (0, 1).

Proof Consider the function f (x)  x k in (0, +∞). Since f  (x)  k(k − 1)x k−2 ≥ 0, then taking αi 

⎛

i1

cik · n

k cj j1



ai ci

k

xi 

 k ai ci

, where

j1

i  1, . . . , n, we obtain by Theorem 11.3 that n

cik , n

ckj

⎞k

n ⎜ cik ai ⎟ ⎜ ⎟ ≥⎜ · ⎟ . n

k ci ⎠ ⎝ i1 cj j1

This ends the proof.

Problems Prove the following inequalities (11.1–11.21) 11.1. 11.2. 11.3.

sin x1 +···+sin xn n ≤ sin x1 +···+x , where x1 , . . . , xn ∈ [0, π ]. n n 2 a+b+c +b2 +c2 a/(a+b+c) b/(a+b+c) ≤a ·b · cc/(a+b+c) ≤ a a+b+c , where a, b, c ∈ N.     3 b−c a  b a−b c · 1 + c−a · 1 + ≤ 1, where a, b, c are the side lengths 1+ a b c

of some triangle and a, b, c ∈ Q. 11.4. a 2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0, where a, b, c are the side lengths of some triangle. 11.5. a 3 b + b3 c + c3 a ≥ a 2 bc + b2 ca + c2 ab, where a > 0, b > 0, c > 0.     11.6. 1 + x1 1 + 1y 1 + 1z ≥ 64, where x > 0, y > 0, z > 0, and x+y+z  1. 11.7.

≤ √a+1 √b + √a+1 √c + √a+1√d + where a > 0, b > 0, c > 0, d > 0. 12 √ √ √ √ a+ b+ c+ d

√ 1√ b+ c

+

√ 1√ b+ d

+

1 √ √ , c+ d

158

11.8.

11 Jensen’s inequality mp p

+

nq q

≥ mn, where m > 0, n > 0, p > 0, q > 0, and

1 p

+

1 q

 1.

 √ √ √ a(a + c − b) + b(a + b − c) + c(b + c − a) ≤ (a 2 + b2 + c2 )(a + b + c), 11.9.

where a, b, c are the side lengths of some triangle.  n   n 2 n

2

2 11.10. ai ci ≥ ai ci . i1 i1   k i1  k−1 n n n



k−1 k k 11.11. ai ci ≤ ai ci , where a1 , a2 , . . . , an , c1 , . . . , cn > i1

i1

i1

0, and 0 < k < 1. n 1  n  q1 n

p p

q ai bi ≥ ai bi , where ai > 0, bi > 0, i  11.12. (a) i1

i1

1, . . . , n, (b)

n

i1

11.13.



p

ai p−1 bi

≥



n

p

ai  p−1 , bi

i1 n

i1 1 + q1 p

 1;

where ai > 0, bi > 0, i  1, . . . , n, and p > 1.

i1

an 2 + a1 −a2a+···+a + · · · + a1 +a2 +···+a ≥ n n−1 −an are the side lengths⎛of some n-gon. ⎞ a1 −a1 +a2 +···+an n

11.14.

p > 0, q > 0, and

i1

dim

n

n

·

i1

bim

n

n

≥ ⎝

n , n−2

where a1 , a2 , . . . , an

m

di bi

i1

n

⎠ , where d1 > 0, . . . , dn > 0, b1 >

0, . . . , bn > 0, and m ≥ 2. 1 2 3 + x3x+x + x1x+x ≥ 23 , where x1 > 0, x2 > 0, x3 > 0; 11.15. (a) x2x+x 3 1 3 xn x1 x2 n−1 (b) x2 +xn + x3 +x1 + · · · + xnx+x + x1 +x ≥ 2, where n ≥ 4, x1 > n−2 n−1 > 0. 0, · · · , x n      z , where x, y, z are the side 11.16.  xy + yz + xz − xy − yz − xz  ≤ 2 1 − xxy+yz+x 2 +y 2 +z 2 lengths of some triangle. a b c d + a+c+d + a+b+d + a+b+c ≥ 43 , where a > 0, b > 0, c > 0, d > 0; 11.17. (a) b+c+d an a1 a2 n ≥ n−1 , (b) a2 +a3 +···+an + a1 +a3 +···+an + · · · + a1 +a2 +···+a n−1 where n ≥ 2, a1 > 0, . . . , an > 0. a b c d + c+d + a+d + a+b ≥ 2, where a > 0, b > 0, c > 0, d > 0. 11.18. b+c 2 2 2 a +b b2 +c2 +a 2 a3 b3 c3 11.19. a + b + c ≤ 2c + 2a + c 2b ≤ bc + ca + ab , where a > 0, b > 0, c > 0. a3 b3 c3 a+b+c 11.20. a 2 +ab+b2 + b2 +bc+c2 + c2 +ca+a 2 ≥ 3 , where a > 0, b > 0, c > 0.  a+b  a a  b b 11.21. a+b ≤ c · d , where a > 0, b > 0, c > 0, d > 0. c+d 11.22. Let (xn ) be a sequence with positive terms such that 1  x0 ≥ x1 ≥ x2 ≥ · · · ≥ xn ≥ · · ·. Prove that there exists n such that for every such sequence (xn ), one has x2 x02 x12 + + · · · + n−1 ≥ 3.999. x1 x2 xn 11.23. Perpendiculars M A1 , M B1 , MC1 are drawn from the inner point M of the triangle ABC to the sides BC, C A, AB, respectively. For which point M AB the of the triangle ABC is the value of the expression MBCA1 + MC BA1 + MC 1 smallest?

Problems

159

11.24. Let a1 , . . . , an , b1 , . . . , bn > 0 and J ⊆ {1, . . . , n}. Prove that if for every J one has m 2  n  n



2 m+1

bi ≥ ai , where m ∈ N, then ai bi ≥ ai . i∈J

i1

i∈J

i1

11.25. Let a1 > 0, . . . , an > 0, b1 > 0, . . . , bn > 0. Prove (a) Radon’s inequality: p

a1

p−1 b1

p

+ ··· +

where p( p − 1) ≥ 0, p p a1 n + · · · + ap−1 ≤ (b) p−1 b1

an

p−1 bn

p

p−1

b1

(a1 + · · · + an ) p , (b1 + · · · + bn ) p−1

≤

(a1 + · · · + an ) p , (b1 + · · · + bn ) p−1

(a1 +···+an ) p , (b1 +···+bn ) p−1

bn

a1

≥

p

+ ··· +

an

p−1

bn

where < 0, p  p   p( pp − 1) p+1 p+1 a1 p+1 n > a1 · ab11 · b p + · · · + abn p + . . . + an · ab11 +···+a ≥ (c) p +···+bn n   p+1   p+11 p 1 (a1 +···+an ) p+1 n b1 · ab11 + p+1 + · · · + bn ab11 +···+a . (b1 +···+bn ) p , p+1 +···+bn where p > 0. 11.26. (a) Given the interval G and a function f (x) defined on the interval I, suppose that for x1 and x2 from the interval I, where x1 + x2 ∈ G, one   all2numbers f (x1 )+ f (x2 ) ≤ . has f x1 +x 2 2 Prove that if the x1 , . . . , xn are numbers from the interval I such that x1 + x1 , . . . , xn + xn ∈ G, where the numbers x1 , . . . , xn are the numbers   f (xn ) n . ≤ f (x1 )+···+ x1 , . . . , xn written in some other order, then f x1 +···+x n n Prove that an a1 + · · · + (n−1)a ≤ 1, where n ≥ 2, ai > 0, i  1, . . . , n, (b) (n−1)a 1 +a2 n +a 1 1 1 n , where n ≥ 2, xi > 0, i  1, . . . , n, (c) √1+x +· · ·+ √1+x ≥ min 1, √1+λ n 1 n and x1 . . . xn  λ (λ > 0), 1−

x1 +...+xn

n

n) n (d) (1−xx11)···(1−x ≥ , where n ≥ 2, xi > 0, i  1, . . . , n, x1 +···+xn ···xn n and xi + x j ≤ 1, i, j ∈ {1, . . . , n}. 1 1 1 27 11.27. Prove that 1+x 2 + 1+y 2 + 1+z 2 ≤ 10 , where x + y + z  1. 1 1 1 1 2−α , where α ≥ 2, 11.28. Prove that (1+a) α + (1+b)α + (1+c)α + (1+d)α ≥ 2 a > 0, b > 0, c > 0, d > 0, and abcd  1.

160

11 Jensen’s inequality

Proofs 11.1. For all a, b ∈ [0, π ] we have sin a+sin b ≤ sin a+b . 2 2 b  sin a+b cos a−b ≤ sin a+b . We have sin a+sin 2 2 2 2 1 Taking α1  · · ·  αn  n and using inequality (11.4), we deduce that sin x1 +···+sin xn n ≤ sin x1 +···+x . n n b ≤ log a+b , it follows that if 11.2. Since for all a > 0, b > 0 one has log a+log 2 2 a b c we set α1  a+b+c , α2  a+b+c , α3  a+b+c , then by inequality (11.4), we have α1 log a + α2 log b + α3 log c ≤ log(α1 a + α2 b + α3 c), thus a a/(a+b+c) · 2 +b2 +c2 . bb/(a+b+c) · cc/(a+b+c) ≤ a a+b+c Let us consider the function f (x)  x ln x in (0, +∞). Since f  (x)  x1 > 0, we have that (11.7) holds for the function f (x). Taking α1  α2  α3  13 , it follows that 13 (a ln a + b ln b + c ln c) ≥ a+b+c  1 . Therefore, a a · bb · cc ≥ a+b+c , or a a/(a+b+c) · + b + c) · ln a+b+c 3 (a 3 3 a+b+c b/(a+b+c) c/(a+b+c) ·c ≥ 3 . b , x2  1 + c−a , x3  1 + a−b , 11.3. Let f (x)  log x, x1  1 + b−c a b c a b c α1  ,α  , α3  , hence a+b+c 2 a+b+c a+b+c       1 b−c c−a a−b a log 1 + + b log 1 + + c log 1 + ≤ a+b+c a b c  a    b c−a c a−b log a+b+c 1 + b−c + 1 + + 1 + , a+b+c b a   a+b+ca−b c c a c−a b · 1 + · 1+ c ≤ 1. and therefore 1 + b−c a b 11.4. Consider the function f (x)  x 2 . One can easily prove that for all numbers m and n, one has α f (m) + β f (n) ≥ f (αm + βn), where α ≥ 0, β ≥ 0, and α + β  1. According to Theorem 11.2 inequality (11.7) holds, and taking , α2  b+c−a , α3  a+b−c , we obtain x1  c, x2  b, x3  a, α1  a+c−b a+b+c a+b+c a+b+c c2

a+c−b b+c−a a+b−c + b2 + a2 a+b+c a+b+c a+b+c 2  ac + c2 − bc + b2 + bc − ab + a 2 + ab − ac ≥ , a+b+c

whence     c2 (a + c)2 − b2 + b2 (b + c)2 − a 2    2 + a 2 (a + b)2 − c2 ≥ a 2 + b2 + c2 , or a 2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0. 11.5. Hint. Take f (x)  x 2 , x1  a + b − c, x2  a + c − b, x3  b + c − a,

Proofs

161

α1 

a b c , α2  , α3  . a+b+c a+b+c a+b+c

Alternative proof. Note that   a 3 b + b3 c + c3 a − a 2 bc + b2 ca + c2 ab  ab(a − c)2 + bc(b − a)2 + ac(b − c)2 ≥ 0.   11.6. Let us consider the function f (t)  ln 1 + 1t on (0, +∞). 1 > 0, inequality (11.7) holds. Taking Since f  (t)  t12 − (t+1) 2

α1  α2  α3  13 ,t1  x, t2  y, t3  z, we obtain        1 3 ln 1 + x1 + ln 1 + 1y + ln 1 + 1z ≥ ln 1 + x+y+z ; hence from x + y + 3 3       3 z  1, it follows that 1 + x1 1 + 1y 1 + 1z ≥ 1 + x+y+z  64.   1 1 1 2 11.7. Let us consider the function f (x)  x (0, +∞). We have 2 a + b1 ≥ a+b , √ √ √ √ and so by Theorem 11, taking x1  a + b, x2  a + c, √ √ √ √ √ √ a + d, x4  b + c, x5  b + d, √ √ 1 x6  c + d, α1  · · ·  α6  , 6

x3 

we obtain that the given inequality holds. 11.8. Let us consider the function f (x)  ln x in (0, +∞). Since f  (x)  −1 < 0, x2 inequality (11.6) holds. Taking α  1p , β  q1 , a  m p , b  n q , we obtain that the given inequality holds.√ 11.9. Let us consider the function f (x)  x in (0, +∞). One can easily prove that for the function f (x), the condition (11.10) holds. , α2  a+b−c , α3  c+b−a , x1  a, x2  b, x3  c, Taking α1  a+c−b a+b+c a+b+c a+b+c we deduce inequality (11.8) in the given form. Hint. For a  9, b  4, c  1 the inequality does not hold. 11.10. Let c12 + · · · + cn2  0. Consider the function f (x)  x 2 . Since f  (x) > 0, on taking αi 

ci2 , xi  2 c1 + · · · + cn2



ai ci

2 ,

where i  1, . . . , n, using Theorem 11.3, we obtain n

i1

ci2 · c12 + · · · + cn2



ai ci

2 ≥

 n

i1

ci2 ai · 2 2 c1 + · · · + cn ci

hence the given inequality holds. If c1  · · ·  cn  0, then we obtain an obvious inequality.

2 ;

162

11 Jensen’s inequality

11.11. The proof is similar to Problem 11.2, using f  (x) < 0. 11.12. (a) The proof is similar to Problem 11.2, on substituting k by p, and ci by q

bip . (b) According to Problem 11.12a, we have ⎞ p ⎞ 1p  p−1 p  n  p−1  p−1 n p

a i p ⎝ p−1 ⎠ ⎠ ⎝ bi ≥ ai , p i1 i1 i1 bi ⎛

n

⎛

and therefore,  n  n  p−1  n p

ap

i bi ≥ ai , p−1 i1 bi i1 i1 or  n p

ai p−1

i1

bi

≥

n

p ai

i1 n

 p−1 .

bi

i1

11.13. Consider the function f (x)  x1 in (0, +∞). Taking xi  a1 +· · ·+ai−1 −ai +· · ·+an , αi  (11.7), we obtain n

1 ai · ≥ n

a1 + · · · + an a1 + · · · + ai−1 − ai + · · · + an i 1

i1

ai , where i a1 +···+an

 1, . . . , n

a1 + · · · + an ,   ai a1 + · · · + ai−1 − ai + · · · + an

whence n

i1

n ai (a1 + · · · + an )2   ≥ ≥ , a1 + · · · + ai−1 − ai + · · · + an n−2 (a1 + · · · + an )2 − 2 a12 + · · · + an2

since   n a12 + · · · + an2 ≥ (a1 + · · · + an )2 (see Problem 2.2).  n k n



k k−1 11.14. In Problem 11.2, taking ci  1, we obtain n ai ≥ ai , i1 i1 ⎞ ⎛ k n n



i1

or n

i1

aik

n

dim

n

ai

≥ ⎝ i1n ⎠ . Now taking k 

m , 2

ai  di2 , it follows that

⎛

⎞ m2 n di2

≥ ⎝ i1n ⎠ . n

In a similar way, we deduce that

i1

bim

n

⎞ m2 ⎛

n bi2

≥ ⎝ i1n ⎠ .

Proofs

163

Multiplying the obtained inequalities term by term and using the inequality of Exercise 11.10, ⎛ ⎛ ⎞ ⎞ n

i1

we obtain

bim

n

n

·

i1

dim

n

n

≥⎝

i1

bi2 ·

n

i1 n2

di2

m/2

⎠

n

≥⎝

i1

m

bi di n

⎠ .

11.15. Consider the function f (x)  x1 in (0, +∞). (a) Taking αi  x1 +xx2i +x3 , i  1, 2, 3, we obtain x1 1 x2 1 x3 1 · + · + · x1 + x2 + x3 x2 + x3 x1 + x2 + x3 x1 + x3 x1 + x2 + x3 x1 + x2 1 , ≥ x1 x2 · + x3 ) + x1 +x2 +x3 (x1 + x3 ) + x1 +xx23 +x3 (x1 + x2 ) x1 +x2 +x3 (x 2 and hence. x1 x2 x3 (x1 + x2 + x3 )2 . + + ≥  x2 + x3 x1 + x3 x1 + x2 2 x1 x2 + x2 x3 + x1 x3  x +x +x 2 x 2 +x 2 +x 2 From the inequality 1 32 3 ≥ 1 32 3 , or x12 + x22 + x32 ≥x1 x2 + x2 x3 + x1 x3 , we have x12 + x22 + x32 3 1 (x1 + x2 + x3 )2  · +1≥ , 2 x x + x x + x x 2 2 x1 x2 + x2 x3 + x1 x3 1 2 2 3 1 3 

1 2 3 and therefore x2x+x + x1x+x + x1x+x ≥ 23 . 3 3 2 xi (b) αi  (x1 +···+xn ) , i  1, . . . , n, and we have that

x1 1 x2 1 xn 1 · + · + ··· + · x1 + · · · + xn x2 + xn x1 + · · · + xn x1 + x3 x1 + · · · + xn x1 + xn−1 ≥

1

      , x1 x2 +xn +x2 x3 +x1 +···+xn x1 +xn−1 x1 +···+xn

whence x1 x2 xn (x1 + · · · + xn )2   + + ··· + ≥ x2 + xn x1 + x3 x1 + xn−1 x1 x2 + x1 xn + x2 x3 + x2 x1 + · · · + xn x1 + xn−1 (x1 + · · · + xn )2  ≥2   2 x1 x2 + x2 x3 + · · · + xn−1 xn + xn x1

(see the solution to Problem 7.6).

164

11 Jensen’s inequality

Hence, it follows that x1 x2 xn + + ··· + ≥ 2. x2 + xn x1 + x3 x1 + xn−1 11.16. Let A  xy + yz + xz − xy − yz − xz ≥ 0. Consider the function f (x)  x1 in (0, +∞). , α2  y+x−z , α3  z+x−y . Let us take α1  y+z−x y+z+x y+z+x  y+z+x 2 y+z−x x+y−z z+x−y 1 1 Hence, x+y+z ≥ x 2 +y2 +z2 , or 3 − A ≥ x(x+y+z) + x + z 2 +y 2 +z 2 , y x+y+z  x y+x z+yz (x+y+z)2 whence A ≤ 3 − x 2 +y 2 +z 2  2 1 − x 2 +y 2 +z 2 .   One can easily prove that x y + x z + yz > 21 x 2 + y 2 + z 2 , and thus it follows that A < 1 . 11.17. (b) We have a1 1 a2 1 an 1 · + · + ··· + · a1 + · · · + an a2 + · · · + an a1 + · · · + an a1 + a3 + · · · + an a1 + · · · + an a1 + · · · + an−1 ≥

1

     ; a1 a2 +···+an +a2 a1 +a3 +···+an +···+an a1 +···+an−1 a1 +···+an

therefore, a1 an + ··· + a2 + · · · + an a1 + · · · + an−1 n (a1 + · · · + an )2 ≥  2 , ≥ 2 2 n−1 (a1 + · · · + an ) − a1 + · · · + an   since n a12 + · · · + an2 ≥ (a1 + · · · + an )2 . 11.18. We have 1 b 1 c 1 a · + · + · a+b+c+d b+c a+b+c+d c+d a+b+c+d d +a 1 1 d · ≥ a(b+c)+b(c+d)+c(d+a)+d(a+b) . + a+b+c+d a+b a+b+c+d Therefore, a b c d (a + b + c + d)2 + + + ≥ ≥ 2, b+c c+d d +a a+b ab + ac + bc + bd + cd + ac + ad + db

since (a + b + c + d)2 − 2(ab + ac + bc + bd + cd + ac + ad + bd) (a − c)2 + (b − d)2 ≥ 0. 11.19. Consider the function f (x)  x1 in (0, +∞).



Proofs

165

Taking α1  α6 

a , 2(a+b+c)

α2  α3 

b , 2(a+b+c)

α4  α5 

c , 2(a+b+c)

x1  ac , x2  bc , x3  ab , x4  ac , x5  bc , x6  ab , we obtain 1 1 1 a b b c · + · + · + 2(a + b + c) ac 2(a + b + c) bc 2(a + b + c) ab 2(a + b + c) 1 1 1 c a 1 · a + · b + · b ≥ 2c+2a+2b  1. 2(a + b + c) 2(a + b + c) c c a 2(a+b+c) Similarly, one can obtain the second inequality:       a 3 b3 c3 1 2 1 1 2 1 1 1 1 1 2 1 a · a · c · + + b2 · ac + + c2 · + b2 · ac + +  bc bc ab ab bc ca ab 2 2 2 b b a a c c 2 2 2    2 + b2 2 + c2 2 + b2 a a c 1 1 1 ≥ · + · + · 2 2abc 2 2abc 2 2abc 

a 2 + b2 a 2 + b2 a 2 + c2 a 2 + c2 b2 + c2 b2 + c2 · + · + · 2c 2ab 2b 2ac 2a 2bc

≥

a 2 + b2 a 2 + c2 b2 + c2 + + . 2c 2b 2a

11.20. Consider the function f (x) 

1 x

in (0, +∞). Taking

α1 

a2 , a 2 +b2 +c2

α2 

b2 , a 2 +b2 +c2

α3 

c2 , a 2 +b2 +c2

x1 

a 2 +ab+b2 , a

x2 

b2 +bc+c2 , b

x3 

c2 +ca+a 2 , c

we obtain a2 · a 2 + b2 + c2 ≥

1 a 2 +ab+b2 a

+

b2 · a 2 + b2 + c2

1 a 3 +a 2 b+ab2 +b3 +b2 c+bc2 +c3 +c2 a+a 2 c a 2 +b2 +c2

1 b2 +bc+c2 b

+

c2 · a 2 + b2 + c2

1 c2 +ac+a 2 c

,

or

2  2 a + b2 + c2 a3 b3 c3 a+b+c  ≥  + + ≥ . a 2 + ab + b2 b2 + bc + c2 c2 + ac + a 2 3 (a + b + c) a 2 + b2 + c2

11.21. Consider the function f (x)  − ln x on (0, +∞). It is a convex function, and therefore,

166

11 Jensen’s inequality

 − ln

c b d a · + · a+b a a+b b

 ≤−

a c b d ln − ln , a+b a a+b b

or 

a+b c+d

a+b ≤

 a a  b b · . c d

Remark If a1 > 0, . . . , an > 0, b1 > 0, . . . , bn > 0, then    a1  an a1 + · · · + an a1 +···+an a1 an ≤ · ... · . b1 + · · · + bn b1 bn 11.22. Consider the function f (x) xi−1 , i  1, . . . , n, x0 +x1 +···+xn−1 we obtain



1 x

on (0, +∞). Taking αi



⎞ ⎛ x2 x2  xn−1 x0 1 1 ⎠  + 1 + · · · + n−1  ⎝ · x + ··· + · x · x0 + x1 + · · · + xn−1 1 x1 x2 xn x0 + x1 + · · · + xn−1 x0 + x1 + · · · + xn−1 x n x0 n−1   x0 + x1 + · · · + xn−1 2 ≥  kn . x1 + x2 + · · · + xn−1 + xn

x02

Now let us prove that there exists a positive integer n 0 , such that for n > n 0 we have kn ≥ 3.999. 3,999 + 1, we have Indeed, if n > 0,001 

x0 + x1 + · · · + xn−1

2

  − 3, 999 x1 + x2 + · · · + xn−1 + xn ≥ 0, 001(n − 1)xn − 3, 999xn ≥ 0,

and one can take n 0  4000. 11.23. Consider the function f (x)  x1 in (0, +∞). Taking α1  BC AC AB , α2  AB+BC+AC , α3  AB+BC+AC , x 1  M A1 , x 2  AB+BC+AC M B1 , x3  MC1 , we obtain BC CA AB + +  M A1 M B1 MC1



BC CA AB 1 1 1 + + · · · BC + AC + AB M A1 BC + AC + AB M B1 BC + AC + AB MC1

(BC + AC + AB) ≥

4 p2 2 p2 (BC + AC + AB)2   , BC · M A1 + AC · M B1 + AB · MC1 2S S

and thus the smallest value of the expression 2

BC M A1

+

CA M B1

+

AB MC1

is equal to

, and equality holds when M A1  M B1  MC1 . 11.24. Without loss of generality one can assume that a1 ≥ a2 ≥ · · · ≥ an . According to Problem 14.10a, we have 2p S

n

i 1

ai bi  b1 (a1 − a2 ) + (b1 + b2 ) (a2 − a3 ) + · · · + (b1 + · · · + bn−1 ) (an−1 − an ) + (b1 + · · · + bn ) an .



Proofs

167

It follows that n

ai bi ≥ a1m (a1 − a2 ) + (a1 + a2 )m (a2 − a3 ) + · · ·

i 1

+ (a1 + · · · + an−1 )m (an−1 − an ) + (a1 + · · · + an )m an  A.

Since the function f (x)  x m is convex on [0, +∞), we have  ≥

a1 (a1 − a2 ) (a1 + a2 )(a2 − a3 ) (a1 + · · · + an−1 )(an−1 − an ) (a1 + · · · + an )an + + ··· + + a1 a1 a1 a1

m 

a12 + a22 + · · · + an2 a1

m .

Hence, we deduce that ⎛ ⎝

n

⎞2 ai bi ⎠ ≥ A2 ≥

2m  a12 + a22 + · · · + an2 a12m−2

i1

 

a12 + a22 + · · · + an2 a12

m−1

 m+1 ≥ a12 + a22 + · · · + an2

m+1  . ≥ a12 + a22 + · · · + an2

 Thus, it follows that

n

2 ai bi

i1

 m+1 ≥ a12 + a22 + · · · + an2 .

11.25. (a) Consider the function f (x)  x 1− p in (0, +∞). Since f  (x)  p( p − 1)x − p−1 ≥ 0, it follows that for the function f (x) one has inequality (11.7). ai , i  1, . . . , n, we obtain Taking αi  a1 +···+a n  1− p  1− p b1 an bn a1 · + ··· + · a1 + · · · + an a1 a1 + · · · + an an   a1 b1 an bn 1− p ≥ · + ··· + · , a1 + · · · + an a1 a1 + · · · + an an or p

a1

p−1

b1

p

+ ··· +

an

p−1

bn

≥

(a1 + · · · + an ) p . (b1 + · · · + bn ) p−1

(b) The proof is similar to Problem 11.25a, using f  (x) < 0. i , i  1, . . . , n, then if i  2, . . . n, we obtain (c) Defining Bi  ab11 +···+b +···+ai

168

11 Jensen’s inequality

bi  Bi (a1 + · · · + ai ) − Bi−1 (a1 + · · · + ai−1 ), and therefore, bi p+1 Bi

a1 + · · · + ai Bi−1 − (a1 + · · · + ai−1 ) · p+1 . p Bi Bi



(11.11)

By to the Problem 9.13, we have 

p p+1



 p+1 p

Bi−1 p Bi

 p+1 1

p p+1

Bi−1

+

p+1 p

≥

p+1

1 p, Bi

and thus it follows that Bi−1 p+1 Bi

p+1 1 . p − p−1 p Bi p Bi

≥

(11.12)

p+1 ai i−1 i i · B p − bp+1 ≥ a1 +···+a − a1 +···+a ,i  p p p p B1 p B1−1 Bi i a1 a1 p ≥ p , and summing all n inequalities, p B1 p B1

From (11.11) and (11.12) we obtain 2, . . . , n, and we obtain p+1 · p



p+1 p

·

a1 p B1

−

b1 p+1 B1







an a1 p + ··· + p B1 Bn

−

b1 p+1

B1

+ ··· +



bn p+1

Bn



a1 a1 + · · · + an . p ≥ p p B1 p Bn

(11.13)

Therefore, from (11.13) it follows that   b1 a1 p+1 an bn > p+1 + · · · + p+1 , · p + ··· + p p B1 Bn B1 Bn

(11.14)

and according to Problem 11.25a, it follows that  b1 p+1 B1

+ ··· +

bn p+1

Bn

 

a1 p B1

p+1

a1 p b1



p+!p

1p + · · · + 

an p Bn

p+1

an p bn



p+!p

1p ≥ 

a1 p B1 p+1

a1 p b1

+ ··· + + ··· +

an p Bn

p+!p

p+1

an p bn

1p .

(11.15) From (11.14) and (11.15) we obtain  a1 ·

a1 b1



p + · · · + an ·

a1 + · · · + an b1 + · · · + bn



p <

p+1 p

p

p+1

a1

p

b1

p+1

+ ··· +

an p bn

 .

(11.16)

Proofs

169

From (11.13), it follows that  a1 + · · · + an p a1 · + · · · + an · b1 + · · · + bn       p+1 a1 + · · · + an p+1 p a1 ≥ + · · · + bn b1 · p+1 b1 b1 + · · · + bn 

a1 b1

+

p



1 (a1 + · · · + an ) p+1 . . p + 1 (b1 + · · · + bn ) p

(11.17)

Remark 1. If p < −1 and a1 ,…, an , b1 ,…, bn are arbitrary positive numbers, then 

a1 ·

a1 b1

p



+ · · · + an ·

a1 + · · · + an b1 + · · · + bn

p



<

p+1 p

p

p+1

a1

p

b1

p+1

+ ··· +

an p bn



.

2. If p > 1 and a1 ,…, an are arbitrary positive numbers, then  a + · · · + a p  a + · · · + a p n 1 n 1 n p · a1 + · · · + + n p−1 n  p p p < · (a1 + · · · + anp ). p−1 3. If p > 1 and b1 ,…, bn are arbitrary positive numbers, then 1

1



b11 + · · · + (b1 · . . . · bn ) n <

p p−1

p

· (b1 + · · · + bn ).

4. If b1 ,…, bn are arbitrary positive numbers, then 1

1

b11 + · · · + (b1 · . . . · bn ) n < e · (b1 + · · · + bn ). x +x 

11.26. (a) Consider the numbers xi  i 2 i , i  1, . . . , n. Note that min(xi , xi ) ≤ xi ≤ max(xi , xi ) and min(xi + x j , xi + x j ) ≤ xi + x j ≤ max(xi + x j , xi + x j ), for all i, j ∈ {1, . . . , n}. Therefore, xi ∈ I and xi + x j ∈ G. If we prove that    f (x1 ) + · · · + f (xn ) x1 + · · · + xn ≤ , (11.18) f n n   then since f xi ≤

f (xi )+ f (xi ) , i  1, . . . , n, and f (x1 ) + · · · + f (xn )  2     f (xn )  x1 +···+xn n n ≤ f (x1 )+···+ f (xn ), n  x1 +···+x , we have f x1 +···+x . n n n

f (x1 ) + · · · + Assume that inequality (11.18) holds for m numbers, and let us prove that it holds for 2m numbers.

170

11 Jensen’s inequality

Indeed, we have  f

  + x2m x1 + x2 + · · · + x2m−1

2m



⎛

⎞ x  +x  + · · · + 2m−12 2m ⎠  f⎝ m  x  +x     x +x f 1 2 2 + · · · + f 2m−12 2m ≤ m f (x 

f (x 1 )+ f (x 2 ) 2

)+ f (x  )

2m−1 2m + ··· + 2 m   ) f (x1 ) + f (x2 ) + . . . + f (x2m−1 ) + f (x2m

≤ 

x 1 +x 2 2

2m

.

Now let us prove that if inequality (11.18) holds for k (k > 2) numbers, then it holds for k − 1 numbers.  ∈ I and xi + x j ∈ G, where i, j ∈ {1, . . . , k − 1}. Let x1 , . . . , xk−1 x  +···+x 

  Note that min(x1 , . . . , xk−1 ) ≤ 1 k−1 k−1  x  ≤ max(x1 , . . . , xk−1 ), and     ∈ I, 2x , x + x ∈ G, for all i  1, . . . , k − 1, whence therefore, x i    x1 +···+xk−1 +x  f (x  )+···+ f (xk−1 )+ f (x  ) ≤ 1 . Thus it follows that k k     x1 +···+xk−1 f (x1 )+···+ f (xk−1 ) ≤ f . Since inequality (11.18) for k−1 k−1

f

two numbers holds, from this statement it follows that it holds for n numbers. Remark If G  R, then we obtain Jensen’s inequality. (b) Let us prove that if n ≥ 2, yi > 0, i  1, . . . , n, and y1 · . . . · yn  1 , then (n−1)n 1 1 + ··· + ≤ n − 1. 1 + y1 1 + yn

(11.19)

1 Indeed, let us consider the function f (x)  1+3 on R, and G  x   2 (−∞, 0]. Note that if x1 + x2 ∈ G, then − x1 +x2 − − 1+31 x1 − 1+31 x2  

 x1 x2 2 3 2 −3 2   x1 +x2 (1+3x1 )(1+3x2 ) 1+3 2 3

x1 +x2 2

−1

1+3

2

≤ 0, and hence f

x + x f (x1 ) + f (x2 ) 1 2 ≤ . 2 2

If yi y j ≤ 1 for all i and j, then xi + x j ∈ G, where yi  3xi , i  1, . . . , n,  and thus for x1  x2 , . . . , xn−1  xn , xn  x1 , the condition of Problem 11.25a holds. Therefore, 1 1 n + ··· + ≤  n − 1. √ 1 + y1 1 + yn 1 + n y1 · . . . · yn

Proofs

171

If there exist numbers i and j such that yi y j > 1, then 1 1 1 1 + ··· + ≤ + + 1 + y1 1 + yn 1 + yi 1 + yj

k i,k  j

ai+1 For yi  (n−1)a , i  1, . . . , n, i follows that

1 < 1+ 1 + yk

k i,k  j

1 < 1+(n − 2)  n − 1. 1 + yk

an+1  a1 , from inequality (11.19), it

a1 an + ··· + ≤ 1. (n − 1)a1 + a2 (n − 1)an + a1 (c) Consider the function f (x)  if x1 + x2 ∈ G, then √

1 + 3x1 +

√

1 + 3x2

2

√1 1+3x

in R, and G  [2, +∞). Note that

  2 + 3x1 + 3x2 + 2 (1 + 3x1 )(1 + 3x2 )  x1 +x2 ≥ 2 + 3x1 + 3x2 + 2 1 + 3 2 ,

whence 2  √ √ x1 +x2 1+3 2 1 + 3x1 + 1 + 3x2   x1 +x2 x1 +x2 1+3 2 ≥ 4 + 3x1 + 3x2 + 2 · 3 2 ≥ 4(1 + 3x1 )(1 + 3x2 ), since   x +x  x x1 +x2  x1 +x2  x 2 2 1 2 1 4 + 3x1 + 3x2 + 2 · 3 2 1+3 2 − 4(1 + 3x1 )(1 + 3x2 )  3 2 − 3 3 2 − 3 2 ≥ 0.

It follows that f

x +x f (x1 ) + f (x2 ) 1 2 ≤ . 2 2

If yi y j ≥ 9 for all i and j, then xi + x j ∈ G, where yi  3xi , i  1, . . . , n,   xn , xn  x1 , the conditions of and therefore for x1  x2 , . . . , xn−1 Problem 11.25a hold. Hence 1 1 n n + ··· + √ ≥ √ . √ √ n 1 + y1 · . . . · yn 1 + y1 1 + yn 1+λ If there exist numbers i and j such that yi y j < 9, then √

1 1 1 1 + ··· + √ ≥√ +√ > 1, 1 + y1 1 + y1 1 + yn 1 + y2

172

11 Jensen’s inequality

since 2   √ 1 + y1 + 1 + y2 ≥ 4 + y1 + y2 + 2 y1 y2 > (1 + y1 )(1 + y2 ). (d) Consider the function f (x)  ln

on (0, 1), and G  (0, 1]. Note x +x   1− 1 2 2 that if x1 , x2 ∈ (0, 1) and x1 + x2 ∈ G, then f x1 +x  ln x1 +x22 ≤ 2 2 2  1−x 1−x ln x 1 +ln x 2 f (x1 )+ f (x2 ) 2−(x1 +x2 ) (1−x1 )(1−x2 ) 2 1 2  , since ≤ , or (x −x 1 2 ) (1− 2 2 x1 +x2 x1 x2 x1 − x2 ) > 0, whence f

1−x x

x + x f (x1 ) + f (x2 ) 1 2 ≤ . 2 2

Thus, from Problem 11.25a, it follows that f

x + ··· + x f (x1 ) + · · · + f (xn ) 1 n ≤ , n n

or (1 − x1 ) · . . . · (1 − xn ) ≥ x1 · . . . · xn



1−

x1 +···+xn n x1 +···+xn n

n .

11.27. Without loss of generality one can assume that x ≤ y ≤ z. Let us consider three cases. (a) If x ≤ − √13 , then y + z ≥ 1 +

√1 > 3 . Therefore, z > 3 . It follows that 2 4 3 1 1 1 16 27 + + < 1 + 1 + < . 2 2 2 1+x 1+y 1+z 25 10 1 1 1 3 3 27 (b) If y ≥ √13 , then z ≥ √13 , whence, 1+x 2 + 1+y 2 + 1+z 2 ≤ 1 + 4 + 4 < 10 .  (c) If x, y ∈ − √13 , √13 , then consider the function f (t)  1+t1 2 in  − √13 , √13 .

−1) Note that f  (t)  2(1+t(1+t)(3t < 0, and thus by Jensen’s inequality, it 2 )2 1 1 2 2 1 1 1 follows that 1+x 2 + 1+y 2 ≤ 2  2 . Therefore, 1+x 2 + 1+y 2 + 1+z 2 ≤ 1+( x+y 1+( 1−z 2 ) 2 ) 2 1 27 2 + 1+z 2 ≤ 10 , since the last inequality is equivalent to the inequality 1+( 1−z 2 ) (3z − 1)2 (3z 2 − 4z + 5) ≥ 0. 11.28. Let us prove that if α ≥ 2 and a > 0, b > 0, then 2

2

1 1 22−α + ≥ α α α (1 + a) (1 + b) 1 + (ab) 2 On setting x  as

1 , 1+a

y

(11.20)

1 , we see that inequality (11.20) can be rewritten 1+b

Proofs

173



α

α

α

α

x 2 y− 2 + x − 2 y 2



α α α α x 2 y 2 + (1 − x) 2 (1 − y) 2 ≥ 22−α . α

α

α

α

Without loss of generality one can assume that y 2 (1 − y) 2 ≥ x 2 (1 − x) 2 . Thus by to the Cauchy–Bunyakovsky–Schwarz inequality, it follows that 

α α α α  α α α α x 2 y − 2 + x − 2 y 2 x 2 y 2 + (1 − x) 2 (1 − y) 2 2   α α α α α ≥ x 2 + y 2 (1 − y) 2 x − 2 (1 − x) 2

 α α 2 ≥ x 2 + (1 − x) 2 , and by Jensen’s inequality, it follows that    α2 2 2 x+(1−x)  22−α . 2 α

α

α

α

α

α



α

α

x 2 + (1 − x) 2

α

2

≥

α

Therefore, (x 2 y − 2 + x − 2 y 2 )(x 2 y 2 + (1 − x) 2 (1 − y) 2 ) ≥ 22−α . Note that 1 1 1 1 22−α 22−α + + + ≥ α + α α α α α (1 + a) (1 + b) (1 + c) (1 + d) 1 + (ab) 2 1 + (cd) 2 

22−α 1 + (ab)

α 2

+

22−α 2−α .  1  α2  2 1 + ab

Problems for Independent Study 1. Prove that if α, β, γ are the angles of some triangle, then (a) cos α + cos β + cos γ ≤ 23 , (b) sin21α/2 + sin21β/2 + sin21γ /2 ≥ 12, √ (c) cos1α/2 + cos1β/2 + cos1γ /2 ≥ 2 3, (d) sin1α/2 + sin 1β/2 + sin 1γ /2 ≥ 6. 2. Prove that if α, β, γ are the angles of some acute triangle, then √ √ √ 1+8 cos2 γ 1+8 cos2 β cos2 α (a) 1+8 + + ≥ 6, sin α sin β sin γ 1 1 1 (b) cos α + cos β + cos γ ≥ 6. 3. Prove that if for a convex quadrilateral ABC D one has





sin 2A sin 2B sin 2C sin 2D  41 , then ABC D is a rectangle. x1 x2 n + x1x+x ≥ n2 , where 3 ≤ n ≤ 6 and 4. Prove that x2 +x3 + x3 +x4 + · · · + xxnn−1 +x1 2 xi > 0, i  1, · · · , n. 5. Prove that among all convex n-gons inscribed in a given circle, the one with greatest area is the regular n-gon.

174

11 Jensen’s inequality

6. Prove that among all convex n-gons inscribed in a given circle, the one with the greatest perimeter is the regular n-gon. an a1 a2 n + 1+xa +· · ·+ 1+xa ≤ n+x , where x > 0, a1 ≥ 0, · · · , an ≥ 0, 7. Prove that 1+xa 1 2 n and a1 + · · · + an  1. 8. Prove that sin α sin β + sin β sin γ + sin γ sin δ + sin δ sin α ≤ 2, where α > 0, β > 0, γ > 0, δ > 0, and α + β + γ + δ  π. 9. Let x1 , x2 , · · · , xn be arbitrary real numbers. (a) Prove that, if λ ≥ 2 or λ < 0, then it holds true the following inequality

⎛ (±x1 ± x2 ± · · · ± xn )λ ≥ 2n ⎝

n

⎞λ/2 xi2 ⎠

,

i1

(b) Prove that if 0 < λ < 2, then one has the following inequality: n λ

2 2 |±x1 ± x2 ± · · · ± xn |λ ≤ 2n xi (on the left-hand side, the summai1

tion is over all combinations of plus and minus signs).  2 2 2  2 2 a −b a +b ≥ − a+b , where a ≥ 21 , b ≥ 21 . 10. Prove that 2 2 2  a n a

n 2 +1 xi + x1i ≥ ( na−1) , where a > 0, x1 > 0, . . . , xn > 0, and 11. Prove that n

i1

xi  1.

i1

12. (a) Prove that a1 x1 + · · · + an xn ≥ x1a1 · . . . · xnan , where a1 + · · · + an  1 and xi > 0, ai ≥ 0, i  1, . . . , n, n n



(b) Let ai j ≥ 0 and ai j  1, j  1, 2, . . . , n, ai j  1, i  1, 2, . . . , n. i1

j1

Prove that (a11 x1 + a12 x2 + · · · + a1n xn ) (a21 x1 + a22 x2 + · · · + a2n xn ) · . . . · (an1 x1 + an2 x2 + · · · + ann xn ) ≥ x1 x2 · . . . · xn , where x1 ≥ 0, · · · , xn ≥ 0. (c) Prove that (ux + vy + wz)(vx + wy + uz)(wx + uy + vz) ≥ (y + z − x)(z + x − y)(x + y − z), where u + v + w  1, u ≥ 0, v ≥ 0, w ≥ 0, x ≥ 0, y ≥ 0, z ≥ 0. (1−x)(1−y) (1−y)(1−z) (1−z)(1−x)    z x y (1−y)(1−z) (1−z)(1−x) · · ≥ 256 , 13. Prove that (1−x)(1−y) z x y 81 where x > 0, y > 0, z > 0 and x + y + z  1 14. Find the largest and smallest values of the expression (a) (b)

a a+b n−1

j1

+

b b+c

xj x j +x j+1

− −

a , a+c

where a > 0, b > 0, c > 0,

x1 , x1 +xn

where n ≥ 3, x1 > 0, . . . , xn > 0.

Problems for Independent Study

175 √

√ x +···+ x

xn x1 15. Prove that √1−x + · · · + √1−x ≥ 1√n−1 n , where n ≥ 2, xi > 0, i  n 1 1, . . . , n, and x1 + · · · + xn  1. x1 +···+xn xn x1 + · · · + √1−x ≥ n · √ xn1 +···+xn . Hint. Prove that √1−x 1

n

1− 3

n

(x1 +···+xn ) xn x1 , where 0 ≤ xi < 1, i  16. Prove that 1−x 2 + · · · + 1−x 2 ≥ n · (x1 +···+xn )2 −(x12 +···+xn2 )2 n 1 1, . . . , n. xi 1 Hint. Consider the function f (x)  1−x i 2 in [0, 1), taking αi  x +···+x , 1 n 1, . . . , n. 17. Prove that (a1 b1 + a2 b2 + · · · + an bn )k + (a1 b2 + a2 b3 + · · · + an b1 )k + · · · + +(a1 bn + a2 b1 + · · · + an bn−1 )k ≤ (b1 + b2 + · · · + bn )k (a1k + a2k + · · · + ank ), where n ≥ 2, ai > 0, bi > 0, i  1, · · · , n, k > 1.  an an n 18. For which values of λ does the inequality a n +λa11·...·an +· · ·+ a n +λa1n·...·an ≥ √1+λ n 1 hold for all positive numbers a1 , . . . , an ?

Chapter 12

Inequalities of Sequences

In this section we consider inequalities related to sequences, in particular sequences given by recurrence relations, which are proved in various ways. While most inequalities can be proved by well-known proof techniques, there are not many tools for inequalities involving sequences. Below we provide some examples of problems on inequalities of sequences and demonstrate the proof techniques with which such problems can be attacked.

Problems x2

12.1. Prove that if xn+1  xn + nn2 , n  1, 2, . . ., and 0 < x1 < 1, then the sequence (xn ) is bounded. √ 1 + an−1 , n  2, 3, . . . , 10, then 0 < a10 − 2 < 12.2. Prove that if a1  1, an  an−1 2 10−370 . 12.3. Consider the sequence (u n ) such that u 1  1, u n+1  u n + u1n , n  1, 2, . . .. Prove that 14.2 < u 100 < 14.22. 12.4. Consider the sequence (u n ), such that u 1  1, u n+1  u n + u12 , n  1, 2, . . .. n Prove that 30 < u9000 < 30.01. 12.5. Prove that if u 0  0.001, u n+1  u n (1 − u n ), n  0, 1, . . ., then u 1000 < 1 . 2000  2 a 12.6. Consider the sequence a1  1, an+1  a2n + 4n + a1n , n  1, 2, . . .. Prove that this sequence is unbounded. 12.7. Given that a0  an  0, ai > 0, i  1, . . . , n − 1, n ≥ 2, and as−12+as+1 > as cos πk , k ∈ N, prove that n ≥ k.

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_12

177

178

12 Inequalities of Sequences

12.8. Let x1 ∈ [0, 1), and for n  1, 2, . . ., suppose that ⎧ if xn  0, ⎨ 0,   xn+1  ⎩ x1 − x1 if xn  0. n n n , where F1  F2  1 and Prove that x1 + x2 + · · · + xn < FF21 + FF23 + · · · + FFn+1 Fn+2  Fn+1 + Fn , n  1, 2, . . . . 12.9. Let n ≥ 2 be a positive integer. Find the smallest possible value of the sum a0 + a1 + · · · + an if a0 , a1 , . . . , an are nonnegative numbers such that a0  1 and ai ≤ ai+1 + ai+2 , i  0, 1, . . . , n − 2. 12.10. Given a sequence of positive numbers x1 , x2 , . . . , xn , . . . such that xnn  n−1  j 1 xn , n  1, 2, . . ., prove that 2 − 2n−1 ≤ xn < 2 − 21n .

j0

12.11. Let a0 , a1 , a2 , . . . , an , . . . be an infinite sequence of positive numbers. √ Prove that the inequality 1 + an > n 2.7 × an−1 holds for infinitely many numbers n. 12.12. Consider the following sequence an  max(x n (1 − x) + (1 − x)n x), n  [0;1]

1, 2, . . .. Prove that an+1 ≥ 21 an , n  1, 2, . . .. 12.13. Given that a0  an+1  0 and |ai−1 − 2ai + ai+1 | ≤ 1, i  1, . . . , n, prove that ak ≤ k(n+1−k) , k  1, . . . , n. 2 12.14. Find the smallest value of the number C such that there exists a sequence (an ) of positive numbers such that the inequality a1 +· · ·+am+1 ≤ Cam holds for all positive integers m.

Proofs 12.1. Let us prove by induction that xn ≤

(1 −

√

If n  1, then we have x1 ≤

nx1 √ , n  1, 2, 3, . . . . x1 )n + x1

√ x1 √ (1− x1 )+ x1

 x1 .

1 √ Assume that the given inequality holds for n  k, that is, xk ≤ (1−√xkx1 )k+ . x1 Let us prove that the given inequality holds for n  k + 1, that is, xk+1 ≤ x2 √ (k+1)x1 √ . Since x k+1  x k + k2 and x k ≤ √ kx1 √ , we have (1− x1 )(k+1)+ x1 k (1− x1 )k+ x1

Proofs

179

kx1 x2 √ √ + √ 1 √ 2  k(1 − x1 ) + x1 (k(1 − x1 ) + x1 ) √ √ k 2 (1 − x 1 ) + k x1 + x1 (k + 1)x1  x1 √ √ 2 ≤ √ √ , as (k(1 − x1 ) + x1 ) (1 − x1 )(k + 1) + x1 √ √ k 2 (1 − x1 ) + k x1 + x1 k+1 −  √ √ √ √ k(1 − x1 ) + 1 k 2 (1 − x1 )2 + 2k(1 − x1 ) x1 + x1 √ √ x1 (1 − x1 )2 > 0.  √ √ √ √ (k(1 − x1 ) + 1)(k 2 (1 − x1 )2 + 2k(1 − x1 ) x1 + x1 )

xk+1 ≤

1 √ Therefore, for every positive integer n we have 0 < xn ≤ (1−√xnx1 )n+ ≤ x1 nx√ x√ 1 1  1− x1 , and hence the sequence (xn ) is bounded. n(1− x1 ) √ 12.2. Prove that an − 2 > 0, where

n 2, 3, . . .. √ √ 2 2 We have an  0.5 an−1 + an−1 ≥ an−1 · an−1  2, and hence an − 2 ≥ 0. There is never √ equality, since all terms of the sequence are rational numbers; hence an − 2 > 0. √ Now let us consider the sequence b n  an − 2. √ √ √ a 2 −2 2a +2 2 − 2  n−1 2an−1 n−1  We have bn  an − 2  0, 5 an−1 + an−1

√ (an−1 − 2)2 2an−1

2 bn−1 , 2an−1

b2

√ . n  2, 3, . . ., and thus bn < 2n−1 2 √ √ b2 1 √ , we have b3 < Since b2  a2 − 2  23 − 2 < 10 and bn < 2n−1 2 1√ , 102 2 2



b4 <

b10 <

√1 √ , 104 (2 2)2 2 2

b22 √ 2 2

<

and continuing in a similar way, we deduce that

1 1 1 1 1 1 · √ · √ · · · √  256 · √ . 10 1028 (2 2)27 (2 2)26 2 2 (2 2)255

Now let us prove that √ √ 255 1 1 1 114 382 · < or 10 < (2  2 2. 2) √ 10256 (2 2)255 10370 √ Since 210 > 103 , it follows that 10114 < 2380 < 2382 2. √ Hence b10 < 101256 · (2√12)255 < 101370 , and thus it follows that a10 − 2 < 10−370 .

12.3. We have u 2n+1  u 2n + 2 + u 22  u 21 + 2 +

1 . u 2n

Thus, it follows that

1 1 1 , u 23  u 22 + 2 + 2 , . . . , u 2n  u 2n−1 + 2 + 2 . 2 u1 u2 u n−1

Summing these equalities, we obtain u 2n  2n +

1 1 1 + + · · · + 2 (n ≥ 3). u 22 u 23 u n−1

(12.1)

180

12 Inequalities of Sequences

Since u n−1 > u n−2 > · · · > u 2  2, we have Therefore, 2n ≤ u 2n <

1 u 22

+

+ ··· +

1 u 23

1 u 2n−1

< n4 .

9 n (n ≥ 2). 4

(12.2)

From the (12.2), it follows that for n   conditions (12.1) and  100 we have 1 1 200 + 49 21 + 13 + · · · + 99 < u 2100 < 200 + 21 21 + 13 + · · · + 99 . 1 Let us estimate from below the following sum; 21 + 13 + · · · + 99 . We have   1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + 2 3 4 5 6 8 10 7 9 11  

1 1 1 1 1 232 652 1 + + ··· + + + + ··· + > 1.675 + 0.346 + + > 3.989. + 12 13 33 34 35 99 23 · 23 65 · 67

1 1 1 + + ··· +  2 3 99

a2

a2

In this proof, we have used that b11 + · · · + bnn ≥ 0, . . . , bn > 0. Now let us estimate from above the following sum: According to Problem 7.1b we have

1 2

+

1 3

where b1 >

+ ··· +

1 . 99

 

1 1 1 1 1 1 + + + ··· + + + ··· + 25 48 49 50 98 99

   1 1 1 1 1 1 1 1 1 + + ··· + + + + ··· + + + + ··· + < 12 13 24 49 25 48 97 50 98

1 1 1 1 + + ··· +  + 12 13 99 12

<

1 1 + ··· + 13 24



(a1 +···+an )2 , b1 +···+bn

+

25 25 25 13 1 + + +  < 2.2. 12 36 36 36 6

It follows that    1 1 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + + + ··· + < 2 3 4 5 6 8 10 7 9 11 12 13 99 

1 1 1 + + ··· + < 4.22. < 2.02 + 12 13 99

1 1 1 + + ··· +  2 3 99

  1 1 < u 2100 < 200 + 21 21 + 13 + · · · + 99 , we have Since 200 + 49 21 + 13 + · · · + 99 14.22 < 200 + 49 · 3.989 < u 2100 < 200 + 21 · 4.22 < 14.222 , hence 14.2 < u 100 < 14.22. 12.4. We have u 3n+1  u 3n + 3 + u33 + u16 , where n  1, 2, . . .. n

n

Therefore, u 32  u 31 +3+ u33 + u16 , u 33  u 32 +3+ u33 + u16 , . . ., u 3n+1  u 3n +3+ u33 + u16 . n n 1 1 2 2 Summing these

obtain

equalities, we u 3n+1  3n + 3 u13 + · · · + u13 + u16 + · · · + u16 ; hence n n 1 1 √ √ u 3n+1 > 3n+ u33  3(n+1), or u n+1 > 3 3(n + 1), thus u 9000 > 3 3 · 9000  30. 1

Proofs

181

On the other hand,  

1 1 1 1 < + · · · + u 3n+1  3n + 3 3 + · · · + 3 + un u 6n u1 u 61

  1 1 1 1 < 3(n + 1) + 1 + 3  + ··· + + + · · · + 3·2 3·n 32 · 22 32 · n 2  

1 1 1 1 1 . + ··· + +  3(n + 1) + 1 + + · · · + 2 n 9 22 n2 Since

1 2

+

1 3

+ ··· +

1 n

< ln n (see Problem 12.3) and

1 1 1 1 1 1 + + ··· + 2 < + + ··· + 1·2 2·3 (n − 1)n 22 32 n  



1 1 1 1 1 1 + + ··· +  1 − < 1,  1− − − 2 2 3 n−1 n n

we must have u 3n+1 < 3(n + 1) + 1 + ln n + 19 ; hence u 39000 < 27002 + ln 9000 <  1 3 . 27002 + ln 225 < 27027 < 30 + 100 Therefore, u 9000 < 30.01 1 1 1  u n (1−u  u1n + 1−u . 12.5. Since u n+1  u n (1 − u n ), we have u n+1 n) n 1 1 1 1 1 1 1 1 Therefore, u 1  u 0 + 1−u 0 , u 2  u 1 + 1−u 1 ,…, u 1000  u 999 + 1−u1 999 . Summing these equalities, we obtain 1 1 1 1  + + ··· + u 1000 u0 1 − u0 1 − u 999 1 1  103 + + ··· + 1 − u0 1 − u 999 3 > 10 + 1 + · · · + 1  2000; 1 hence u 1000 < 2000 . In this proof, we have used the inequalities 0 < u n < 1, which can be proved in the following way:

u n+1  u n (1 − u n ) ≤

u n + (1 − u n ) 2

2 

1 < 1 (n  0, 1, 2, . . .). 4

If u n ≤ 0, then from the equality u n  u n−1 (1 − u n−1 ) it follows that u n−1 ≤ 0. Continuing in a similar way, we deduce that u 0 ≤ 0, which leads to a contradiction; hence u n > 0. 12.6. Assume that the sequence (an ) is bounded, that is, m < an < M (n  1, 2, 3, . . .), where M > 0, since an > 0. 1 2 − an+1 an  a1n or a1n  an+1 + a 2 a12 . We have an+1 Hence

1 a1



1 a2

+

1 , 1 a12 a22 a2



1 a3

+

1 , a22 a32

…,

1 an

n n+1



1 an+1

+

1 2 . an2 an+1

182

12 Inequalities of Sequences

Summing these equalities, we deduce that 1 1 1 1 + a 21a 2 + a 21a 2 + · · · + a 2 a12 > 4 + 4 + · · · + 4  1  an+1 n n+1 1 2 2 3 M M M   

n , or M4

M 4 > n,

n

which leads to a contradiction, and hence the sequence (an ) is bounded. 12.7. Assume that n < k. Thus, πk < πn , and it follows that cos πk > cos πn . Hence we obtain a2 > 2a1 cos πn , a1 +a3 > 2a2 cos πn , …, an−2 > 2an−1 cos πn . Let us multiply both sides of the first inequality by sin πn , and both sides of , and so on, ending by multiplying both sides the second inequality by sin 2π n of the last inequality by sin (n−1)π . n Summing all these inequalities, we obtain 0 > 0, which is a contradiction. Therefore, our assumption is incorrect, since n ≥ k. 12.8. Without loss of generality, one can assume that xn  0. We have   1 k  2, . . . , n, whence xk ∈ (0, 1), k  1, 2, . . . , n, and xk  xk−1 xk 

1  xk+1 + x1

≤

1 xk+1 +1

 f (xk+1 ) k  1, 2, . . . , n −1 (∗), where f (x) 

1 . x+1

k

Lemma If the function f (x) is decreasing on an interval I, and the function x + f (x) is increasing there, then the function g(x)  x + f (x) + f ( f (x)) + · · · + f ( f (. . . f (x) . . .)) is increasing on I, where m ∈ N.    m

Note that the function f ( f (. . . f (x) . . .)), k ∈ N is increasing on I, and hence    2k

the function f ( f (. . . f (x) . . .)) + f ( f (. . . f (x) . . .)) is increasing on I, and       2k

2k+1

the function g(x)  x + f (x) + f ( f (x)) + f ( f ( f (x))) + . . . is the sum of increasing functions. Thus, it follows that this function is increasing on I. 1 on I  [0, 1] satisfies the One can easily verify that the function f (x)  x+1 assumptions of the lemma. Therefore, by the lemma and (∗), we have x1 + x2 + · · · + xn ≤ f (x2 ) + x2 + x3 + · · · + xn ≤ x3 + f (x3 ) + f ( f (x3 ))+ + x4 + · · · + xn ≤ . . . ≤ xn + f (xn ) + f ( f (xn )) + · · · + f ( f (. . . f (xn ) . . .)) ≤    n−1

F1 F2 Fn ≤ 1 + f (1) + f ( f (. . . f (1) . . .))  + + ··· + ,    F2 F3 Fn+1 n−1

as 1 

F1 F2

and f

Fi Fi+1





Fi+1 , Fi+2

i  1, 2, . . . , n − 1.

Proofs

183

12.9. Let us prove the following lemma by induction. Lemma If n ≥ 2, c0  0, cn ≥ 0, and ci ≤ ci+1 + ci+2 , i  1, 2, . . . , n − 2, then c0 + c1 + · · · + cn ≥ 0. Indeed, for n  2 and n  3, we have c0 + c1 + c2 ≥ 2c0  0 and c0 + c1 + c2 + c3 ≥ 2c0 + c3 ≥ 0. Assume that for n ≤ k the lemma holds. Let us prove that the lemma holds for n  k + 1 (k ≥ 3). Consider the following two cases. (a) c1 ≥ 0. Since 0 ≤ c1 ≤ c2 + c3 , for the numbers 0, c2 , c3 , . . . , ck+1 the assumptions of the lemma hold, and therefore 0 + c2 + c3 + · · · + ck+1 ≥ 0, whence c0 + c1 + c2 + · · · + ck+1 ≥ c2 + c3 + · · · + ck+1 ≥ 0. (b) c1 < 0. We then have c2 ≥ −c1 > 0, and for the numbers 0, c3 , c4 , . . . , ck+1 the assumptions of the lemma hold. Hence 0 + c3 + c4 + · · · + ck+1 ≥ 0. On the other hand, c0 + c1 + c2 + c3 + · · · + ck+1 ≥ 2c0 + c3 + · · · + ck+1 ≥ 0. This ends the proof of the lemma. i  0, 1, . . . , n, where F0  0, F1  1, Note that the numbers ci  ai − FFn−i n and Fi+2  Fi+1 +Fi , i  0, 1, . . . , n−2, satisfy the assumptions of the lemma. + · · · + FFn0 . On Therefore, c0 + c1 + · · · + cn ≥ 0, or a0 + a1 + · · · + an ≥ FFnn + FFn−1 n

the other hand, the numbers ai  FFn−i satisfy the assumptions of the problem, n and therefore, the smallest value of the sum a0 + a1 + · · · + an is equal to Fn +Fn−1 +···+F0 . One can prove by induction that F0 + F1 + . . . + Fn  Fn+2 − 1. Fn  1 and xn > 1 for n  2, 3, . . .. Indeed, for n > 1 we have 12.10. Note that x 1 x n −1 xnn  xnn −1 , or xnn (2 − xn )  1. Let xn  2 − yn . Then 0 < yn < 1 for n  2, 3, . . . and yn (2 − yn )n  1. We 1 for n  2, 3, . . .. We have yn  (2−y1 n )n > 21n need to prove that 21n < yn ≤ 2n−1 (for n  2, 3, . . .). By Bernoulli’s inequality, we have 1  yn (2 − yn )n  yn (1 + (1 − yn ))n ≥ yn (1 + n(1 − yn )), whence yn ≤ n1 , 1 , for n  2, 3, . . .. and therefore yn  2n 1−1 yn n ≤ 2n 1−1 nyn ≤ 2n−1 ( 2 ) ( 2) √ 12.11. We argue by contradiction. Assume that the inequality 1 + an > n 2.7 × an−1 holds for finitely many numbers n. Thus,√there exists a positive integer n 0 n such that for all  n≥ n 0 , one has 1 + an > 2.7 × an−1 . 1 n We have lim 1 + n  e; thus there exists a positive integer n 1 such that for n→∞ n n ≥ n 1 , 1 + n1 > 2.7 (e  2.718281828 . . .). Hence, for n ≥ max(n 0 , n 1 ), √ an 1 a , and therefore n+1 + n+1 < an−1 . It we have 1 + an ≤ an−1 · n 2, 7 < n+1 n n−1 n am−1 am+k−2 am am am+1 1 1 1 m+k−1 . follows that m+1 < m − m+1 , m+2 < m+1 − m+2 , . . . , m+k < m+k−1 − am+k am−1 1 1 1 Summing these inequalities, we obtain m+1 + m+2 + · · · + m+k < m −  am+k−1 1 1 + 21 + · · · + k1 < am−1 < am−1 , and hence m+1 , and it follows that the m+k m m

184

12 Inequalities of Sequences

sequence xk  1 + 21 + · · · + k1 is bounded, which leads to a contradiction, since x2n − xn > 21 , where n ∈ N. 1 1 1 Indeed, x2n  xn + n+1 + · · · + 2n > xn + n · 2n  xn + 21 . 12.12. We proceed by induction. We have a1  21 , a2  41 , and therefore a2 ≥ 21 a1 . Let am+1 ≥ 21 am , where m ∈ N, and let us prove that am+2 ≥ 21 am+1 . Let x0 ∈ [0, 1] be such that am+1  x0m+1 (1 − x0 ) + (1 − x0 )m+1 x0 . Then am+2 ≥ x0m+2 (1 − x0 ) + (1 − x0 )m+2 x0  (x0m+1 (1 − x0 ) + (1 − x0 )m+1 x0 )(x0 + (1 − x0 )) a a − x0 (1 − x0 )(x0m (1 − x0 ) + (1 − x0 )m x0 ) ≥ am+1 − x0 (1 − x0 )am  m+1 + m+1 − x0 (1 − x0 )am 2 2

2 1 am+1 am am+1 am+1 ≥ ≥ + − x0 (1 − x0 )am  + am − x0 , 2 4 2 2 2

and therefore, am+2 ≥ 21 am+1 . Hence, an+1 ≥ 21 an , n  1, 2, . . .. 12.13. Let us set ai+1 − ai  bi , i  0, 1, . . . , n, and bi+1 − bi  ci , i  0, 1, . . . , n − 1. Then |ci | ≤ 1, i  0, 1, . . . , n − 1. We have bi+1  c0 + · · · + ci + b0 , i  0, 1, . . . , n − 1, and ai+1  b0 + · · · + bi  (i + 1)b0 + ic0 + (i − 1)c1 + · · · + ci−1 . Therefore, 0  an+1  (n + 1)b0 + nc0 + · · · + cn−1 , n 1 c0 − · · · − n+1 cn−1 and whence, b0  − n+1  

(i + 1)n (i + 1)(n − i + 1) c0 + · · · + 1 − ci−1 ai+1  i − n+1 n+1 n−i 1 ci − · · · − cn−1 , − n+1 n+1 ≤ 0, for k  0, . . . , i − 1. Thus, ai+1 is where we note that i − k − (i+1)(n−k) n+1 maximal if c0  · · ·  cn−1  −1, in which case ai+1  (i + 1)b0 − i − (i − 1) − · · · − 1 i(i + 1) (i + 1)(n − i) 1 + ··· + n −  .  (i + 1) · n+1 2 2 Therefore, ak ≤ k(n+1−k) , k  1, . . . , n. 2 12.14. Let the sequence of the positive numbers (an ) be such that a1 + · · · + am+1 ≤ Cam , (1) m  1, 2, . . . , and let us prove that C ≥ 4. Indeed, for m  2, 3, . . ., we have Sm+1 ≤ C(Sm − Sm−1 ), where Sn  a1 + · · · + an (n ∈ N). √ √ Therefore, C Sm ≥ Sm+1 + C Sm−1 ≥ 2 Sm+1 · C Sm−1 , whence C ≥ √ 2 Sm+1 ·Sm−1 . Sm

Proofs

185

  √ n √ √ ·S1 S1 n n+2 ·Sn Therefore, C ≥ 2 SSn+1 · · · · · 2 SS23 ·S1  2n · SSn+2 > 2 · , and S2 n+1 ·S2  √ √ hence C ≥ 2 · n q, where q  SS21 . Thus letting n → +∞, it follows that √ C ≥ 2, and we obtain that C ≥ 4. If we prove that for C  4 there exists a sequence of positive numbers (an ) such that inequality (1) holds, then we obtain that the smallest possible value of the number C is 4. For the sequence an  2n−1 , we have a1 + · · · + am+1  1 + · · · + 2m  2m+1 − 1 < 4 · 2m−1  4am , and hence a1 + · · · + am+1 < 4am . This ends the proof.

Problems for Independent Study u 2 +2

n 1. Consider the sequence (u n ) such that u 1  109 , u n+1  2u , n  1, 2, . . .. n √ −13 Prove that 0 < u 36 − 2 < 10 . 2. Let a1 , a2 , . . . , an be real numbers such that a1  0, |a2 | |a1 +1|, |a3 | |a2 +1|, …, |an | |an−1 + 1|. Prove that a1 +a2 n+···+an ≥ − 21 . 3. Consider the sequence a1 , a2 , . . . , an , . . . such that a1  1, an+1  an + a12 , n n  1, 2, . . .. Is this sequence bounded? 4. Suppose the sequence (an ) is nondecreasing and a0  0. Given that the sequence bn  an −an−1 , n  1, 2, . . ., is not increasing, prove that the sequence cn  ann , n  1, 2, . . ., is not increasing. xn4 +9 , n  1, 2, . . .. Prove 5. Consider the sequence (xn ) such that x1  2, xn+1  10x n 4 5 that 5 < xn ≤ 4 , for all n > 1.  2 a 6. Consider the sequence a1  1, an+1  a2n + 4n + a1n , n  1, 2, . . .. Prove that a250 < 10. 7. Consider the sequence x1  2, xn+1  xn2 − xn + 1, n  1, 2, . . .. Prove that x11 + x12 + · · · + x1n < 1.  8. Consider the sequences (an ) and (bn ) such that a1  1, an+1  an + an2 + 1, bn  a2nn , n  1, 2, . . .. Prove that the sequence (bn ) is bounded. √ √ 9. Given that R1  1, Rn+1  1+ Rnn , n  1, 2, . . ., prove that n ≤ Rn ≤ n +1.

a2

10. Given that an+1  an + nn , prove that there exists a number A such that 0 < n( A − an ) < A3 . 11. Consider the following sequence: a0  2, an+1  an + a1n , n  0, 1, . . .. Prove that 12 < a70 < 12.25. 12. Consider the following sequences: (an ) (an > 0, n  1, 2, . . .) and x1  1, x2  2, xn+2 + an xn+1 + xn  0, n  1, 2, . . .. Prove that the sequence (xn ) contains an infinite number of positive terms and an infinite number of negative terms.

186

12 Inequalities of Sequences

1 13. Consider the following sequence: a1  1, an  an−1 + an−1 , n  2, 3, . . .. √ √ Prove that 2n − 1 ≤ an ≤ 3n − 2. xn4 +9 14. Consider the following sequence: x1  2, xn+1  10x , n  1, 2, . . .. n −10 −10 Prove that 1 − 10 < x100 ≤ 1 + 10 . 2 − 2, n  1, 2, . . .. 15. (a) Consider the following sequence: a1 ≥ 2, an+1  an

Prove that

1 a1

+

1 a1 a2

+

1 a1 a2 a3

(b) Let a > 2 and a0  1, a1  a, Prove that

1 a0

+

1 a1

+ ··· +

1 ak

a2 1 < a21 − 41 − 1. a1 a2 ...an

a2 an+1  a 2 n − 2 an , n  1, 2, . . .. n−1

+ ··· + <

1 2



√ 2 + a − a2 − 4 .

16. For which values of x1 are all terms of the sequence xn+1  2xn2 − 1, n  1, 2, . . ., negative? n+1 1 17. Consider the following sequence: xn  . Prove that xn k −1 > kxn−1 , where i i2

n, k ∈ N and n > 1, k > 1.  18. Consider the following sequence: a1  1, an+1  (a1 + · · · + an )2 + 1, n  1, 2, . . .. Prove that a11 + a12 + · · · + a1n < 2.5. u n−1 , n  2, 3, . . .. 19. Consider the following sequence: u 1  1, u n  n! + n−1 n Prove that u11 + u12 + · · · + u1n < 1. 20. Consider a monotonic sequence (xn ) of positive numbers. Prove that xx21 + · · · +

n−1 ) 1) + xxn1 ≥ n + (x2x−x + · · · + (xnx−x , where n ≥ 2. 1 x2 n−1 x n xn x2 Hint. Let x1  1 + α1 , . . . , xn−1  1 + αn−1 . Then xxn1  (1 + α1 ) . . . (1 + αn−1 ).   21. Let 0 ≤ ai ≤ a, i  1, 2, . . ., and ai − a j  ≥ i+1 j , for all i < j. Prove that a ≥ 1. Hint. Let ai1 ≤ . . . ≤ ain , where {i 1 , . . . , i n }  {1, . . . , n}. Then by inequality (8.4) (see Chapter 8) and the problem, it follows that a ≥    the assumptions  of(n−1) 2     ain − ai1  ain − ain−1 + · · · + ai2 − ai1 ≥ n(n+1) . 22. Let the sequence (u n ) be such that |u m+n − u m − u n | ≤ mn , for all m > n, m, n ∈ N. Prove that u n  nu 1 , n  1, 2, . . .. Hint. Prove that lim (u m+n − u n )  u m , then u m+1  lim (u m+1+n − u n )  n→∞ n→∞  lim (u m+1+n − u n+1 ) + lim (u n+1 − u n )  u m + u 1 .

xn−1 xn

n→∞

2

2

n→∞

Chapter 13

Algebraic Inequalities in Number Theory

Inequalities arise not only in algebra, but very often in number theory as well. In this chapter we consider inequalities in number theory and prove them using algebraic inequalities. In order to explain how a large number of problems may be attacked, we provide a list of problems and their proof techniques.

Problems Prove the following inequalities (13.1–13.3). √  √  13.1. mn < 2 1 − 4n1 2 , where mn < 2, m, n ∈ N.  √   √  √   13.2. n d + 1 sin π dn  ≥ 1, where n, d ∈ N and d ∈ / N. 13.3. Let S, p, q ∈ N, and suppose that S is divisible by q and leaves a remainder p−1) if q > p, S < pq. 1 on division by p. Prove that S ≤ pq − q(q− p 13.4. Let m, k ∈ N and suppose that all prime divisors of the number m are k+1 less than or equal to an integer n such that m ≤ n 2 . Prove that m can be represented as the product of k positive integers each of which is less than or equal to n. 13.5. Let n ∈ N and let σ (n) denote the sum of all divisors of the number n (note that σ (3)  4, σ (6)  12, σ (12)  28). Prove that if σ (a) > 2a and b ∈ N, then σ (ab) > 2ab. 13.6. Find all numbers αsuch that  for1every positive integer n there exists a positive . integer m such as α − mn  < 3n 13.7. A number is called an interesting number if it is equal to the product of two primes. What is the greatest number of consecutive interesting numbers? n 13.8. Prove  there is an infinite number of positive integers n such that S(2 ) >  n+1that , where S(a) is the sum of the digits of the positive integer a. S 2 13.9. Let the positive integer n be represented as a sum of natural numbers. We denote the number of all such representations by P(n) (for example, P(4)  © Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_13

187

188

13.10.

13.11.

13.12. 13.13.

13 Algebraic Inequalities in Number Theory

5, since 4  4, 4  3 + 1, 4  2 + 2, 4  2 + 1 + 1, 4  1 + 1 + 1 + 1). Prove that P(n + 1) + P(n − 1) ≥ 2P(n), n  2, 3, . . . . Let the positive integer a be divided (long division) by a positive integer b yielding the number c0 .c1 c2 c3 . . .. Given that cn+1  cn+2   · · ·  cn+k  9 b , (b) k ≤ log b . for some n ∈ {0, 1, . . .}, k ∈ N, prove that, (a) k ≤ 10 Find for every positive integer n the greatest number f (n) such that from the numbers 1, 2, . . . n one can choose f (n) numbers such that no two of them have ratio equal to 2. Let n be a positive integer. Prove that n can be represented as the sum of the squares of m positive integers, where 3 log4 n + 5 ≤ m ≤ n − 14. Letb a and  b be distinct positive integers greater than or equal to 3. Prove that a − ba  > 0.62ba−1 .

Proofs √ 2−

√

2 2 2n √ −m n ( 2n+m )

1 ≥ n √2n+m , since 2n 2 − m 2 is a ) √ √ ( positive integer. On the other hand, m < 2n, and therefore 2 − mn ≥ √ 1√ √1 > n √2n+  √1 . Hence, we have obtained that 2 − mn > n ( 2n+m ) ( √  2n ) 2 2n 2 √1 , or m < 2 1 − 4n1 2 . n 2 2n 2

√ √ 13.2. Let dn  k, whence k < dn < k + 1. Consider the following two cases. √ (a) If k < dn < k + 21 , then

13.1. We have

m n



2n−m n



  √   √ √   √    n d + 1 sin(π dn)  n d + 1 sin π dn − k  ≥ √  √   √  dn 2 − k 2  √ 2n d + 2 ≥ n d + 1 2 dn − k  2n d + 2 √ ≥ √ > 1. dn + k dn + k

Here we have used that sin x ≥ π2 x, where 0 ≤ x ≤ π2 (this can be proved using derivatives).   √  √  √  (b) If k + 21 < dn < k + 1, then n d + 1 sin π k + 1 − dn  ≥ √ 2n √ d+2 k+1+ dn

> 1.

13.3. By to the assumption of the problem, we have that S can be represented in the following way: S  p(q − i) + 1  pq − ( pi − 1)  pq − q j, where i, j ∈ N. We have pi − 1  q j and q > p, and hence i > j, where i ≥ j + 1 and q j ≥ p( j + 1) − 1, or (q − p) j ≥ p − 1. Therefore, since q > p, we have p−1) . S  pq − q j ≤ pq − q(q− p 13.4. Consider all possible representations of the number m  a1 · · · · · ak , where a1 ≤ a2 ≤ . . . ≤ ak and a1 , a2 , . . . , ak ∈ N. Note that the number of such

Proofs

189

representations is finite. Let m  a1 a2 · · · ak be the representation among these representations such that ak is the smallest possible. If ak ≤ n, then this ends the proof. Assume that ak > n. Then ak is not a prime number, since m|ak . Let p be the smallest prime divisor of ak . Therefore, √ p ≤ ak . We have m  ( pa1 )a2 · · · · · ak−1 · apk , and hence pa1 ≥ ak , where a1 ≥ √ √ ak > n. √ k−1 k+1 Therefore, m  a1 a2 · · · · · ak−1 ak > n · n  n 2 , which leads to a contradiction. 13.5. Let the numbers 1  a1 < · · · < ak  a be all the divisors of a. Then the numbers ba1 < · · · < bak are the divisors of ab, whence σ (ab) ≥ ba1 + · · · + bak  bσ (a) > 2ab.   13.6. Without loss of generality one can assume that α ∈ [0, 1), since α − mn     {α} − m−[α]n . n Note that α  0 satisfies the condition of the problem, and it is sufficient to take m  0. Let us prove that if 0 < α < 1, then there exists a positive integer 1 n such that for every integer m we have α − mn  ≥ 3n . p If α ∈ Q, then α  q , where 0 < p < q, p, q ∈ Z. Assume that α satisfies the condition of the problem. If n  q, q + 1, . . . , 2q, then the numbers qα to the union of thesets  . . . , 2qα  2 p belong    1p, (q +1 1)α, p − 3 , p + 3 , p + 1 − 13 , p + 1 + 13 , . . . , 2 p − 13 , 2 p + 13 . Since the number of these numbers is equal to q + 1, and the number of intervals is equal to p + 1, we have according to the Dirichlet’s principle that there exist i, j ∈ {q, q + 1, . . . , 2q} i  j, such that the numbers αi , α j belong to the same interval. Hence we obtain that α < 23 . On the other hand, there exists a number l ∈ {q, q + 1, . . . , 2q} such that the numbers αl and α(l + 1) belong to different intervals. Therefore α > 13 . Hence if 13 < α < 23 , then for n  1 there does not exist an integer m such that |α − m| < 13 . This leads to a contradiction. If α is an irrational number, then there exists n 0 ∈ N such that l + 25 < αn 0 < l + 35 , where l ∈ Z (see Problem 14.25).    m0  Assume that m 0 ∈ Z and α − n 0  < 13 . If l  m 0 , then |αn 0 − m 0 |  |αn 0 − l + l − m 0 | ≥ |l − m 0 | − |αn 0 − l| > 1 − 35  25 > 13 . If l  m 0 , then |αn 0 − m 0 |  |αn 0 − l| > 25 > 13 . Thus, only integers satisfy the condition of the problem.     Alternative solution. Let n ∈ N, k, m ∈ Z and α − m  < 1 and α − k  < 1 . 6n

If k  2m, then |k − 2m|≥ 1; thus 1 1 1 + 6n  2n , 3n

n

1 2n

 ≤  mn −

3n

  k  ≤ α − 2n

2n

   k  m + n − α < 2n

190

13 Algebraic Inequalities in Number Theory

  which leads to a contradiction. Let k  0, 1, 2, . . ., α − m2kk  < 3·21 k , and m k ∈ Z. Then m 0  m21  · · ·  m2kk  · · ·, and hence |α − m 0 |< 3·21 k . Therefore α  m 0 ∈ Z.   If α ∈ Z, then it is obvious that for m  αn we have m ∈ Z and α − mn   1 0 < 3n . 13.7. Let us provide an example of three consecutive interesting numbers. 33  3 · 11, 34  2 · 17, 35  5 · 7. Let us prove that there do not exist four consecutive interesting numbers. We argue by contradiction. Assume that there exist four consecutive interesting numbers, Then one of these numbers is divisible by 4. Therefore, that number is equal to 4. Thus, it follows that one of the numbers is equal to either 5 or 3, neither of which is an interesting number. This is thedesired contradiction. 13.8. Assume that n satisfies the inequality S(2n ) > S 2n+1 . Therefore, there exists a positive integer n 0 such that for all n ≥ n 0 , n ∈ N,     S 2n ≤ S 2n+1

(1)

  Note that S(2n )  S 2n+1 is impossible, for otherwise, we would have that 2n  2n+1 − 2n is divisible by 9, which leads to a contradiction. Let us prove the following lemmas.   Lemma 1 If n ≥ n 0 , then S 2n+6 ≥ S(2n ) + 27. Indeed, on dividing the numbers 2n , 2n+1 , 2n+2 , . . . , 2n+6 by 9, we obtain as remainders seven consecutive terms of the following periodic sequence: 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, . . .. Hence from (1), it follows that                    S 2n+6 − S 2n  S 2n+6 − S 2n+5 + S 2n+5 − S 2n+4 + . . . + S 2n+1 − S 2n ≥ ≥ 1 + 2 + 4 + 8 + 7 + 5  27.

Using Lemma 1 and induction, one can easily prove the following Lemma 2.   Lemma 2 S 2n 0 +6k ≥ S(2n 0 ) + 27k, where k ∈ N.   Lemma 3 S 2n 0 +6k < 9m + 18 k, where k ∈ N, and m is the number of digits of 2n 0 . Indeed, we have 2n 0 < 10m , where 2n 0 +6k < 10m+2k ; hence the number of the digits of 2n 0 +6k is not greater than (m + 2k), and therefore, S 2n 0 +6k < 9(m + 2k). Note that Lemmas 2 and 3 contradict each other. This leads to a contradiction. 13.9. Let x1 , x2 , . . . , xk ∈ N, x1 ≥ x2 ≥ · · · ≥ xk , and n  x1 + x2 + · · · + xk . Therefore, n + 1  x1 + x2 + · · · + xk + 1; hence one can create a one-to-one correspondence between partitions of the number n and the partitions of the number n + 1 that contain 1. It follows that 

   x1 , x2 , . . . , xk ←→ x1 , x2 , . . . , xk , 1 .

Proofs

191

Hence, the number p(n + 1) − p(n) is equal to the number of the partitions of n + 1 not containing 1. If n  y1 + y2 +· · ·+ yl , where y1 ≥ y2 ≥ · · · ≥ yl ≥ 2, then n +1  (y1 + 1)+ y2 + · · · + yl , and the number of partitions of the number n + 1 not containing 1 is greater than or equal to the number of similar partitions of the number n. Therefore, p(n + 1) − p(n) ≥ p(n) − p(n − 1), or p(n + 1) + p(n − 1) ≥ 2 p(n), n  2, 3, . . .. 13.10. Let   b  a . . . c c . . . n+1 n+2 .. . − −

10an+1 cn+1 b 10an+2 cn+2 b

..

.

We have that 10an+1 − 9b  an+2 , 10an+2 − 9b  an+3 , . . . , 10an+k − 9b  an+k+1 , ≤ where 0 < an+i < b, i  1, 2, . . . , k,an+k+1 ≥ 0, whence 9b 10 an+1 , an+2 , . . . , an+k < b. b (a) Assume that k > 10 . Then among the numbers an+1 , an+2 , . . . , an+k there are two equal numbers. Therefore, b b −1 <  10 −1, hence A ≤ 10 A  max an+i − min an+i ≤ b−1− 9b 10 1≤i≤k

1≤i≤k

k − 1. It follows that there exist numbers p < q such that an+ p  an+q . Hence, we obtain 10an+ p − 9b  an+ p+1 , 10an+ p+1 − 9b  an+ p+2 , . . . , 10an+q−1 − 9b  an+q  an+ p , and summing these equations, we deduce that an+ p + . . . + an+q−1  (q − p)b, which leads to a contradiction. Hence, we obtain b . that k ≤ 10 (b) Define an+i  b − bn+i i  1, 2, . . . , k + 1. It then follows that 1 ≤ bn+1 , bn+2 , . . . , bn+k < b and 1 ≤ bn+k+1 ≤ b. Note that bn+2  10bn+1 , bn+3  10bn+2 , . . . , bn+k+1  10bn+k  , and  therefore, bn+k+1  10k bn+1 , where 10k ≤ 10k bn+1 ≤ b, or k ≤ log b . . . . 9, then Remark If b  10k and a  99  k

a b

 0, 99 . . . 9.  log b

13.11. Consider the set X  {4α q | 4α q ≤ n, α ∈ Z0 , q odd}. It is clear that the ratio of two elements of the set X is never equal to 2, and therefore f (n) ≥ |X |, where we denote by |X | the cardinality of the set X. Let Υ ⊂ {1, 2, . . . , n} and |Υ | > |X |, and let us put every number a from X into correspondence with the pair (a, 2a). We have that 2a ∈ / X and |{(a, 2a)}|  |X |. Therefore, for some a ∈ Υ , we have 2a ∈ Υ . It follows that f (n) ≤ |X |; hence f (n)  |X |.

192

13 Algebraic Inequalities in Number Theory

Let n  2k1 +2k2 +. . .+2km , where k1 > k2 > . . . > km ≥ 0 and k1 , k2 , . . . , km are integers. Consider the following sets: X 0  {2k1 + 2k2 + · · · + 2ki |i ∈ {1, 2, . . . , m}, ki even}, X i  {2r1 + 2r2 + · · · + 2rs |s ≥ i, r1 > r2 > . . . > rs ≥ 0, rs even and r1  k1 , r2  k2 , . . . , ri−1  ki−1 , ri < ki , } i  1, 2, · · · , m. Obviously, X 0 , X 1 , . . . , X m are mutually disjoint sets and X  X 0 ∪ X 1 ∪ · · · ∪ X m . Therefore, |X |  |X 0 | + |X 1 | + · · · + |X m |. Let us denote by λ the number of even numbers among the numbers k1 , k2 , . . . , km and the number of odd numbers by μ. We have |X 0 |  λ. Let us prove that ⎧ k +1 2 i −1 ⎪ ⎪ if ki is odd, ⎨ 3 |X i |  ki +1 ⎪ −2 ⎪ ⎩2 if ki is even. 3 Indeed, let ki be an even number. We have that 2ri + ... + 2rs ≤ 2ki −1 + ... + 20  2ki − 1. Note that |X i | is one less than the number of numbers from 1, 2, 3, · · · , 2ki whose canonical decomposition contains an even power of 2, that is, 2ki −2ki −1 + 2ki −2 − 2ki −3 + · · · + 22 − 2 + 1 − 1 

2ki +1 − 2 (−2)ki +1 − 1 −1 . −2 − 1 3

If ki is odd, then the proof can be obtained in a similar way. Thus, it follows k1 km  23 n + (−1) +···+(−1) . that f (n)  λ + 23 n − 23 λ − μ3  23 n + λ−μ 3 3 2 2 2 2 2 2 . . + 1, n  3 + 3 + 1 + . . . + 1 and n  22 + 22 + 13.12. We have that n  1 + . n

n−18

32 + 12 + . . . + 12 . Therefore, n can be represented as the sum of the squares n−17

of m positive integers, where m ∈ {n − 16, n − 14, n}. 2 , then n  On the other hand, if n  4k + 4k + 4k + 4k + x12 + . . . + xm−4 2 2 k+1 4 + x1 + . . . + xm−4 , where k  0, 1, . . . ; hence n can be represented as the sum of the squares of m − 3 positive integers. · · + 12 , it follows that n can Thus, taking into consideration that n  12 + · n

be represented as the sum of the squares of m positive integers, where m ∈ {n, n − 3, n − 6, . . .} and m ≥ c0 +c1 +· · ·+ck , where ck ck−1 . . . c0 represents n written in the base-4 number system.

Proofs

193

  Therefore, c0 +c1 +· · ·+ck ≤ 3(k + 1) and 4k ≤ n < 4k+1 ; hence k  log4 n . In order to complete the proof, it is sufficient to note that {n, n − 3, n − 6, . . .} ∪ {n − 14, n − 17, n − 20, . . .} ∪ {n − 16, n − 19, n − 22, . . .} ⊃     ⊃ n − 14, n − 15, n − 16, . . . , 3 log4 n + 5 .

13.13. Let us prove the following lemmas. n  Lemma 1 If n ∈ N, then 1 + n1 < 3. By mathematical induction, one can easily prove that if k ∈ N and n  1, 2, . . . , k, then   1 n n n2 1+ <1+ + 2 (1) k k k  n If n  1, then we have 1 + k1  1 + k1 < 1 + k1 + k12 . Assume that for n  m ≤ k − 1, inequality (1) holds, and let us prove that it holds for n  m + 1.   m+1   m     2 Note that 1 + k1  1 + k1 · 1 + k1 < 1 + k1 · 1 + mk + mk 2 < 1 + + (m+1) , since m 2 < (m + 1)k. k2 This ends the proof of inequality (1). k  For n  k we obtain 1 + k1 < 3. m+1 k

2

Lemma 2 If a, b ∈ N and a > b ≥ 3. Then ba−1 > a b−1 . n  For n ∈ N and n ≥ 3, using Lemma 1, we deduce that n ≥ 3 > 1 + n1 . Therefore, 1

1

n n > (n + 1) n+1 . 1

(2)

1

Thus, from (2) it follows that b b > a a , and therefore ba > a b > b · a b−1 , whence ba−1 > a b−1 . From Lemma 2 it follows that without loss of generality one can assume that a > b. Let a  b + c, where c ∈ N. Consider the following two cases: (a) b ≥ 4. We obtain         c b  b    − bc   a − ba   (b + c)b − bb+c   bb  1 + b   c−1      c b c  bc c  bc c  bc  c−1 + bc−2 1 + + ... + 1 + ≥ bb |b − (1 + ) c |bc−1   bb b − 1 +  b b b b b           c  bc  c  bc     ba−1 b − 1 + a b − ba  > ba−1 b − 1 + , . b b

therefore,

(3)

194

13 Algebraic Inequalities in Number Theory

For c > 1, using Problem 2.1 for (1 + c/b), . . . , (1 + c/b), 1, we deduce   c−1  b  c−1  c−1 b   1+ c +...+ 1+ c +1 that ( b ) c ( b ) > c 1 + bc , or 1 + c−1 > 1 + bc c . b  b b  Hence, we have 1 + bc c ≤ 1 + b1 < 3, and therefore,    b  c c b − 1 + b  > 1.   From (3), it follows that a b − ba  > ba−1 . (b) If b  3, then by induction one can easily prove that 2.38 · 3a−1 > a 3 , where a  4, 5, . . .. This ends the proof.

Problems for Independent Study Prove the following inequalities (1–4).   √ √ √    p p ≤ ≤ 1. , where a, p, q ∈ N, a ∈ / N and a − 1.  a − p+aq   p+q q q 2. (a) S(mn) ≤ S(m)S(n), where n, m ∈ N. (b) S(m + n) ≤ S(m) + S(n), where n, m ∈ N (see Problem 13.8). 3. S(1981n ) ≥ 19, where n ∈ N. 4. S(1998n ) > 106 for all positive integers n starting from some number. > 5. Prove that for every number M there exists a positive integer n such that SS(n) (n 2 ) M. 6. Prove that 52l can be represented as the sum of the squares of m numbers, where l ∈ N and 1 ≤ m ≤ 2l . 7. Find the smallest positive integer n such that for every partition of the positive integers 1, 2, . . . , n into two groups, in one of them there are three numbers forming a geometric progression. 8. Let a1 , a2 , . . . , ak be integers such that 1 < a1 < a2 < · · · < ak ≤ n and a1 · a2 · · · ak is not divisible by ai2 , i  1, 2, . . . , k. Prove that k ≤ π(n), where π (n) is the number of primes less than or equal to n. S(n) is bounded if and 9. Let k be a given positive integer. Prove that the sequence S(kn) α β only if k  2 · 5 , where α, β ∈ {0, 1, . . .}. 10. Let x ≥ 1 and let A(x) be the least common multiple of the numbers 1, 2, . . . , [x]. Prove that   (a) A x2 · [x]! ≥ A(x), where x ≥ 2, ([ x2 ]!)2 x (b) A(x) < 5 , ) ≥ A(n) · (2n)! , where n ≥ 2, n ∈ N, (c) A(2n) · A( 2n 3 √  (n!)2 (d) A(2n) > A(n) A 2n , where n ∈ N, n ≥ 100,

Problems for Independent Study

195

(e) there exists a prime number belonging to (n, 2n), where n ∈ N, n > 1. Remark If we assume that belonging to (n, 2n),  is no primenumber √there √  prove that A(2n)|A(n)A 2n (i.e., A(n)A 2n divides A(2n)). 11. Prove that d(1) + d(2) + · · · + d(n) < 2.78, where d(k) is the number of the positive 12 22 n2 integer divisors of k.   ∞ 2  ∞  ∞   ∞  1  1  1  d(k) . Remark   2 2 2 2 k i j k k1

i1

j1

k1

12. Find all numbers α such that the following holds: for every positive  condition  1 integer n there exists an integer m such as α − mn  ≤ 3n . 13. Let a1 ≤ a2 ≤ · · · ≤ an be positive integers such that no sum of some of them is equal to the sum of some of the others. Prove that (a) a1 + a2 + · · · + ak ≥ 2k − 1, k  1, 2, . . . , n, 1 . (b) a11 + a12 + · · · + a1n ≤ 1 + 21 + · · · + 2n−1 14. Let a1 , . . . , an are distinct positive integers. Prove that a13 + · · · + an3 ≥ (a1 + · · · + an )2 . Remark Let a1 < · · · < an and ai  i + bi . Then b1 ≤ · · · ≤ bn .

Chapter 14

Miscellaneous Inequalities

In this chapter we consider miscellaneous inequalities, and we mostly use various proof techniques not included in the previous chapters.

Problems Prove the following inequalities (14.1–14.5).    14.1. a12 + b21 + · · · + an2 + b2n ≥ (a1 + · · · + an )2 + (b1 + · · · + bn )2 .  √ 2 2 a2 +b2 ≤ − ab ≤ √(a−b) , where a, b > 0. 14.2. (a−b) 2(a+b) 2 2(a+b) 1

1

1

1

14.3. x2n − x1n ≤ (x2 − α) n − (x1 − α) n , where 0 ≤ α ≤ x1 ≤ x2 , n ∈ N.    14.4.  n a1n + · · · + akn − n bn1 + · · · + bnk  ≤ |a1 − b1 | + · · · + |ak − bk |, where n ≥ 2, n ∈ N, a1 > 0, . . . , ak > 0, b1 > 0, . . . , bk > 0. 14.5. 1+1√3 + √5+1√7 + · · · + √9997+1√9999 > 24. 14.6. Given that a2 − 4a + b2 − 2b + 2 ≤ 0, c2 − 4c + d 2 − 2d + 2 ≤ 0 and e2 − 4e + f 2 − 2f + 2 ≤ 0, find the greatest possible value of the expression (a − c)(f − d ) + (c − e)(b − d ). 14.7. Prove that if an increasing sequence of positive numbers a1 , a2 , . . . , an , . . . is unbounded, then there exists a sufficiently large index k such that the following inequalities hold: (a) (b)

a1 a2 a1 a2

k + aa23 + · · · + aak+1 < k − 21 , a2 ak + a3 + · · · + ak+1 < k − 1985.



n 

14.8. Minkowski’s inequality: Prove that (ai + bi ) i1  n  1r  r bi , where ai ≥ 0, bi ≥ 0, i  1, . . . , n.

r

 1r

 ≤

n 

i1

air

 1r +

i1

© Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_14

197

198

14 Miscellaneous Inequalities

14.9. Young’s inequality: Prove that if the function f (x) is increasing and contina b uous on [0, +∞) and f (0)  0, then f (x)dx + f −1 (y)dy ≥ ab, where the 0

0

function f −1 (x) is the inverse function of the function f (x) and a > 0, b > 0. 14.10. (a) Prove that x1 y1 + · · · + xn yn  x1 (y1 − y2 ) + (x1 + x2 )(y2 − y3 ) + . . . + (x1 + · · · + xn−1 )(yn−1 − yn ) + (x1 + . . . + xn )yn . (b) Prove that if x1 ≥ x2 ≥ · · · ≥ xn and y1 ≥ y2 ≥ · · · ≥ yn , then x1 yn + x2 yn−1 + · · · + xn y1 ≤ x1 yi1 + x2 yi2 + · · · + xn yin ≤ x1 y1 + · · · + xn yn , where i1 , i2 , . . . , in is some permutation of the numbers 1, 2, . . . , n. 14.11. Prove that (a) x1 + · · · + xn ≥ n, where n ≥ 2, x1 > 0, . . . , xn > 0, and x1 · · · xn  1, (b) a1 (a12 +1) +· · ·+ an−1 (a1 n +1) + an (a11 +1) ≥ 1+a1n···an , where n ≥ 2, (n−3)(ai −1) ≥ 0, ai > 0, i  1,

. . . , n, +1)(S+1) , where (c) 3 + (A + M + S) + A1 + M1 + S1 + MA + MS + AS ≥ 3(A+1)(M AMS+1 A > 0, M > 0, S > 0,

1

1 1 9 + 1+b + 1+c , where a > 0, b > 0, c > 0, ≥ 1+abc (d) a1 + 1b + 1c 1+a (e)

√a a+b

+

√b b+c

+

√c c+a

≥

√ √ √ a+ b+ c √ , 2

where a > 0, b > 0, c > 0.

14.12. Schur’s inequality: Prove that ϕ(x1 ) + · · · + ϕ(xn ) ≥ ϕ(y1 ) + · · · + ϕ(yn ), where y1 ≥ y2 ≥ · · · ≥ yn , x1 + · · · + xk ≥ y1 + · · · + yk , k  1, . . . , n − 1, x1 + · · · + xn  y1 + · · · + yn , and for every x in I, one has ϕ  (x) > 0 (x1 , . . . , xn , y1 , . . . , yn ∈ I ). 14.13. Prove that b1 + · · · + bn ≥ a1 + · · · + an , where a1 ≥ · · · ≥ an > 0, b1 ≥ a1 , b1 b2 ≥ a1 a2 , . . . , b1 · . . . · bn ≥ a1 · . . . · an . √ 14.14. Let α, β, γ be angles of an acute triangle. Prove that 2 < sin α2 + sin β2 + sin γ2 ≤ 23 . √ √ 14.15. Let angles of an obtuse triangle. Prove that 0 < sin α + sin β + √ α, β, γ be √ sin γ < 1 + 4 8. √ 14.16. Prove that 1 + √12 + √13 + · · · + √1n > 2 n − 2, where n ∈ N. 14.17. Prove that a > 0, b > 0, c > 0, d > 0, if S1  a + b + c + d , S2  ab + bc + cd + ad + ac + bd > 0, S3  abc + bcd + cda + abd > 0, S4  abcd > 0.

n 14.18. Prove that n! < n ne , where n ≥ 8 and n ∈ N.



1 14.19. Prove that 1 + 21 · · · 1 + 21n < 3, where n ∈ N. 14.20. Prove that if g is a convex function on [0, a1 ] (see Chapter 11) and a1 ≥ a2 ≥ · · · ≥ a2m ≥ a2m+1 ≥ 0, then g(a1 ) − g(a2 ) + g(a3 ) − · · · + g(a2m+1 ) ≥ g(a1 − a2 + a3 − · · · + a2m+1 ). 14.21. Let the absolute value of the polynomial p(x) on [−1, 1] be less than or equal to 1. Prove that (a) |a| + |b| + |c| ≤ 3, where p(x)  ax2 + bx + c, (b) a2 + b2 + c2 ≤ 5, where p(x)  ax2 + bx + c, (c) |a| ≤ 4, where p(x)  ax3 + bx2 + cx + d .

Problems

199

14.22. Let the polynomial p(x)  xn + an−1 xn−1 + · · · + a0 (n > 1) have n negative roots. Prove that a1 p(1) ≥ 2n2 a0 . 14.23. The complete graph on n vertices has each of its edges colored in one of two given colors. We denote by t(n) the number of triangles whose sides are colored in the same color. Prove that ⎧ ⎪ , if n  2k, ⎪ k(k−1)(k−2) ⎪ 3 ⎨ t(n) ≥ 23 k(k − 1)(4k + 1), if n  4k + 1, ⎪ ⎪ ⎪ ⎩ 2 k(k + 1)(4k − 1), if n  4k + 3. 3 14.24. Consider a graph having n (n ≥ 3) vertices. The number of its edges is 2 greater than n4 . Prove that the number of triangles formed by these edges is n not less than 2 . 14.25. Prove that if α, (α > 0) is an irrational number and 0 < a < b < 1, then there exists a positive integer n such that a < {nα} < b. 14.26. Let x1 + · · · + xn  0 and xi ∈ [m, M ], i  1, . . . , n. Prove that (a) (b)

n  i1 n  i1

xi2 ≤ −mMn,

xi4 ≤ −mMn · m2 + M 2 + mM .

m 14.27. Prove that x2 + y2 ≥ 2m xm ym + (xm − ym )2 , where m ∈ N. 14.28. Prove that cos(α − β)cos(β − γ )cos(γ − α) ≥ 8 cosα cosβ cosγ , where α, β, γ arethe angles of some triangle.  14.29. Prove that sin x + sin22x + · · · + sinnnx  < 3. 2 a2 +a3 1 14.30. Prove that a1 +a · 2 · · · an−12+an · an +a ≤ a1 +a32 +a3 · a2 +a33 +a4 · · · an +a31 +a2 , where 2 2 0 < a1 ≤ a2 ≤ · · · ≤ an and n ≥ 3. 14.31. Prove that if x, y, z ≥ 0 and x2 + y2 + z 2  1, then (a) 1 ≤ (b) 1 ≤

x 1−yz x 1+yz

+ +

√ y z + 1−xy ≤ 323, 1−zx √ y z + 1+xy ≤ 2. 1+zx

14.32. For which values of λ does the inequality √

a a2

+ λbc

+√

b b2

+ λca

+√

c c2

+ λab

≥√

hold for arbitrary positive numbers a, b, c? 14.33. Prove that for every triangle with sides a, b, c, one has √

√ √ a2 + ab + b2 + b2 + bc + c2 + c2 + ca + a2 √ ≤ 5a2 + 5b2 + 5c2 + 4ab + 4bc + 4ca.

3 1+λ

(14.1)

200

14 Miscellaneous Inequalities

14.34. Prove that x13 +···+xn3 x1 ···xn x14 +···+xn4 x1 ···xn

(a) (b)

≥ ≥

(1−x1 )3 +···+(1−xn )3 , where n (1−x1 )···(1−xn ) (1−x1 )4 +···+(1−xn )4 , where n (1−x1 )···(1−xn )

≥ 3, 0 ≤ xi ≤ 21 , i  1, . . . , n, ≥ 4, 0 ≤ xi ≤ 21 , i  1, . . . , n.

14.35. Let n ≥ 2 and let a1 , . . . , an be distinct positive integers. Prove that a12015 + · · · + an2015 12015 + · · · + n2015 ≥ . 2000 12000 + · · · + n2000 a1 + · · · + an2000  √   √  14.36. Prove that 1 + x2 + 1 + y2 + (1 − x)2 + (1 − y)2 ≥ 1 + 5 (1 − xy), where 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. 14.37. Prove that if xi > 0, i  1, . . . , n and x1 · · · xn  1, then √1 1+x1 √1 1+x1

(a) (b)

+ ··· + + ··· +

14.38. Prove that n  xiα −xi i1

x1 +···+xn +xiα −xi

√1 1+xn √1 1+xn

≤

√

2 2

· n, where n  2, 3, . . .,

< n − 1, where n ≥ 4.

≥ 0, where n ≥ 2, xi > 0, i  1, . . . , n, x1 · · · xn ≥ 1,

and α ≥ 1.

Proofs 14.1. Consider the points A1 (0, 0), A2 (a1 , b1 ), A3 (a1 + a2 , b1 + b2 ), . . . , An+1 (a1 + · · · + an , b1 + · · · + bn ).

Since A1 A2 + A2 A3 + · · · + An An+1 ≥ A1 An+1 , it follows that

    a12 + b21 + a22 + b22 + · · · + an2 + b2n ≥ (a1 + · · · + an )2 + (b1 + · · · + bn )2 .

 14.2. Let us rewrite the expression 

 a2

+ b2 2

−

√ ab 

a2 +b2 2

a2 +b2 2

−

√ ab in the following way:

  √ √ a2 +b2 − ab + ab 2 (a − b)2 .    √ √ a2 +b2 a2 +b2 + ab 2 + ab 2 2



√ + ab ≤ a + b. The left-hand side of this  2 √ 2 2 2 a2 +b2 inequality obviously holds, since + ab > a +b + ab  (a+b) . 2 2 2

It is left to prove that

a+b √ 2

≤

a2 +b2 2

In side, let us prove the following inequality:  order to prove the right-hand √ 2 2 a2 +b2 a+b a+b  ≤  (a−b) − 2 ≤ 2 − ab, or (a−b) √ . a+b 2 2 2 4

a +b 2

+ a+b 2

4

2

+ ab

Proofs

201

 Note that this inequality holds, because 14.3. We have

a2 +b2 2

≥

√ ab.

1 1 x2 − x1 x2n − x1n   n−1  n−2  1 n−1 1 1 1 + x2n · x1n + · · · + x1n x2n

x2 − x1 ≤       1 n−1 1 n−2 1 1 n−1 + (x2 − α) n · (x1 − α) n + · · · + (x1 − α) n (x2 − α) n 1

1

 (x2 − α) n − (x1 − α) n .

  14.4. Let us rewrite the expression n a1n + · · · + akn − n bn1 + · · · + bnk in the following way:   an + · · · + an − bn − · · · − bn   n n 1 1 k k n n n n ,  a1 + · · · + ak − b1 + · · · + bk   An−1 + An−2 B + · · · + Bn−1   where A  n a1n + · · · + akn , B  n bn1 + · · · + bnk . Therefore, |A − B| 

       + · · · + (ak − bk ) akn−1 + akn−2 bk + · · · + bn−1  (a1 − b1 ) a1n−1 + · · · + bn−1 1 k

An−1 + An−2 B + · · · + Bn−1   n−1   n−1   an−1 + · · · + bn−1  a   k + · · · + bk   1 |a | + · · · + ≤ |a1 − b1 | ·  1n−1 − b ·    k k A  An−1 + · · · + Bn−1  + · · · + Bn−1  ≤ |a1 − b1 | + · · · + |ak − bk |,

as A > ai > 0, B > bi > 0 (i  1, . . . , n). 14.5. We have  1 1 1 √ +√ √ √ + ··· + √ 1+ 3 9997 + 9999 5+ 7 1 1 1 1 1 1  +√ √ + √ +√ √ √ √ +√ √ + ··· + √ 1+ 3 1+ 3 9997 + 9999 9997 + 9999 5+ 7 5+ 7 1 1 1 1 1 1 > +√ √ ++√ √ + ··· + √ √ √ √ +√ √ +√ 1+ 3 7+ 9 9997 + 9999 9999 + 10001 3+ 5 5+ 7 √ √ √ √ √ √ √ √ 3−1 5− 3 7− 5 10001 − 9999 10001 − 1 100 − 1 + + + ··· +  > > 48,  2 2 2 2 2 2 

2·

whence 1+1√3 + √5+1√7 + · · · + √9997+1√9999 > 24. 14.6. We have that the points A(a, √ b), B(c, d ), C(e, f ) are inside the circle with center O(2, 1) and radius 3.

202

14 Miscellaneous Inequalities

Consider

the

vectors

→

→

→

→

BA{a − c, b − d }, BC{e − c,⎛ f − d} →

→

∧

⎞and

→

BD{f − d , c − e}. Note that BC · BD  0, and therefore, ⎝BA, BD⎠  ⎛ ⎞ ⎛ ⎞ ∧ ∧ → → → →  ◦  90 − β , or ⎝BA, BD⎠  |90◦ + β|, where ⎝BA, BC ⎠  β.   → →  It follows that BA · BD  BA · BD · |cos(90◦ ± β)|  BA · BD sin β  BA · BC sin β  2SABC . √ √ Hence, |(a − c)(f − d ) + (c − e)(b − d )|  2SABC ≤ 2 · 9 4 3  9 2 3 , since √ among all triangles inscribed in a circle of radius 3, the one with the largest area is an equilateral triangle (see the proof of Problem 2.7). It is left to note that the expression (a − c)(f − d )+(c − e)(b − d ) can assume √ the value 9 2 3 . √

Thus, the greatest possible value of the given expression is 9 2 3 . k  1 + αk , then we have 14.7. (a) If we define aa21  1 + α1 , aa23  1 + α2 , . . . , aak+1 −1 < α1 < 0, . . . , −1 < αk < 0. 1 . We have (1 + α1 ) · · · (1 + αk )  aak+1 Since the sequence (ak ) is unbounded, there exists a number k0 such that 1 < 21 , and therefore, 21 > (1 + α1 ) · · · (1 + αk ) ≥ 1 + α1 + k ≥ k0 aak+1 k · · · + αk (see Problem 10.6), and hence aa21 + aa23 + · · · + aak+1 < k − 21 . ak1 a1 (b) Let the numbers k1 , k2 , . . . , k2·1985 be such that a2 + · · · + ak +1 < k1 − 21 , ak1 +1 a + · · · + ak k2+1 < (k2 ak1 +2 2 1  2 · 1985). 2 (S

ak

1

s − k1 ) − 21 , . . . , aks−1 +2 + · · · + aak k+1 < (ks − ks−1 ) − s−1

+1

s

Summing these inequalities, we obtain a1 s + aa23 + · · · + aak k+1 < ks − S · 21  ks − 1985; hence for k ≥ kS , it follows a2 s k m that aa21 + aa23 + · · · + aak+1 < k − 1985, since aam+1 < 1. n n n    r r−1 r−1 ai (ai + bi ) + bi (ai + bi ) . 14.8. (ai + bi )  i1

i1

i1

By Problem 11.12, we have n 

ai (ai + bi )

r−1

≤

i1

 n 

air

i1

1/r  n (r−1)/r 

r−1 r/(r−1) · (ai + bi ) i1

and n  i1

bi (ai + bi )

r−1

≤

 n  i1

bri

1/r  n (r−1)/r 

r−1 r/(r−1) · . (ai + bi ) i1

Proofs

203

It follows that n 

(ai + bi ) ≤ r

i1

 n 

(r−1)/r ⎛ (ai + bi )

⎝

r

i1

n 

1/r air

+

i1

 n 

1/r ⎞ bri

⎠,

i1

or  n 

1/r (ai + bi )

≤

r

 n 

i1

1/r air

+

 n 

i1

14.9. For b < f (a) we have that

a

ab  S1 + S3 , and therefore, ab ≤

a

f (x)dx +

0

For the case b ≥ f (a) the proof is similar. 14.10. (a)

.

i1

b

f (x)dx  S1 + S2 ,

0

1/r bri

b

f −1 (y)dy  S3 , where

0

f

−1

(x)dx.

0

x1 (y1 − y2 ) + (x1 + x2 ) (y2 − y3 ) + · · · + (x1 + · · · + xn−1 ) (yn−1 − yn ) + (x1 + · · · + xn ) yn  x1 y1 + (x1 + x2 − x1 ) y2 + · · · + ((x1 + · · · + xn ) − (x1 + · · · + xn−1 )) yn  x1 y1 + x2 y2 + · · · + xn yn . (b) We have



x1 yi1 + · · · + xn yin  yi1 (x1 − x2 ) + yi1 + yi2 (x2 − x3 ) + · · · + yi1 + · · · + yin−1 (xn−1 − xn )

+ yi1 + · · · + yin xn ≤ y1 (x1 − x2 ) + (y1 + y2 )(x2 − x3 ) + · · · + (y1 + · · · + yn−1 )(xn−1 − xn ) + (y1 + · · · + yn )xn  x1 y1 + · · · + xn yn .

The second inequality can be proved in a similar way. , where a1 > 0, . . . , an > 0. 14.11. (a) Let x1  aa21 , x2  aa23 , . . . , xn−1  aan−1 n an In this case, xn  a1 . Using Problem 14.10(b) for the numbers a1 , a2 . . . , an and a11 , a12 , . . . , a1n , we obtain a1 · a11 + a2 · a12 + · · · + an · a1n ≤ a1 · a12 + a2 · a13 + · · · + an · a11 , or n ≤ x1 + x2 + · · · + xn . (b) Note that   1 1 + ··· + (1 + a1 · · · an ) a1 (a2 + 1) an (a1 + 1) 1 a2 · · · an a1 a2 · · · an−1 1 + ··· + + + ··· +  A.  a1 (a2 + 1) an (a1 + 1) a2 + 1 a1 + 1 For n ≥ 3 we obtain A ≥

a2 a2 a3 an 1 +· · ·+ an (a11 +1) + aa11+1 + a2 +1 +· · ·+ aan1+1 a1 (a2 +1)

 B.

204

14 Miscellaneous Inequalities

One can easily prove that a1 , b and order; hence

1 , a , (a, b b+1 a+1

> 0) have the same

1 ab 1 a 1 + ≥ · + ·b a(b + 1) a + 1 a a+1 b+1

(1)

(see Problem 14.10(b)). By (1), we obtain       a2 1 a3 1 a1 1 + + + + ··· + + B≥ a1 + 1 a2 + 1 a2 + 1 a3 + 1 an + 1 a1 + 1     a1 1 an 1 + + ··· + +  n.  a1 + 1 a1 + 1 an + 1 an + 1   Therefore, (1 + a1 · · · an ) a1 (a12 +1) + · · · + an (a11 +1) ≥ n. For n  2, we have a1 , a2 ≤ 1; hence   1 1 a1 a2 1 +  + + (1 + a1 a2 ) a1 (a2 + 1) a2 (a1 + 1) a1 + 1 a2 + 1 a1 (a2 + 1) a1 a2 1 1 1 ≥ + + +  2. + a2 (a1 + 1) a1 + 1 a2 + 1 a1 + 1 a2 + 1 3+A+M +S+ 1 +

1

+1+

A

+M +S

1 A M S M S A (c) It is sufficient to prove that  A(M1+1) + M (S+1) + (A+1)(M +1)(S+1) 1 ; then it is sufficient to use inequality of Problem 14.11(b) for n  3. S(A+1) 1 1 1 1 1 1 (d) Let a(1+b) + b(1+c) + c(1+a) ≤ a(1+c) + b(1+a) + c(1+b) . Then since 1 1 1 1 1 1 + b(1+c) + c(1+a) ≤ a(1+a) + b(1+b) + c(1+c) (see Problem 14.10(b)). a(1+b) Hence    1 1 1 1 1 1 + + + + a b c 1+a 1+b 1+c   1 1 9 1 + + ≥ ≥3 a(1 + b) b(1 + c) c(1 + a) 1 + abc

(see Problem 14.10(b)). (e) It is sufficient to note that √2ab , √2bc , √2ac and a+c a+b b+c have opposite order. Thus it follows that  √

√ 1 , √1 , √1 c+a a+b b+c

2 b 1 c a2 b2 c2 2ab +√ ·√ +√  + + +√ a+b b+c c+a c+a a+b a+b b+c b+c a

2ac 1 1 a2 1 b2 c2 2ab 2bc +√ ·√ ·√ ≥ ·√ + + +√ +√ a+b b+c c+a c+a a+c b+c a+b a+b a+b     1 a+b+c 1 2ac a+b ab b+c bc 2bc ·√  ·√ +√ + + + + +√ 2 4 a+b 4 b+c a+c c+a b+c b+c √ √ √ 2 c +a  √ √ √ a+ b+ c ca a+b+c , + + ≥ + ab + bc + ca  √ 4 c+a 2 2

Proofs

205 or √

a a+b

+√

b b+c

+√

c ≥ c+a

√

a+

√ √

b+

2

√

c

.

14.12. Let us begin by proving that ϕ(u) ≥ ϕ(x)+(u−x)ϕ  (x). Consider the function f (x)  ϕ(u) − ϕ(x) + (x − u)ϕ  (x). Then f  (x)  (x − u)ϕ  (x), and hence in the case x > u, it follows that f (x) is increasing, and for x < u it follows that f (x) is decreasing. Thus, it follows that f (x) ≥ f (u)  0, or ϕ(u) ≥ ϕ(x) + (u − x)ϕ  (x), and therefore ϕ(x1 ) ≥ ϕ(y1 ) + (x1 − y1 )ϕ  (y1 ), . . . , ϕ(xn ) ≥ ϕ(yn ) + (xn − yn )ϕ  (yn ). We obtain ϕ(x1 ) + · · · + ϕ(xn ) ≥ ϕ(y1 ) + · · · + ϕ(yn ) + (x1 − y1 )ϕ  (y1 ) + · · · + (xn − yn )ϕ  (yn ). Now let us prove that (x1 − y1 )ϕ  (y1 ) + · · · + (xn − yn )ϕ  (yn ) ≥ 0. By Problem 14.10(a), we have

(x1 − y1 )ϕ  (y1 ) + · · · + (xn − yn )ϕ  (yn )  (x1 − y1 ) ϕ  (y1 ) − ϕ  (y2 ) 

+ (x1 − y1 + x2 − y2 ) ϕ (y2 ) − ϕ  (y3 ) + · · · + (x1 − y1 + x2 − y2 + · · · + xn − yn )ϕ  (yn ) ≥ 0,

since x1 + x2 + · · · + xk ≥ y1 + y2 + · · · + yk and ϕ  (yk−1 ) − ϕ  (yk ) ≥ 0 (k  1, . . . , n − 1) (ϕ  (x) is increasing). We deduce that ϕ(x1 ) + · · · + ϕ(xn ) ≥ ϕ(y1 ) + · · · + ϕ(yn ). 14.13. Consider the numbers ln a1 , . . . , ln an and ln b1 , . . . , ln bn−1 , ln bn , where 1 ···an , and the function ϕ(x)  ex Then from Problem 14.12, bn  ba1 ···b n−1

it follows that ϕ(ln b1 ) + · · · + ϕ ln bn ≥ ϕ(ln a1 ) + · · · + ϕ(ln an ), or b1 + · · · + bn−1 + bn ≥ a1 + · · · + an . 1 ···an ≤ bn , and therefore, b1 +· · ·+bn ≥ a1 +· · ·+an . On the other hand, bn  ba1 ···b n−1

14.14. Consider the function f (x)  − sin 2x in 0, π2 .

We have f  (x)  14 sin 2x > 0 if x ∈ 0, π2 . Let α ≥ β ≥ γ . Then we have α ≥ π3 , α +β ≥ π3 + π3 , α +β +γ  π3 + π3 + π3 , and

by the inequality of Problem 14.12, we have f (α)+f (β)+f (γ ) ≥ therefore, f π3 + f π3 + f π3 , or sin α2 + sin β2 + sin γ2 ≤ 23 . On the other hand, π2 > α, π2 + π2 > α + β, π2 + π2 + 0  α + β + γ , whence √



f π2 + f π2 + f (0) > f (α) + f (β) + f (γ ), or 2 < sin α2 + sin β2 + sin γ2 . √ √ √ 14.15. The proof of the inequality sin α + sin β + sin γ > 0 is√obvious. Let us prove the second inequality. Consider the function f (x)  − sin x in (0, π ). 2 2 x+cos x √ Therefore f  (x)  24sin > 0 if x ∈ (0, π ). sin x sin x Since in the case α ≥ β ≥ γ one has α > π2 , α + β > π2 + π4 , α + β + γ  π π π + + , it follows by the inequality of Problem 14.12 that f (α)+f (β)+f (γ ) > 2 4 4

√ √

√ √ f π2 + f π4 + f π4 , or sin α + sin β + sin γ < 1 + 4 8.  Alternative proof. Note that x + y ≤ 2(x2 + y2 ). Therefore for α > π2 , we have     √ sin α + sin β + sin γ < 1 + sin β + sin γ ≤ 1   √ √  β +γ 4 + 2(sin β + sin γ ) ≤ 1 + 4 sin < 1 + 2 2  1 + 8. 2

206

14 Miscellaneous Inequalities

14.16. We have 1 1 2 2 1 2 1 + √ + √ + ··· + √ > √ +√ √ + ··· + √ √ n 2 3 1+ 2 2+ 3 n+ n+1 √  √    √ √ √ √ √  2 2 − 1 + 2 3 − 2 + · · · + 2 n + 1 − n  2 n + 1 − 2 > 2 n − 2.

14.17. Consider the polynomial P(x)  (x − a)(x − b)(x − c)(x − d ). Since P(x)  x4 − S1 x3 + S2 x2 − S3 x + S4 and P(a)  0, we must have a4  S1 a3 − S2 a2 + S3 a − S4 .

(1)

If a ≤ 0, then S1 a3 − S2 a2 + S3 a − S4 < 0, and a4 ≥ 0, which leads to a contradiction with (1). Therefore, a > 0. In a similar way one can prove that b > 0, c > 0, d > 0. 14.18. Consider the sequence an  n n!n n . (e) Thus, it follows that an+1 e 1  <

1 n+1 an 1+ n 1+

1 n

√

n (see Problem 9.32(d)), and therefore, an+1 < √n+1 · an . We have √ √ √ n−1 n−1 n−2 · an−1 < √ ·√ an < √ · an−2 n n n−1 √ √ √ e n−1 n−2 1 a1 ·√ · · · √ · a1  √  √ . < ··· < √ n n n n−1 2

Hence an < √en , and therefore, if n ≥ 8, then an < √e8 < 1. 14.19. By induction, one can easily prove that 2n ≥ n2 − 1, n ∈ N. It follows that           1 1 1 1 1 1 + 1 ··· 1 + n ≤ 1 + · 1+ 2 ··· 1 + 2 2 2 2 2 −1 n −1 2 2 n 3n 3 2 ···  < 3.  · 2 1·3 (n − 1) · (n + 1) n+1 14.20. Let a, b, c ∈ [0, a1 ] and a ≥ b ≥ c. If b ≤ a+c , then the following conditions 2 hold: a ≥ a−b+c, a+c  (a−b+c)+b and a−b+c ≥ b. Therefore, by Problem 2.8, it follows that g(a) + g(c) ≥ g(a − b + c) + g(b), or g(a) − g(b) + g(c) ≥ g(a − b + c). , then a ≥ b, a + c  b + (a − b + c) and b ≥ a − b + c. Therefore, If b ≥ a+c 2 g(a) + g(c) ≥ g(a − b + c) + g(b).

Proofs

207

It follows that g(a) − g(b) + g(c) ≥ g(a − b + c). Hence using Problem 7.5, we obtain the proof of the given inequality (taking f (x)  g(x) − g(0)). 14.21. (a), (b) Let p(1)  a + b + c  m, p(−1)  a − b + c  n. Then |m| ≤ 1, |n| ≤ 1, and |c|  |p(0)| ≤ 1. , a  m+n − c, and therefore, Note that b  m−n 2 2 m + n   m − n     + |c| |a| + |b| + |c|   − c +  2 2   m + n   m − n      + 2|c| ≤ max(|m|, |n|) + 2|c| ≤ 3. ≤ + 2 2  We have m2 + n2 + 2c2 − mc − nc 2 m2 + c2 n2 + c2 m2 + n2 + 2c2 + +  m2 + n2 + 3c2 ≤ 5. ≤ 2 2 2

a2 + b2 + c2 

(c) Consider the polynomial F(x)  a + bx + cx2 + dx3 . Then by Lagrange’s formula, 1 we have (x + 1)(x − 1)(x − 2) (x + 2)(x − 1)(x − 2) + F(−1) · −12 6 (x + 2)(x + 1)(x − 1) (x + 2)(x + 1)(x − 2) + F(2) · . + F(1) · −6 12

F(x)  F(−2) ·

Hence a  F(0)  F(−2) + 23 F(−1) + 23 F(1) − 16 F(2). −6

On the other hand, F(x)  x3 p 1x , x  0, and therefore F(2)  1

1

8p 2 , F(−2)  −8p − 2 , F(1)  p(1), F(−1)  −p(−1).



Thus, a  43 p − 21 − 23 p(−1) + 23 p(1) − 43 p 21 , and it follows that   |a| ≤ 43 p − 21  + 23 |p(−1)| + 23 |p(1)| + 43 p 21  ≤ 43 + 23 + 23 + 43  4. Note that the polynomial F(x)  4x3 − 3x satisfies the assumption of the problem (if we set x  cosα, then F(x)  cos3α). 14.22. Let the roots of the polynomial P(x) be the numbers −b1 , −b2 , . . . , −bn , where bi > 0 i  1, 2, . . . , n. We have P(x)  (x + b1 ) · · · (x + bn ), and hence a0  b1 · · · bn and a1  a0 b11 + · · · + b1n . Thus, we need to prove that 1 If

P(x) is a polynomial of degree n, and x1 , x2 , . . . , xn , xn+1 are distinct real numbers, then (x − x2 ) · · · (x − xn+1 ) (x − x1 )(x − x3 ) · · · (x − xn+1 ) + P(x2 ) · (x1 − x2 ) · · · (x1 − xn+1 ) (x2 − x1 )(x2 − x3 ) · · · (x2 − xn+1 ) (x − x1 ) · · · (x − xn ) . + · · · + P(xn+1 ) · (xn+1 − x1 ) · · · (xn+1 − xn )

P(x)  P(x1 ) ·

208

14 Miscellaneous Inequalities



 1 1 + ··· + (1 + b1 )(1 + b2 ) · · · (1 + bn ) ≥ 2n2 , n  2, 3, . . . . b1 bn

By Cauchy’s inequality, it follows that b11 + · · · + b1n ≥ √n b bn ···b , and 1 + bi  1 2 n  1 1 bi n +bi ≥ n · (n−1)n−1 , i  1, . . . , n. Therefore, + ... + n − 1  n − 1 

n−1

  n−1 1 1 1 + ··· + (1 + b1 ) · · · (1 + bn ) ≥ n2 1 + b1 bn n−1   1  2n2 . ≥ n2 1 + (n − 1) · n−1

In order to complete the proof, it is sufficient to use Bernoulli’s inequality. 14.23. We call two segments having a common vertex a bird, and the segments themselves wings. The number of non-single-color triangles is equal to Cn3 − t(n). Each non-single-color triangle contains birds with wings of distinct colors. Therefore, the number of birds with wings of distinct colors is equal to 2(Cn3 − t(n)). On the other hand, if through some vertex, k segments of the first color and n − 1 − k segments of the second color pass, then the number of birds with wings of distinct colors containing that vertex (that is, both wings contain that vertex) is equal to k(n − 1 − k). Note that ⎧

⎨ n−1 2 , if n is even, 2 k(n − 1 − k) ≤ ⎩ n n − 1 , if n is odd. 2 2

Thus, it follows that ⎧ ⎨ (n−1)2 n , if n is odd, 4 3 2 Cn − t(n) ≤

2 ⎩ n n − 1 , if n is even. 2 2



Hence, we obtain if n  2k, then t(n) ≥ if n  4k +

2 n(n−1)(n−2) − n (n−2)  k(k−1)(k−2) , 6 8 3 n(n−1)(n−2) n(n−1)2 2 − 8  3 k(k − 1)(4k + 1), 1, then t(n) ≥ 6 2 n(n−1)(n−2) − n(n−1)  23 k(k + 1)(4k − 1). 3, then t(n) ≥ 6 8

if n  4k + 14.24. We proceed by induction. If n  3, 4, then the statement of the problem is true. Assume that the statement holds for n ≤ k − 1 points. Let us prove that it holds for n  k points (k ≥ 5). Consider the following two cases.

Proofs

209

(a) Every edge is part of a triangle. k2 , and thereIn this case, the number of triangles formed is greater than 12 k  fore not less than 2 , k  5, 6, . . .. (b) There exists an edge AB that does not belong to any triangle. Let us denote all other vertices of the graph by A1 , A2 , . . . , Ak−2 . Hence, each vertex Ai is not connected to both A and B (i  1, 2, . . . , k − 2). Therefore, the 2 number of edges with vertices A1 , A2 , . . . , Ak−2 is greater than k4 − (k − 2 1)  (k−2) , and hence the number of triangles formed by these vertices 4   k  is at least k−2  2 − 1. 2 Let one of the triangles be A1 A2 A3 . If none of the vertices Ai is connected to A or B, then we will not consider the segment A1 A2 . Then the points 2 A1 , A2 , . . . , Ak−2 are connected to (k−2) segments. Therefore, there exist at 4 k    least 2 − 1 triangles, and with A1 A2 A3 , there exist k2 triangles. It is left to consider the case in which A is connected to the vertices A1 , A2 , . . . , Am , and B is connected to the rest of them. If the segment Ai Aj is an edge for some (i, j), i < j ≤ m or m < i < j, then this ends the proof. Otherwise, the number of edges with vertices A1 , A2 , . . . , Ak−2 is not greater than m(k − 2 − m), 2 . which leads to a contradiction, since m(k − 2 − m) ≤ (k−2) 4 14.25. Note that the numbers {nα} are distinct numbers. We argue by contradiction. Assume that {iα}  {jα}, i  j. Then iα  jα + k, k ∈ Z, and hence k ∈ Q, which leads to a contradiction. α  i−j Let m ∈ N and m1 < min(a, b − a, 1 − b). There exist positive integers i > j such that |{iα} − {jα}| < m1 (it is sufficient to divide the segment [0, 1] into m+ 1equal parts and use Dirichlet’s principle). Therefore,  (i − j)α − [iα] + jα  < 1 , and hence {(i − j)α} < 1 , or {(i − j)α} > m m 1 − m1 . Let {(i − j)α} < m1 , (i − j)α  x + k, k ∈ Z, x  {(i − j)α}. Consider the numbers x, 2x, 3x, . . . . One of these numbers belongs to (a, b), so suppose a < lx < b, l ∈ N. We have that l(i − j)α  lx + kl; hence {l(i − j)α}  lx ∈ (a, b). Let {(i − j)α} > 1 − m1 , (i − j)α  x + k, k ∈ Z, x  {(i − j)α}. We have 1 − x < m1 . Consider the numbers 1 − x, 2(1 − x), . . . . One of these numbers belongs to (1 − b, 1 − a), so let 1 − b < l(1 − x) < 1 − a, l(i − j)α  lx + lk  −l(1 − x) + l + lk. Therefore, l(k + 1) + a − 1 < l(i − j)α < l(k + 1) + b − 1, or a < {l(i − j)} < b. n n   14.26. (a) We have xi2 ≤ −nmM . (xi − m)(M − xi ) ≥ 0, and therefore i1

(b) Note that M ≥ 0 ≥ m, whence m3 )(M − xi )) ≥ 0.

n  i1

i1

(M (xi − m)(M 3 − xi3 ) − m(xi3 −

210

14 Miscellaneous Inequalities

Thus, we deduce that (M −m) m, then we have follows that

n  i1

n  i1

xi4

n  i1

xi4



xi4 ≤ −n M 4 m − m4 M . Therefore, if M 



≤ −mMn m2 + M 2 + mM . If m  M  0, then it



 0  −mMn m2 + M 2 + mM .

14.27. We have





Cm1 x2m−2 y2 + x2 y2m−2 + . . . + Cmm−1 x2 y2m−2 + y2 x2m−2 2 ≥ Cm1 xm ym + Cm2 xm ym + · · · + Cmm−1 xm ym  (2m − 2)xm ym .



x2 + y2

m

− x2m − y2m 

 k m−k 2 m−k 2 k Here we have used that x2 · y2 + x · y ≥ 2 x2m · y2m ≥ 2xm · ym , k 

m 1, · · · ,mm − 1. ≥ Therefore, x2 + y2 − x2m − y2m ≥ (2m − 2)xm ym , or x2 + y2 m m 2 m m m (x − y ) + 2 x y . This ends the proof. 14.28. If α ≤ π2 , β ≤ π2 , γ ≤ π2 , then we have sin α cos(β − γ )  √ √ 2γ sin 2β 1 . 2γ + sin 2β) ≥ sin 2γ sin 2β, and thus cos(β − γ ) ≥ sin sin 2 (sin α Therefore, √

cos(α − β) cos(β − γ ) cos(γ − α) ≥

sin 2α sin 2β · sin γ

√ √ sin 2γ sin 2β sin 2α sin 2γ · sin α sin β

 8 cos α cos β cos γ .

Let α ≤ β < π2 < γ . If γ ≤ π2 + α or γ ≥ π2 + β, then cos(α − β) cos(β − γ ) cos(γ − α) ≥ 0 > 8 cos α cos β cos γ . If π2 + α < γ < π2 + β, then 0 < − cos(γ − α) < − cos γ and 0 < cos(α − β) cos(γ − β) < cos(β − α) < 4 cos(β − α) + 4 cos(α + β)  8 cos α cos β; hence cos(α − β) cos(β − γ ) cos(γ − α) > 8 cos α cos β cos γ . 14.29. Consider the following two cases.   (a) if sin 2x  0, we have x  2π n, n ∈ Z and sin x + sin22x + · · · + sinnnx   0 < 3.   1 (b) if sin 2x  0, then there exists a positive integer m such that m+1 < sin 2x  ≤ 1 . m For n ≤ m we have     sin x + sin 2x + · · · + sin nx  ≤ |sin x| + |sin 2x| + · · · + |sin nx|  2 n  2 n  x   ≤ |sin x| + |sin x| + · · · + |sin x| ≤ m|sin x| ≤ 2msin  ≤ 2    2 n

(see Problems 1.13 and 7.4(a)).

(1)

Proofs

211

  Now let us estimate the following expression:  sin(m+1)x + · · · + sinnnx  for n > m+1 m. By Problem 14.10, we have that   sin(m + 1)x sin nx 1 1 + ··· +  sin(m + 1)x − m+1 n m+1 m+2   1 1 + +(sin(m + 1)x + sin(m + 2)x) − m+2 m+3   1 1 + · · · + (sin(m + 1)x + · · · + sin(n − 1)x) − n−1 n 1 + (sin(m + 1)x + · · · + sin nx) n Thus, it follows that

     sin(m + 1)x 1 sin nx  1  |sin(m ≤ + 1)x| + · · · + −  m+1 n  m+1 m+2   1 1 + |sin(m + 1)x + sin(m + 2)x| − m+2 m+3   1 1 1 + · · · + |sin(m + 1)x + · · · + sin(n − 1)x| − + |sin(m + 1)x + · · · + sin nx| n−1 n n   1 1 1 1 1 1 1   < 1.   ≤  − + ··· + − + sin x  m + 1 m+2 n−1 n n (m + 1)sin x  2

2

Hence    sin(m + 1)x sin nx   < 1. + ··· +  m+1 n 

(2)

Here we have used the following inequality: 1 |sin kx + sin(k + 1)x + · · · + sin px|≤  x  . sin  2 Let us prove this inequality. We have |sin kx + sin(k + 1)x + · · · + sin px|   2 sin x sin kx + · · · + 2 sin x sin px 2 2    2sin 2x                  cos k − 21 x − cos k + 21 x + cos k + 21 x − cos k + 1 + 21 x + · · · + cos p − 21 x − cos p + 21 x    2sin 2x          1 1 cos k − 2 x − cos p + 2 x 2 1     .  ≤   sin x  2sin x  2sin x  2

2

2

From inequalities (1) and (2), it follows that

212

14 Miscellaneous Inequalities

    sin x + sin 2x + · · · + sin nx  < 3.  2 n  14.30. We need to prove that (a1 + a2 + a3 )(a2 + a3 + a4 ) · · · (an + a1 + a2 ) A ≥ (a1 + a2 )(a2 + a3 ) · · · (an + a1 )

 n 3 . 2

(1)

Let us prove (1) by mathematical induction. (a) For n  3, we have 2 a1 + a2 + a3  a1 +a + 2

a2 +a3 2

+

a3 +a1 2

 2 ≥ 3 3 a1 +a · 2

a2 +a3 2

·

a3 +a1 . 2

(a1 +a2 ) (a2 +a3 ) (a3 +a1 ) 2 2 2

Therefore,

· (a1 + a2 + a3 )(a2 + a3 + a1 )(a3 + a1 + a2 ) ≥  (a1 + a2 )(a2 + a3 )(a3 + a1 ) (a1 + a2 )(a2 + a3 )(a3 + a1 ) 33

 3 3 . 2

(b) Let n ≥ 4 and suppose that inequality (1) holds for n − 1 numbers. Let us first prove inequality (1) if a1  a2 or an−1  an . If a1  a2 , we have (2a2 + a3 )(a2 + a3 + a4 ) · · · (an−1 + an + a2 )(an + 2a2 ) 2a2 (a2 + a3 ) · · · (an−1 + an )(an + a2 ) (2a2 + a3 )(an + 2a2 ) (a2 + a3 + a4 ) · · · (an−1 + an + a2 )(an + a2 + a3 )  · 2a2 (an + a2 + a3 ) (a2 + a3 ) · · · (an−1 + an )(an + a2 )  n−1  n−1  n (2a2 + a3 )(an + 2a2 ) 3 3 3 3 ≥ ≥ · ≥ , · 2a2 (an + a2 + a3 ) 2 2 2 2

A

since the inequality (2a2 + a3 )(an + 2a2 ) ≥ 3a2 (an + a2 + a3 ) is equivalent to the inequality (a2 − a3 )(a2 − an ) ≥ 0. If an−1  an , the proof is similar. Now let us prove that a1 < a2 and an−1 < an . If we substitute the sequence of numbers a1 , a2 , . . . , an−1 , an by the numbers a1  a1 + x, a2 , . . . , an−1 , an − x  an , where 0 ≤ x ≤ min(a2 − a1 , an − an−1 ), then the value of A does not increase, and therefore, A≥

(a1 + a2 + a3 + x)(a2 + a3 + a4 ) · · · (an−2 + an−1 + an − x)(an−1 + an + a1 )(an + a1 + a2 ) , (a1 + x + a2 )(a2 + a3 ) · · · (an−1 + an − x)(an + a1 )

or (a1 + a2 + a3 )(an−2 + an−1 + an ) (a1 + a2 + a3 + x)(an−2 + an−1 + an − x) ≥ . (a1 + a2 )(an−1 + an ) (a1 + x + a2 )(an−1 + an − x) (2) Note that the last inequality can be rewritten in the following way: ax2 +bx+c ≤ 3 )(an−2 +an−1 +an ) − 1 > 0, c  0, and therefore, in order 0, where a  (a1 +a(a2 +a 1 +a2 )(an−1 +an ) to prove (2) it is sufficient to prove it for the values x  0 and x  a4 − a1

Proofs

213

(n  4), x  a5 + a4 − a2 − a1 (n  5), and x  a2 − a1 (n ≥ 6), since min(a2 − a1 , an − an−1 ) ≤ a2 − a1 ≤ a4 − a1 ≤ a4 − a1 + a5 − a2 . If x  0, the proof of (2) is obvious. If n  4, then x  a4 − a1 , and we need to prove that (a1 +a2 +a3 )(a2 +a3 +a4 ) 3 +a4 )(a2 +a3 +a1 ) ≥ (a2 +a , or (a4 − a1 )(a3 − a2 ) ≥ 0. (a1 +a2 )(a3 +a4 ) (a2 +a4 )(a1 +a3 ) If n  5, then x  a5 + a4 − a2 − a1 , and the proof of (2) is obvious. For n ≥ 6, then x  a2 − a1 , and we need to prove that ax + b ≤ 0. Therefore,   (a1 + a2 + a3 )(an−2 + an−1 + an ) − 1 (a2 − a1 ) (a1 + a2 )(an−1 + an )  (a1 + a2 + a3 )(an−2 + an−1 + an ) + an +an−1 +an−2 −a1 −a2 −a3 − a1 + a2  (a1 + a2 + a3 )(an−2 + an−1 + an ) ≤ 0, + an−1 + an or (a3 an−2 + a3 (an−1 + an ) + an−2 (a1 + a2 ))(a2 − a1 ) (a1 + a2 )(an−1 + an )   (an−1 + an − a1 − a2 )(a3 an−2 + a3 (an−1 + an ) + an−2 (a1 + a2 )) ≤ 0, + an−2 − a3 − (a1 + a2 )(an−1 + an )

(a3 an−2 + a3 (an−1 + an ) + an−2 (a1 + a2 ))(2a2 − an − an−1 ) + (an−2 − a3 )(a1 + a2 )(an−1 + an ) ≤ 0, 2a2 an−2 (a1 + a2 + a3 ) + (an−1 + an )(a2 a3 − a3 an−2 − a3 an−1 − a3 an − a3 a1 ) ≤ 0, 2a2 an−2 (a1 + a2 + a3 ) ≤ (a3 an−1 + a3 an )(an−2 + an−1 + an + a1 − a2 ), where the last inequality holds because 2a2 an−2 ≤ a3 an−2 + a3 an−2 ≤ a3 an−1 + a3 an and a1 + a2 + a3 ≤ an−2 + an−1 + an + a1 − a2 . Taking x  min(a2 − a1 , an − an−1 ), we obtain for the numbers a1 , a2 , . . . , an−1 , an the case a1  a2 or an−1  an , such that a1 ≤ a2 ≤ n (a +a2 +a3 )(a2 +a3 +a4 )···(an +a1 +a2 ) · · · ≤ an−1 ≤ an Thus, it follows that A ≥ 1 (a ≥ 23 .   1 +a2 )(a2 +a3 )···(an +a1 ) This ends the proof of the given inequality. 14.31. (a) Note that yz < 1, xz < 1, xy < 1; therefore y y x z x + 1−zx + 1−xy ≥ x + y + z ≥ x2 + y2 + z 2  1, and hence 1−yz + 1−zx + 1−yz z ≥ 1. 1−xy Note that yz ≤

y2 +z 2 , 2

xz ≤

z 2 +x2 , 2

xy ≤

x2 +y2 ; 2

therefore,

y z x x y z + + ≤ 2 +z 2 + 2 +x 2 + 2 2 y z 1 − yz 1 − zx 1 − xy 1 − 1− 2 1 − x +y 2 2 2x 2y 2z  + + . 1 + x2 1 + y2 1 + z 2

214

14 Miscellaneous Inequalities

On the other hand, note that 

2 2 − 1 + x2 1 + y2



x y − 1 + x2 1 + y2

 −

2(x − y)2 (x + y)(1 − xy) ≤ 0, (1 + x2 )2 (1 + y2 )2

and thus from inequality (8.5.1) (see Chapter 8) we obtain

√  2y 2z 6(x + y + z) 3 3 3 2x 2 + y2 + z 2 )  + + ≤ ≤ 3(x . 1 + x2 1 + y2 1 + z 2 3 + x2 + y2 + z 2 2 2 √ y x z 3 3 + + ≤ (see also 1−yz 1−zx 1−xy 2 y y2 x z x2 z2 + 1+zx + 1+xy  x+xyz + y+xyz + z+xyz , 1+yz

It follows that

(b) We have By inequality (8.4), we have

Problem 5.17). xyz  0.

x y z (x + y + z)2 (1 + xy + yz + zx) + xy + yz + zx + + ≥  1 + yz 1 + zx 1 + xy x + y + z + 3xyz x + y + z + 3xyz x + y + z + (1 − x)(1 − y)(1 − z) + xyz + xy + yz + zx x + y + z + xyz + xy + yz + zx  ≥ > 1; x + y + z + 3xyz x + y + z + 3xyz

y x z hence 1+yz + 1+zx + 1+xy > 1. y x z + 1+zx + 1+xy  If xyz  0, for example x  0, so that y2 + z 2  1 and 1+yz 2 2 2 2 2 y + z ≥ y + z  1, let us√prove that if x, y, z ≥ 0 and x + y + z  1, y x z + 1+zx + 1+xy ≤ 2. then 1+yz 1 Note that if a ≥ 0, then 1+a ≤ 1 − a + a2 , and therefore,

x y z + + ≤ x(1 − yz + y2 z 2 ) + y(1 − xz + x2 z 2 ) + x(1 − xy + x2 y2 ) 1 + yz 1 + zx 1 + xy  x + y + z − 3xyz + xyz(yz + xz + xy). y x z Moreover, we have xy + yz + zx ≤ x2 + y2 + z 2 , whence 1+yz + 1+zx + 1+xy ≤ x + y + z − 2xyz. √ Let us prove that x+y+z−2xyz ≤ 2, where x, y, z ≥ 0 and x2 +y2 +z 2  1. Indeed,√let max(x, y, z)  √ z. Then 3z 2 ≥ x2 + y2 + z 2  1; therefore, 2 2 2 and√thus from the 2z + 2 2z − 1 ≥ 2z +√2 2z 2 − 1 > √ 3z − 12 ≥ 0, 4 2 inequality 4z − 8z + 4 2z − 1  ( 2z − 1) (2z 2 + 2 2z − 1) ≥ 0, 2 √ √ 4 2 +4 2z−1 it follows that z(x + y) − 2√1 z + 4z −8z 4z ≥ 0, or z(x + y)2 − √ √

(x + y) + z 3 − 2z + 2 ≥ 0, x + y + z − (x + y)2 + z 2 − 1 z ≤ 2, √ x + y + z − 2xyz ≤ 2. 14.32. Let us prove that if inequality (14.1) holds for all positive numbers a, b, c, then λ ≥ 8 or λ  0. Indeed, if λ  0, then for a  1, b  c  1n , from (14.1) we obtain



1 1 + λ n2

+√

1 1 + nλ

+√

1

3 ≥√ . 1 + nλ 1+λ

Proofs

215 3 For n → +∞ we have that 1 ≥ √1+λ , and therefore, λ ≥ 8 (if λ < 0, then √ the expression 1 + nλ is not defined for n > − λ1 ). Let us prove that for λ ≥ 8 and a > 0, b > 0, c > 0 inequality (14.1) holds. 1 1 + √1+y ≥ 1. Lemma 1 If x > 0, y > 0, and xy ≤ 9, then √1+x  √ 2 √ √ 1 + x + 1 + y  2 + x + y + 2 (1 + x)(1 + y) ≥ 2 + x + Indeed, we have √ √ √ √ √ y +2+2 xy ≥ (1+x)(1+y), and therefore, 1 + x + 1 + y ≥ 1 + x 1 + y. 1 1 + √1+y ≥ √ 2√ . Lemma 2 If x > 0, y > 0, and xy > 9, then √1+x 1+ xy √ 2 √ √

We need to prove that 1 + x + 1 + y 1 + xy ≥ 4(1 + x)(1 + y). √ 2 √ √ √ √

We have 1 + x + 1 + y 1 + xy ≥ 4 + x + y + 2 xy 1 + xy (see the proof of Lemma 1). √

√ Now, let us prove the following inequality: 4 + x + y + 2 xy 1 + xy ≥ 4(1 + x)(1 + y). Note that it is equivalent to the following obvious inequality:

√ √ √ 2 xy −3 x −  y ≥ 0.  2 1+x2

Define

−

2 1+y2

x 1+x2

−

2

y 1+y2 3

(x+y)(1−xy)  − 2(x−y) ≤ 0. Then x > (1+x2 )2 (1+y2 )2

0, y > 0, z > 0, xyz  λ . Let us prove that for λ ≥ 8 the following inequality holds: √

1

1 3 1 +√ ≥√ . +√ 1 + y 1+x 1+z 1+λ

Let max(x, y, z)  z. Then z ≥ λ. If xy ≤ 9, then by Lemma 1 we have 1 1 3 1 1 1 +√ ≥1≥ √ +√ >√ +√ . √ 1 + y 1 + y 1+x 1+z 1+x 1+λ If xy > 9, then by Lemma 2, we have √

2 √ 1+ zλ

√1 1+x

1 + √1+y ≥√

2 √ , 1+ xy

√1 1+z

1 + √1+λ ≥

(since zλ ≥ λ2 ≥ 64 > 9).

Summing these inequalities, we obtain 2 1 1 1 1 2 +√ +√ +√ ≥ √ √ + √ . 1+y 1 + xy 1+x 1+z 1+λ 1 + zλ √ √ √ √ Since xy ≥ 3 and zλ ≥ λ, we have xy · zλ ≥ 3λ ≥ 24 > 9, and 4  √1+λ . therefore, by Lemma 2, we have √ 2√ + √ 2√ > √ 4√4 Hence (14.1).

√1 1+x

+

√1 1+y

+

√1 1+z

+

√1 1+λ

1+ xy 4 > √1+λ ,

1+ zλ

1+ xyzλ

which ends the proof of inequality

216

14 Miscellaneous Inequalities

Remark In a similar way, one can prove that n ≥ 2, a1 > 0, . . . , an > 0. Then   a1n−1 ann−1 + · · · + ≥ 1. a1n−1 + (n2 − 1)a2 · · · an ann−1 + (n2 − 1)a1 · · · an−1 14.33. Let us first prove the following lemma. Lemma If x, y, z are the lengths of some triangle, then 

√ y + z  y + z 2 + x2 + xy + y2 x2 + xz + z 2 ≤ x2 + x . 2 2

Indeed, we need to prove that 



2 2 2 + y+z . This inequality holds x + xy + y2 x2 + xz + z 2 ≤ x2 + x y+z 2 2 because it is equivalent to the following obvious inequality:  y−z 2  y+z 2 + yz + x(y + z − x) ≥ 0. 2 2 This ends the proof of the lemma. From the lemma, we have 2 √ c2 + ac + a2 √ √  2a2 + 2b2 + 2c2 + ab + bc + ca + 2 a2 + ab + b2 + b2 + bc + c2 √ √ √ √ + 2 a2 + ab + b2 c2 + ac + a2 + 2 b2 + bc + c2 c2 + ac + a2 (a + c)2 ≤ 2a2 + 2b2 + 2c2 + ab + bc + ca + 2b2 + b(a + c) + 2 2 2 (b + c) (a + b) + 2c2 + c(a + b) + + 2a2 + a(b + c) + 2 2  5a2 + 5b2 + 5c2 + 4ab + 4bc + 4ca.

√

a2 + ab + b2 +

√

b2 + bc + c2 +

It follows that √ √ √ a2 + ab + b2 + b2 + bc + c2 + c2 + ac + a2 √ ≤ 5a2 + 5b2 + 5c2 + 4ab + 4bc + 4ca. Remark 1. One can prove that if a, b, c ≥ 0, then √

√ √ a2 + ab + b2 + b2 + bc + c2 + c2 + ca + a2 √ ≤ 5a2 + 5b2 + 5c2 + 4ab + 4bc + 4ca.

2. If a  −1, b  c  1, then 1 +

√

3+1>

√

11.

Proofs

217

14.34.

(a) Let us begin by proving that if the given inequality holds for n  3, then it holds for all n  4, 5, . . . . Indeed, we have x13 + · · · + xn3 x1 · · · xn

⎛ ⎞ 3 3 3 xn3 + x13 + x23 1 1 1 ⎝ x1 + x 2 + x 3 ⎠ · · + ··· + 3 x1 x2 x3 x4 · · · xn xn x1 x2 x3 · · · xn−1  1 1 (1 − x1 )3 + (1 − x2 )3 + (1 − x3 )3 · + ··· ≥ 3 (1 − x1 )(1 − x2 )(1 − x3 ) x4 · · · xn  (1 − xn )3 + (1 − x1 )3 + (1 − x2 )3 1 · + (1 − xn )(1 − x1 )(1 − x2 ) x3 · · · xn−1  1 (1 − x1 )3 + (1 − x2 )3 + (1 − x3 )3 + ··· ≥ 3 (1 − x1 ) · · · (1 − xn )  (1 − xn )3 + (1 − x1 )3 + (1 − x2 )3 + (1 − x1 ) · · · (1 − xn )





(1 − x1 )3 + · · · + (1 − xn )3 , (1 − x1 ) · · · (1 − xn )

x3 +···+x3

3

3

1 ) +···+(1−xn ) and therefore, 1x1 ···xn n ≥ (1−x . (1−x1 )···(1−xn ) Let us prove that the given inequality holds for n  3.

x13 + · · · + xn3



x1 · · · xn



where xi





 xn3 + x13 + x23 1 1 + ··· + · x1 x2 x3 x4 · · · xn xn x1 x2 x3 · · · xn−1     1 1 1 1 ≥3+ +   +   (x1 − x2 )2 + (x2 − x3 )2 + (x3 − x1 )2 2 x1 x2 x2 x3 x3 x1 1 3

x13 + x23 + x33

·

(1 − x1 )3 + (1 − x2 )3 + (1 − x3 )3 , (1 − x1 )(1 − x2 )(1 − x3 )

1 − xi , i



1, 2, 3. Therefore,

(1−x1 )3 +(1−x2 )3 +(1−x3 )3 . (1−x1 )(1−x2 )(1−x3 )

x13 +x23 +x33 x1 x2 x3

≥

(b) It is sufficient to prove the given inequality for n  4 (see the proof of Problem 14.34(a)). Let x1 ≥ x2 ≥ x3 ≥ x4 ; then x1 x1 ≥ x2 x2 ≥ x3 x3 ≥ x4 x4 and x1 ≥ xx2 ≥ xx3 ≥ xx4 , where xi  1 − xi , i  1, 2, 3, 4. It follows that x 1

2

3

x14 + x24 + x34 + x44 x1 x2 x3 x4





4   (x − x4 )4 4(x3 − x4 )2 (x1 − x2 )4 4(x1 − x2 )2 x x x x + +2 1 2 + 3 4 + 3 + x1 x2 x3 x4 x3 x4 x3 x4 x1 x2 x1 x2 x3 x4 x1 x2   x1 x2 (x1 − x2 )4 4(x1 − x2 )2 x3 x4 (x3 − x4 )4 4(x3 − x4 )2 ≥     + +2   +   +     + x1 x2 x3 x4 x3 x4 x3 x4 x1 x2 x1 x2 x3 x4 x1 x2

(1 − x1 )4 + (1 − x2 )4 + (1 − x3 )4 + (1 − x4 )4 , (1 − x1 )(1 − x2 )(1 − x3 )(1 − x4 )

since x1 x2 x3 x4 x x x x + − 1 2 − 3 4 x3 x4 x1 x2 x3 x4 x1 x2  (x1 x1 · x2 x2 − x3 x3 · x4 x4 )



1 1 −   x1 x2 x3 x4 x1 x2 x3 x4

 ≥ 0.

218

14 Miscellaneous Inequalities

Another proof of this inequality can be obtained using the following identity:   x14 + x24 + x34 + x44 1 (x1 − x2 )4 + (x2 − x3 )4 + (x3 − x4 )4 + (x4 − x1 )4  x1 x2 x3 x4 2 x1 x2 x3 x4    (x1 − x3 )2 (x2 − x4 )2 2(x1 − x2 )2 2(x2 − x3 )2 2(x3 − x4 )2 2(x4 − x1 )2 + + + + +2 +2 . + x3 x4 x1 x4 x1 x2 x2 x3 x1 x3 x2 x4

14.35. We need to prove that B  a12015 +· · ·+an2015 −A(a12000 +· · ·+an2000 ) ≥ 0, 2015 2015 where A  112000 +···+n . +···+n2000 If {a1 , . . . , an }  {1, . . . , n}, then B  0. If {a1 , . . . , an }  {1, . . . , n}, then there exist numbers i and j (i, j ∈ {1, . . . , n}) such that ai ≥ n + 1 and ak  j, for k  1, . . . , n. Then ai2015 − Aai2000 > j 2015 − Aj 2000 , as ai2000 > j 2000 and ai15 ≥ (n + 1)15 > A. Therefore, on substituting ai by j, we see that the value of B is decreasing. After several such √ steps we obtain  that B≥ 0. 14.36. We need to √ prove that 1 + x2 + 1 + y2 + (1 − x)2 + (1 − y)2 − (1 − xy) ≥ 5(1 − xy). Note that       1 + x2 + 1 + y2 + (1 − x)2 + (1 − y)2 − (1 − xy)  xy − 1 + x2 − 1 1 + y2 − 1 +

      (1 − x)2 + (1 − y)2 + 1 + x2 · 1 + y2 ≥ (1 − x)2 + (1 − y)2 + 1 + x2 · 1 + y2 ,

√ as a ≥ 1 + a2 −1 ≥ 0, for a ≥ 0. √  √ Let us prove that (1 − x)2 + (1 − y)2 + 1 + x2 · 1 + y2 ≥ 5(1 − xy). If xy ≥ 21 , then    √ √ (1 − x)2 + (1 − y)2 + 1 + x2 · 1 + y2 ≥ 1 + x2 · 1 + y2 √ 3 5 √ ≥ 1 + xy ≥ > ≥ 5(1 − xy). 2 2

If xy < 21 , then by inequality (14.1), we have

    2   (1 − x)2 + (1 − y)2 + 1 + x2 · 1 + y2  1 − 2xy + (1 − x − y)2 + (1 − xy)2 + (x + y)2   2   1 − 2xy + 1 − xy + 1  3 − 4xy + x2 y2 + 2(1 − xy) 1 − 2xy ≥  √ ≥ 3 − 4xy + x2 y2 + 2(1 − xy)(1 − 2xy)  5(1 − xy),

√   √ and therefore, (1 − x)2 + (1 − y)√2 + 1 + x2 · 1 + y2 ≥ 5(1−xy). √ 1+ x 1 1 14.37. (a) For n  2 we have √1+x + √1+x  √1+x1 ≤ 2. 1 2 1 For n  3 let us prove that there exists an acute triangle with angles α, β, γ γ tan β tan α such that x1  tan , x2  tan γ , and then x3  tan . β tan α

Proofs

219

√ Indeed, it is sufficient to take tan α  (1 + x2 + x1 x2 )x1 , tan β    1+x2 +x1 x2 1+x2 +x1 x2 1 , tan γ  · x2 . x1 x    1 sin γ cos β sin α cos γ β cos α + + ≤ √32 , or equivaWe need to prove that sinsin γ sin α sin β lently,     sin β sin(2α) + sin γ sin(2β) + sin α sin(2γ ) ≤ 3 sin α sin β sin γ . According to the Cauchy–Bunyakovsky–Schwarz inequality, we have    sin β sin(2α) + sin γ sin(2β) + sin α sin(2γ ) 

sin2 β + sin2 γ + sin2 α (sin(2α) + sin(2β) + sin(2γ )) ≤  

 2 sin2 β + sin2 γ + sin2 α sin α sin β sin γ ≤ 3 sin α sin β sin γ , √ √ √ and √ therefore, sin β sin(2α) + sin γ sin(2β) + sin α sin(2γ ) ≤ 3 sin α sin β sin γ (b) The proof is by induction. 1 1 1 + √1+b < 1 + 1+ab , where a > 0, b > 0, For n  4, from the inequality √1+a it follows that √

1 1 1 1 1 1 +√ +√ +√ <1+ +1+  3. 1 + x1 x2 1 + x3 x4 1 + x2 1 + x4 1 + x1 1 + x3

Assume that n ≥ 5 and that the inequality holds for n − 1 numbers. Let us prove that it holds for n numbers. We have 1 1 1 1 1 1 + ··· + √ +√ < √ + ··· + √ +1+ √ 1 + xn−1 xn 1 + xn−1 1 + xn−2 1 + x1 1 + xn 1 + x1 1 1 1 < √ + ··· + √ +√ + 1 < n − 2 + 1  n − 1, 1 + xn−2 1 + xn−1 xn 1 + x1

whence √

1 1 + ··· + √ < n − 1. 1 + x1 1 + xn

1 1 + √ 1 < 1 + 1+ab , where a > 0, b > 0. Now let us prove that √1+a √ 1+b √ Define 1 + a  1 + x, 1 + b  1 + y, and note that x > 0, y > 0. We need to prove that (2 + x + y)(1 + x2 y2 + 2x2 y + 2xy2 + 4xy) < (1 + x + y + xy)(2 + x2 y2 + 2x2 y + 2xy2 + 4xy), or equivalently,

2xy + 2x2 y + 2xy2 < x + y + x3 y3 + 2x2 y3 + 2x3 y2 + 3x2 y2 .

(1)

220

14 Miscellaneous Inequalities

Indeed, we have x y + 2x3 y2 ≥ 2x2 y, + 2x2 y3 ≥ 2xy2 , 2 2  x y 3 3 2 2 + + 3x y ≥ 3 · xy > 2xy, x3 y3 > 0. 2 2 4 Summing these inequalities, we obtain inequality (1). 14.38. Consider the following two cases. 1 . Then note that (a) 1 ≤ α ≤ 2 + n−1 xiα − xi xi − xi2−α xi − xi2−α  ≥ , i  1, . . . , n. n x1 + · · · + xn + xiα − xi x1 + · · · + xn xi + x1 +···+x − xi2−α xα−1 i

Summing these inequalities, we obtain n  i1

If 1 ≤ α ≤ 2 −

 xi − x2−α xiα − xi i ≥ . x1 + · · · + xn + xiα − xi x + · · · + xn 1 i1

1 , n−1

n

(1)

then by Problem 3.42(a), it follows that n  xi − xi2−α ≥ 0. x + · · · + xn i1 1

If 2 −

1 n−1

≤α ≤2+

1 , n−1

then by Problem 3.42(b), again it follows that n  xi − xi2−α ≥ 0. x + · · · + xn i1 1

Hence, from (1) and (2) we have

n  i1

xiα −xi x1 +···+xn +xiα −xi

(2)

≥ 0.

1 . (b) α ≥ 2 + n−1 This solution was proposed by N.Nikolov. If we prove that there exists a number γ ≥ 1 such that

xiα

− xi ≥ x1 + · · · + xn + xiα − xi

n 

γ

nxi −

i1 n 

γ

xi

(n − 1)

i1

γ xi

then summing these inequalities, we will obtain

, i  1, . . . , n,

n  i1

xiα −xi x1 +···+xn +xiα −xi

≥ 0.

Proving inequality (3) is equivalent to proving the following inequality:

(3)

Proofs

221 γ

nxi (xiα−1 − 1) (n − 1)xi  ≥n − 1.  γ x1 + · · · + xn γ xi − xi

(4)

i1

It is sufficient to prove that ⎛ ⎞ γ α−1 x nxi 1)xi ⎝ α−1 i ⎠ ≥  (n −  − 1, α−1 − 1 n  x1 + · · · + xn x n · · · x n γ γ n xi − xi 1

(5)

i1

α−1

α−1

since x1 n · · · xn n ≥ 1. Without loss of generality, one can assume that xi  1. Now let us prove that there exists a number γ ≥ 1 such that the following inequality holds   1 1 n − 1 ≥ − 1, (6) x1 + · · · + xn G A n 

γ

xi −1

α−1

α−1

where A  n−1 , G  x1 n · · · xn n . Taking γ  (n−1)(α−1) , we obtain that A ≥ G. n n If G ≥ 1, then 0 ≥ G1 − 1 ≥ A1 − 1 and x1 +···+x ≤ 1, which ends the proof of n inequality (6). n ≥ 1, since by If G ≤ 1 and A ≤ 1, then G1 − 1 ≥ A1 − 1 ≥ 0 and x1 +···+x n x1 +···+xn −1 γ Jensen’s inequality, we have 1 ≥ A ≥ , which completes the n−1 proof of inequality (6). 1

n − 1 ≥ 0 > A1 − 1. If G ≤ 1 and A > 1, then x1 +···+x n G i1

Problems for Independent Study Prove the following inequalities (1–16). √ √ √ 1. 3x2 + 2x + 1 + 3x2 − 4x + 2 ≥ 351 .  2  √ √ √ 2. 2 + 2 − 1 ≤ x2 + y2 ≤ 1 and (2 − 2 + 2)2 ≤ (x − 2/2)2 + (y − √ √ √ √ √ 2 2 + 2/2)2  2 + 2, x ∈ [0, 1]. 2/2) ≤ 2 − 2,where (x + 2/2) + (y √ a12 + (1 − a2 )2 + a22 + (1 − a3 )2 + · · · + an2 + (1 − a1 )2 ≥ n 2/2, where 3. n√≥ 2.  √ √ √ 2 + 4 − 2 2 + · · · + 2n − 2 (n − 1)n ≥ n(n + 1), where n ∈ N. 4. √ 2 2 ≤ a+b − ab ≤ (a−b) , where a ≥ b > 0. 5. (a) (a−b) 8a 2 8b

222

6. 7.

8. 9. 10. 11. 12. 13. 14. 15.

14 Miscellaneous Inequalities

  √ √ (b) (a + c)2 + (b + d )2 ≤ a2 + b2 + c2 + d 2 ≤ (a + c)2 + (b + d )2 + √ |ad −bc| , where a > 0, b > 0, c > 0, d > 0. (a+c)2 +(b+d )2 √ √ √ √ 3 − n + 1 < 1/16 0 < 4n + 2 − n  √ n , where n ∈ N. 2 2 2 2 2 xz + z 2 ; (a) x + xy + y + y + yz + z ≥ x + √ √ √ (b) c a2 − ab + b2 + a b2 − bc + c2 ≥ b a2 + ac + c2 , where a > 0, b > 0, √ √ √ c > 0. 1 − cos(x3 − x2 ) + 1 − cos(x2 − x1 ) ≥ 1 − cos(x3 − x1 ). 3 √ 3 √ 4 x4 + 1dx + 4 x4 − 1dx < 9.0001. 9< 0 x13 x12 +x1 x2 +x22

0 x23 + x22 +x2 x3 +x32

+ 0, . . . , xn > 0. x17 x14 +2x13 x2 +2x1 x23 +x24

··· +

xn3 xn2 +xn x1 +x12

≥ 13 (x1 + . . . + xn ), where n ≥ 3, x1 >

x27 x24 +2x23 x3 +2x2 x33 +x34

x7

1 3 3 n + · · · + x4 +2x3 x +2x 3 4 ≥ 6 (x1 + · · · + xn ), n x1 +x1 n n 1 where n ≥ 3, x1 > 0, . . . , xn > 0. 3n + 4n + · · · + (n + 2)n < (n + 3)n , where n ≥ 6 and n ∈ N. 1 2015 1 < ln 2014 < 2014 . 2015

3 tan x ≥ x , where x ∈ 0, π2 . n n   tan αi ≥ (n − 1) cot αi , where n ≥ 3, 0 < αi < π2 , i  1, . . . , n and

i1 n 

+

i1

cos αi  1. 2

i1 3

16. a + b3 + c3 + 6abc > 41 (a + b + c)3 , where a > 0, b > 0, c > 0. 100  100  100 2 100 17. Compare the numbers 13 + 3 and √12 + 1 − √12 . 18. Prove that if a + b + c + d  0 and a2015 + b2015 + c2015 + d 2015  0, then 3  0. a3 + b3 + c3 + d √ 2 2 + (1 − α) ab ≤ a+b holds for every pair of positive 19. Prove that if α a +b 2 2 20. 21. 22. 23.

numbers a, b, then α ≤ 21 . Prove that if |a1 | + · · · + |an | < ε < 21 , then |1 − (1 + a1 ) · · · (1 + an )| < 2ε. The function f is defined on the set Q and f (0) · f (1) < 0. Prove that there exist rational numbers r1 and r2 , such that f (r1 ) − f (r2 ) > (r1 − r2 )2 . Given that a ≥ 0, b ≥ 0, c ≥ 0 and a + b + c ≤ 1, find the greatest value of the expression |a − b| · |b − c| · |c − a|. Let α, β, γ be the angles of some triangle, and a, b, c be its sides. Prove that √

(a) 0 < sin α + sin β + sin γ ≤ 3 2 3 , √ 1 √ √ (b) 0 < sin α + sin β + sin γ ≤ 3 · 34 4 , (c) 0 < sin α sin β sin γ ≤ (d) 2 < cos α2 + cos β2 + cos (e) cos α cos β cos γ ≤ 18 ,

(f) 0 < cos α2 cos β2 cos γ2 ≤ 2 +b2 +c2 1 (g) 13 ≤ a(a+b+c) 2 < 2,

√ 3 3 , 8 √ γ 3 3 ≤ , 2 2 √ 3 3 , 8

Problems for Independent Study

(h) (i)

1 4 1 4

≤ ≤

223

ab+bc+ca ≤ 13 , (a+b+c)2 (a+b)(b+c)(c+a) 8 ≤ 27 . (a+b+c)2

24. Let α, β, γ be the angles of an acute triangle. Prove that √

(a) 2 < sin α + sin β + sin γ ≤ 3 2 3 , √ 1 √ √ (b) 2 < sin α + sin β + sin γ ≤ 3 · 34 4 , (c) 34 ≤ sin2 α2 + sin2 β2 + sin2 γ2 < 1, (d) 1 < cos α + cos β + cos γ ≤ 23 , (e) 2 < cos2 α2 + cos2 β2 + cos2 γ2 ≤ 49 , √

(f) 21 < cos α2 cos β2 cos γ2 ≤ 3 8 3 , m+2 (g) tanm α + tanm β + tanm γ ≥ 3 2 , where m ≥ 1. 25. Prove that for an obtuse triangle with angles α, β, γ and sides a, b, c the following inequalities hold: √ (a) 0 < sin α + sin β + sin γ ≤ 1 + 2, (b) 0 < sin α sin β sin γ ≤ 21 , (c) 0 < cos α2 cos β2 cos γ2 ≤ α cos β cos γ ≥ 2, (d) 1+cos sin α sin β sin γ (e) (f) (g)

√ 1+ 2 , 4

2 1 +b2 +c2 3 ≤ a(a+b+c) 2 ≤ 8, 3 5 < ab+bc+ca ≤ 13 , 16 (a+b+c)2 (a+b)(b+c)(c+a) 9 8 < (a+b+c)2 ≤ 27 . 32

26. Prove that if 0 < a1 ≤ . . . ≤ an and 0 < b1 ≤ . . . ≤ bn , then (a1 + b1 ) · · · (an + bn ) ≤ (a1 + bi1 ) · · · (an + bin ) ≤ (a1 + bn ) · · · (an + b1 ), where the numbers i1 , . . . , in are a permutation of the numbers 1, . . . , n. 27. Given that the assumptions of Problem 26 hold for the numbers ai , bi , i  1, . . . , n, and g is a nondecreasing convex function (see Chapter §11). Prove that             bn bi1 bin bn b1 b1 + ··· + g ≤g + ··· + g ≤g + ··· + g . g a1 an a1 a1 a1 an 28. Let g be a convex function  n on [0, a1] and g(0) ≤ 0. Prove that n   (−1)i−1 g(ai ) ≥ g (−1)i−1 ai , where 0 < an ≤ . . . ≤ a1 . i1

i1

29.  Prove 0, . . ., an > 0, then n that, if a1 >  n n an−1 a1 + · · · + + aan1 ≥ aa21 + · · · + a2 an

an−1 an

+

an , a1

where n ≥ 2.

30. Suppose that for the polynomial p(x)  ax + bx + c the following assumptions hold: 2

224

14 Miscellaneous Inequalities

0 ≤ p(−1) ≤ 1, 0 ≤ p(0) ≤ 1, 0 ≤ p(1) ≤ 1. Prove that for all x ∈ [0, 1] the following inequality holds: p(x) ≤ 98 . 31. Given that for a trinomial p(x)  ax2 + bx + c on [0, 1] the inequality |p(x)| ≤ 1 holds, prove that |b| ≤ 8. 2 32. Given that for a trinomial    p(x)  ax +bx+c on [−1, 1] the inequality |p(x)| ≤ 1 holds, prove that p (x) ≤ 4 on [−1, 1].   Remark. For every polynomial p(x) the following inequality holds: p (x) ≤ 2n2 max|p(x)|, where x ∈ [a, b] and n is the degree of p(x). b−a [a;b]

33. Given that the degree of the polynomial p(x) is not greater than 2n and that for every integer k ∈ [−n, n] the inequality |p(k)| ≤ 1 holds, prove that for all x ∈ [−n, n] the following inequality holds: |p(x)| ≤ 22n . 34. Let x0 < x1 < · · · < xn be integers. Prove that at least one of the numbers |p(x0 )|, |p(x1 )|, . . . , |p(xn )| is not smaller than 2n!n , where p(x)  xn + a1 xn−1 + · · · + an . 35. Let x1 < . . . < xk , k ≥ 3, be given numbers belonging to [−1, 1]. Prove that 1 1 + ··· + ≥ 2k−2 . |(x1 − x2 ) · · · (x1 − xk )| |(xk − x1 ) · · · (xk − xk−1 )| 36. Prove that if a > 0, b > 0, c > 0, then a+

√

b c a + + ≤ 1. √ √ (a + b)(a + c) b + (b + a)(b + c) c + (c + a)(c + b)

37. Let p(x)  ax3 + bx2 + cx + d and suppose that for all x belonging to [−1, 1] the following inequality holds: |p(x)| ≤ 1. Prove that |a| + |b| + |c| + |d | ≤ 7. 38. Prove that (a) xλ (x − y)(x − z) + yλ (y − x)(y − z) + z λ (z − x)(z − y) ≥ 0, where x > 0, y > 0, z > 0; g(x)(f (x) − f (y))(f (x) − f (z)) + g(y)(f (y) − f (x))(f (y) (b) − f (z)) + g(z)(f (z) − f (x))(f (z) − f (y)) ≥ 0, where f and g are monotonic functions and E(g) ⊆ [0, +∞). 39. Let each of the edges of the complete graph on n vertices be colored in one of three colors. Prove that there exists a single-color connected subgraph having at least 2n vertices. √ √ √ √ 40. Prove that a1 + · · · + an ≤ a1 + C(n) · · · an + C(n), where n ≥ 2, a1 ≥ 0, . . . , an ≥ 0, C(n)  41. Prove that 0 ≤ sin x + . . . +

sin(nx) n

n−1 n−2

n n−1

.

√ < 2 2, where 0 ≤ x ≤

π . 2

Hint. If x  0, then there exists a positive integer m such that √

2 . 2m

√

2 2(m+1)

< sin

x 2

≤

Problems for Independent Study

225

42. Let f : R → R. Prove that f is constant if (a) for all real numbers x and y the following inequality holds: |f (x) − f (y)| ≤ C|x − y|α , where C > 0, α > 1, (b) for all real numbers a, b, c, d forming an arithmetic progression, the following inequality holds: |f (a) − f (d )| ≥ A|f (c) − f (b)|, where A > 3. 1 1 1 4 1 43. Prove that a+3 + b+3 + c+3 + d 1+3 ≤ 23 + 27 · abcd , where a > 0, b > 0, 27 c > 0, d > 0 and a + b + c + d  4. Hint. We have that   1 1 1 3a + 1 3b + 1 1 + + + ≤ · bcd + · acd abcd a+3 b+3 c+3 d +3 16 16 3d + 1 3 4 11 4 23 3c + 1 · abd + · abc ≤ abcd + + abcd  + abcd + 16 16 4 27 108 27 27

(see Problem 2.5).

Selected Inequalities Problems Prove the following inequalities.         1  ≤ ab − 1c  + bc − a1  + ca − 1b , where abc  0. 1. (ab + bc + ca)1 − abc  

2. sin2 β sin α2 sin α + β2 > sin2 α sin β2 sin β + α2 , where 0 < α < β and α + β < π.  a+b ab  3. a−b  ≥ 1, where a  b.

11 b−c 11 c−a 11 + b+c + c+a < 1, where a > 0, b > 0, c > 0. 4. a−b a+b 1 1 1 5. x0 + x0 −x + + · · · + ≥ xn + 2n, where n ∈ N and x0 > x1 > . . . > xn . x1 −x2 xn−1 −xn 1 n  1 6. Sk + Sm − Skm ≤ 1, where k, m ∈ N and Sn  . i i1



1 1 1 1 + 13 + · · · + 2n−1 > 1n 21 + 14 + · · · + 2n , where n ≥ 2, n ∈ N. 7. n+1 1

2

n

8. 2 2 · 2 22 · · · 2 2n < 4, where n ∈ N. 9. (a1 − a2 )(a1 − a3 )(a1 − a4 )(a1 − a5 ) + · · · + (a5 − a1 )(a5 − a2 )(a5 − a3 )(a5 − a4 ) ≥ 0. m+1

n+1

. 10. am + an ≥ mm + nn , where m, n ∈ N and a  mmm +n +nn (m+n)! n 2 n 11. 2 · n! ≤ (m−n)! ≤ (m + m) , where m, n ∈ N and m ≥ n.  √   12.  2 − pq  > 3q12 , where p ∈ Z, q ∈ N. √ √ 1 7 − mn > mn , where m, n ∈ N and 7 − mn > 0. 13. √   14. (x + y + z)2 − 3(xy + yz + zx) ≤ max(x, y, z) − min(x, y, z) ≤ 26 (x + y + z)2 − 3(xy + yz + zx). 15. 2 ab + bc + ac ≥ a + b + c + a1 + 1b + 1c , where a > 0, b > 0, c > 0 and abc  1. 16. (n − 0, 5)(1 + a2n ) ≥ a2n−1 + a2n−2 + · · · + a2 + a, where n ∈ N.

226

14 Miscellaneous Inequalities

17. x14 + · · · + xn4 ≥ x13 + · · · + xn3 , where x1 + · · · + xn ≥ n. 18. (−a1 + a2 + a3 + a4 )(a1 − a2 + a3 + a4 )(a1 + a2 − a3 + a4 )(a1 + a2 + a3 − a4 ) 19. 20. 21. 22. 23. 24.

≤ 8(a12 a42 + a22 a32 ). sin α + · · · + sin2 αn ≤ 49 , where α1 > 0, . . . , αn > 0 and α1 + · · · + αn  π. √ √ 1 √ √ √ √ a + b − c + b + c − a + c + a − b ≤ a + b + c, where a, b, c are the sides of some triangle. a3 + b3 + c3 ≥ a2 (2c − b) + b2 (2a − c) + c2 (2b − a), where a > 0, b > 0, c > 0. 1 1 + b3 +c31+abc + c3 +a13 +abc ≤ abc , where a > 0, b > 0, c > 0. a3 +b3 +abc √ 2 2 2 12uvw + u + v + w ≤ ≥ 0, v ≥ 0, w ≥ 0 and u + v + w  1.  1, where u  2   n(n−1) 1 1 · ≥ 4 · , where n ≥ 2, a1 > 0, . . . , an > 0. 2 ai ·aj ai +aj 2

1≤i<j≤n

1≤i<j≤n

25. 7(pq  + qr + rp) ≤2 + 9pqr, where p >0, q > 0, r > 0√and p + q + r  1. a2 + a22 + a32 + a22 + a32 + a42 + · · · + an2 + a12 + a22 ≥ 3(a1 + · · · + an ). 26.  1   √ 27. a12 + 2a22 + 3 a22 + 2a32 + 7 2a12 + a22 ≥ (5a1 + 4a2 + 2a3 ) 3. √ √ √ √ 28.  a2 + ab + b2 + b2 + bc + c2 + c2 + ca + a2 ≥ 3 · |a + b + c|.  3 −b3 b3 −c3 c3 −a3  (a−b)2 +(b−c)2 +(c−a)2 29.  aa+b + b+c + c+a  ≤ , where a > 0, b > 0, c > 0. 6 30. 31. 32. 33. 34. 35.

1 + y1k + z1k ≥ xk xk 1 + 1y + 1z ≥ x + y x 2 2

+ yk + z k , where k ≥ 2, k ∈ N, x > 0, y > 0, z > 0 and + z, xyz  1. x + y ≤ 2, where x > 0, y > 0 and x2 + y3 ≥ x3 + y4 . loga (loga b) + logb (logb c) + logc (logc a) > 0, where c > b > a > 1. sin(nα) ≤ 0, where n ∈ N, 0 < α < π and sin α + . . . + sin(nα) ≤ 0. sin((n + 1)α) ≥ 0, where n ∈ N, 0 < α < π and sin α + . . . + sin(nα) ≤ 0. x13 x63 + · · · + x5 +x5 +x5 +x ≤ 35 , where 0 ≤ xi ≤ 1, i  1, . . . , 6. 5 x5 +x5 +x5 +x5 +x5 +5 +x5 +5 2

3

4

5

6

1

2

3

4

5

36. (ab + ac + ad + bc + bd + cd )2 + 12 ≥ 6(abc + abd + acd + bcd ), where a + b + c + d  0. 37. (y3 + x)(z 3 + y)(x3 + z) ≥ 125xyz, where x ≥ 2, y ≥ 2, z ≥ 2. 38. (a + 3b)(b + 4c)(c + 2a) ≥ 60abc, where c ≥ b ≥ a ≥ 0. √ abc, where 39. (a + b + 3c)(a + 3b + c)(3a + b + c) + 8(ab + bc + ac + abc) ≥ 349 27 √ √ √ a > 0, b > 0, c > 0 and a + b + c  1. k l 40. kp · ql ≤ 1, where p, k, q, l ∈ N and p + q ≤ k + l. √ 41. n x1 · . . . · xn ≥ (n − 1)a, where n ≥ 2, a > 0, x1 > 0, . . . , xn > 0 and 1 1 + · · · + a+x ≤ a1 . a+x1 n



42. tan α1 · · · tan αn+1 ≥ nn+1 , where α1 , . . . , αn+1 ∈ 0, π2 and tan α1 − π4 +· · ·+

tan αn+1 − π4 ≥ n − 1. 1 1 43. n−1+x + · · · + n−1+x ≤ 1, where x1 > 0, . . . , xn > 0 and x1 · · · xn  1. 1 n 4

44. a4 + b4 + c4 − 2(a2 b2 + b2 c2 + c2 a2 ) + 3(abc) 3 ≥ 0, where a ≥ 0, b ≥ 0, c ≥ 0. 45. a4 + b4 + c4 − 2(a2 b2 + b2 c2 + c2 a2 ) + abc(a + b + c) ≥ 0, where a ≥ 0, b ≥ 0, c ≥ 0. 2 2 2 +b +c +a + bc+a + ca+b ≥ 2, where a > 0, b > 0, c > 0 and a + b + c  1. 46. ab+c

Selected Inequalities



227

   2 +cd 2 +da 2 + bb+1 + cc+1 + dd +ab ≥ 4, where a > 0, b > 0, c > 0, d > 0 +1 and abcd  1.



√ 3(ab + bc + ca) 9(a + b + c)2 + ab + bc + ca ≤ 9 (a + b + c)3 + abc , 48. where a > 0, b > 0, c > 0.

2 3 49. 4(x3 +y3 +z 3 +xyz)2 ≥ x2 + y2 + z 2 + x+y+z , where x > 0, y > 0, z > 0. 3

47.

a2 +bc a+1

50. 2, 25(x2 − x + 1)(y2 − y + 1)(z 2 − z + 1) ≥ (xyz)2 − xyz + 1. 51. ab(a2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a2 ) ≤ 41 (a2 +b2 +c2 )2 , where a > 0, b > 0, √ √ √ c > 0. √ a − 1 + b − 1 + c − 1 ≤ abc + c, where a ≥ 1, b ≥ 1, c ≥ 1. 52. 53. (p(xy))2 ≤ p(x2 )p(y2 ), where p(t)  a0 t n + a1 t n−1 + . . . + an and a0 ≥ 0, a1 ≥ 0, . . . , an ≥ 0. .(x1 + · · · + xn ), where x1 < · · · < xn and x1 , . . . , xn ∈ N. 54. x12 + · · · + xn2 ≥ 2n+1 3 1 1 1 1 55. (1+a) + + + (1+d ≥ 1, where a > 0, b > 0, c > 0, d > 0 and 2 (1+b)2 (1+c)2 )2 abcd  1. n  xi √ √ < π2 , where n ≥ 2, x0  1, x1 > 0, . . . , xn > 0 56. 1 < 1+x +···+x · x +···+x 0

i1

57. 58. 59. 60. 61.

i<j

62. 63. 64. 65. 66.

i−1

i

n

and x1 + · · · + xn√ 1.  0, 785n2 − n < n2 − 12 + · · · + n2 − (n − 1)2 < 0, 79n2 , where n ≥ 2, n ∈ N. a2 b2 c2 + (b+c)(b+a) + (c+a)(c+b) ≥ 34 , where a > 0, b > 0, c > 0. (a+b)(a+c) 1 1 1 + 1+bc + 1+ca ≥ 23 , where a > 0, b > 0, c > 0 and a2 + b2 + c2  3. 1+ab a b c + c+2a + a+2b ≥ 1, where a > 0, b > 0, c > 0. b+2c n   ai xi2 2 xi xj ≤ n−2 + , where n ≥ 2, a1 > 0, . . . , an > 0, x1 > n−1 1−ai i1

0, . . . , xn > 0 and a1 + · · · + an  x1 + · · · + xn  1. a3 + b3 + c3 ≥ a + b + c, where a > 0, b > 0, c > 0 and ab + bc + ca ≤ 3abc. a2 x + b2 y + c2 z > d 2 , where x > 0, y > 0, z > 0, 1x + 1y + 1z ≤ 1 and a, b, c, d are the sides of√some quadrilateral. √ √ √ x + y + z ≥ x − 1 + y − 1 + z − 1, where x ≥ 1, y ≥ 1, z ≥ 1 and 1 + 1y + 1z  2. x   1 1 k≥ x1k−1 + · · · + xnk−1 ≥ (n − 1) x11 + . . . + x1n , where 1+x k + · · · + 1+x k  1, n 1 2, k ∈ N and √ x1 ≥ 1, . . . , xn√≥ 1. √ √ a2 + ab + b2 +

b2 + bc + c2 +

c2 + ca + a2 ≤

5a2 + 5b2 + 5c2 + 4ab + 4bc + 4ac,

where   a > 0, b > 0, c > 0.  √ 3 67. 2 3 4 4 . . . n n < 2, where n ≥ 2, n ∈ N.     √ 68. 1 + 2 + 3 + · · · + n < 2, where n ∈ N.

n , where n ∈ N. 69. n! ≤ 7n+9 16 √

2 2

·

√1 2n

<

√

· 43 · · · 2n−1 < 23 · √12n , where n ≥ 2, n ∈ N. 2n √ √ 71. Maclaurin’s inequality: k bk ≥ k+1 bk+1 , k  2, . . . , n − 1, where

70.

1 2

228

14 Miscellaneous Inequalities bk 

1 (a1 · · · ak−1 ak + a1 · · · ak−1 ak+1 + · · · + a1 · · · ak−1 an + · · · + an−k+1 · · · an−1 an ), Cnk

where n ≥ 2, a1 , a2 , . . . , an > 0. 72. a1 + a22 + · · · + ann ≥ an , where n ≥ 2 and ai+j ≤ ai + aj , i, j ∈ N, i + j ≤ n. 73. a1 a24 + a2 a34 + · · · + an a14 ≥ a2 a14 + a3 a24 + · · · + a1 an4 , where n ≥ 3 and a1 < a2 < · · · < an . u2 , where n ∈ N and u1  1, u2  2, un+2  3un+1 − 74. un+2 + un ≥ 2 + un+1 n un , n  1, 2, . . . . 75. ai ≥ 0, i  1, 2, . . . , n, where n ≥ 3 and ai−1 +ai+1 ≤ 2ai , i  2, . . . , n−1. 76. max(a1 , a2 , . . . , an ) ≥ 2, where n ≥ 4 and a1 +· · ·+an ≥ n, a12 +· · ·+an2 ≥ n2 . n−1 n−1 n n · a1 +···+a ≥ a0 +···+a · a1 +···+a , where n ≥ 2 and ai > 0, i  77. a0 +···+a n+1 n−1 n n 2 0, 1, . . . , n, ai−1 ai+1 ≤ ai , i  1, . . . , n − 1. √ n+1 n 78. 1−x > 1−x · x, where n ∈ N, 0 < x < 1. n+1 n 79. sin(x1 + x2 ) + . . . + sin(xn−1 + xn ) + sin(xn + x1 ) > 2, where n  3, 4, . . . and x1 + · · · + xn  π2 . 80. xy + yx > 1, where x > 0, y > 0. 81. a12 + · · · + an2 − n ≥ n−√2nn−1 · (a1 + · · · + an − n), where n ≥ 2 and a1 · · · an  1. 82. x1 arctan x1 + . . . + xn arctan xn ≥ n ln 2, where n ≥ 2 and x1 > 0, . . . , xn > 0, x1 · · · xn  1. p 83.

n  i1

n 

p

ai bi

≥n

2−p

·

ai

i1 n 

bi

, where ai > 0, bi > 0, i  1, . . . , n, p ≥ 2.

i1 x

x

2

n

84. x2 + · · · + xn ≥ 2x + · · · + nx , where x ≥ n ≥ 2, n ∈ N. 1 1 1 + 2−b + 2−c ≥ 3, where a ≥ 0, b ≥ 0, c ≥ 0 and a2 + b2 + c2  3. 85. 2−a a b c d e 86. 4+a 2 + 4+b2 + 4+c2 + 4+d 2 + 4+e2 ≤ 1, where a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0, e ≥ 0 1 1 1 1 1 and 4+a + 4+b + 4+c + 4+d + 4+e  1.  p  n n  a1 +...+ak p p p ≤ p−1 ak , where p > 1 and ai > 87. Hardy’s inequality: k 0, i  1, . . . , n.

k1

k1

n    88. Abel’s inequality:  ak bk  ≤ B(|a1 | + 2|an |), where |b1 + . . . + bk | ≤ i1 B, k  1, . . . , n,  a1 ≤ . . . ≤ an or a1 ≥ . . . ≥ an . n   89. Abel’s inequality:  ak bk  ≤ Ba1 , where |b1 + · · · + bk | ≤ B, k  i1 1, . . . , n, a1 ≥ · · · ≥ an ≥ 0. n k √

  ai2 > 41 1 + 21 + · · · + 1n , where a1 > 0, . . . , an > 0 and ai > k, k  90. i1

i1

1, . . . , n. 91. x11 + · · · + x1n ≤ y11 + · · · + y1n , where n ≥ 2, xi > 0, yi > 0, x1 + · · · + xi ≥ y 1 + · · · + yi , i  1, . . . , n and x1 y1 < x2 y2 < · · · < xn yn . n−1  cos(n−k)x n−1  cos(n+k)x   < 6, where n ≥ 2, n ∈ N. 92.  −  k k k1

k1

Selected Inequalities

229

 n−1   cos(n−k)x n−1 √  cos(n+k)x   93.  −  < 4 2, where n ≥ 2, n ∈ N and 0 ≤ x ≤ k k k1

k1

1

1

1

π . 2

1

94. a1k + · · · + ank ≤ b1k + · · · + bnk , where k ≥ 2, k ∈ N, ai > 0, bi > 0, b1 + · · · + bi ≥ a1 + · · · + ai , i  1, . . . , n and b1 ≤ · · · ≤ bn . 95. x11 +· · ·+ x1n ≤ u11 +· · ·+ u1n , where u1  2, u2  3, uk+1  u1 · · · uk +1, k  2, 3, . . ., and x1 , . . . , xn ∈ N, x11 + · · · + x1n < 1. 96. A(b − c)2 + B(c − a)2 + C(a − b)2 ≥ 0, if at least one of the following conditions holds: a ≥ b ≥ c and B ≥ 0, B + A ≥ 0, B + C ≥ 0, a ≥ b ≥ c and A ≥ 0, C ≥ 0, A + 2B ≥ 0, C + 2B ≥ 0, a ≥ b ≥ c > 0 and B ≥ 0, C ≥ 0, a2 B + b2 A ≥ 0, A + B + C ≥ 0, AB + BC + CA ≥ 0, a ≥ b ≥ c, A ≥ 0, B ≥ 0, b2 B + c2 C ≥ 0, and a, b, c are the sides of some triangle.   1 1 1 ≥ 49 , where x > 0, y > 0, z > 0. 97. (xy + yz + zx) (x+y) 2 + (y+z)2 + (z+x)2 n  98. 1−h < x2i (x2i+1 − x2i−1 ) < 1+h , where 0  x1 < x2 < · · · < x2n+1  1 and 2 2 (a) (b) (c) (d) (e)

i1

xi+1 i ≤ h, i  1, 2, . .. , 2n.    − x  ai − aj  + bi − bj  ≤ ai − bj . 99. 1≤i<j≤n 1≤i,j≤n  n 4  n  n    2  2 2 2 100. Carlson’s inequality: ai ≤ π ai i ai . i1

i1

i1

Hints     1  ≤ ab + bc + ca − 1c − a1 − 1b . 1. (ab + bc + ca)1 − abc sin2 β sin α2 sin(α+ β2 )−sin2 α sin β2 sin(β+ α2 ) 2.  2 sin β−α cos α+β sin(α + β). 2 2 sin α2 sin β2

 a+b  3. ab ·  a−b  − 1 ≥ 0. 4. Let max(a, b, c)  a, and then a−b < 1, b−c ≤ a−c . a+b b+c c+a 5. x0 − xn  (x0 − x1 ) + · · · + (xn−1− xn ), n ≥ 2. 

1 1 1 1 6. Skm  Sk + k+1 ≥ Sk +k· 2k1 +· · ·+k· mk + · · · + 2k1 +· · ·+ (m−1)k+1 + · · · + mk .   



1 1 1 1 1 1 1 1 1 1 7. n+1 1 + 3 + · · · + 2n−1 − n 2 + 4 + · · · + 2n  2n(n+1) 2n n+1 + · · · + 2n −   1 + · · · + 1n .



8. 21 + 222 + · · · + 2nn < 21 + 212 + · · · + 212 + 213 + · · · + · · ·  1 + 21 + 212 + · · ·  2. 9. Let a1 ≤ a2 ≤ . . . ≤ a5 , then (a3 − a1 )(a3 − a2 )(a3 − a4 )(a3 − a5 ) ≥ 0, (a1 −a2 )(a1 −a3 )(a1 −a4 )(a1 −a5 ) ≥ (a1 −a2 )(a2 −a3 )(a2 −a4 )(a2 −a5 ), and (a5 − a1 )(a5 − a2 )(a5 − a3 )(a5 − a4 ) ≥ (a4 − a1 )(a4 − a2 )(a4 − a3 )(a5 − a4 ).

230

14 Miscellaneous Inequalities m

m−1

m−1

+···+a 10. Let m ≥ n, then m ≥ a ≥ n and a − n  mnn · (m − a) ≥ mnn−1 +···+a n−1 · (m − a). (m+n)! 11. (m−n)!  (m − n + 1) · · · (m + n) ≥ (2n)! and k(2m + 1 − k) ≤ m2 + m, k  m − n + 1, . . . , m.  √ √ √   12. If p ≤ 0, then  2 − pq  ≥ 2 > 13 ≥ 3q12 , and if 0 < p < (3 − 2)q, √    |2q2 −p2 | 1 then  2 − pq   q(√2q+p) ≥ q(√2q+p) > 3q12 . It is left to consider the case √ p ≥ (3 − 2)q. 13. m2 ≤ 7n2 − 3.

14. (x + y + z)2 − 3(xy + yz + zx)  21 (x − y)2 + (y − z)2 + (z − x)2 . 15. xy + yz ≥ x + 1z , where x > 0, y > 0, z > 0 and xyz  1. k 2n−k 16. 1 + a2n ≥ |a| , where k ∈ {1, 2, . . . , n}.  3  + |a| 4 3   17. x − x ≥ x (|x| − 1) ≥ |x| − 1. 18. 8(a12 a42 + a22 a32 ) − (−a1 + a2 + a3 + a4 )(a1 − a2 + a3 + a4 )(a1 + a2 − a3 + a4 )(a1 + a2 + a3 − a4 )  (a12 + a42 − a22 − a32 )2 + 4(a1 a4 − a2 a3 )2 . 19. sin2 α + sin2 β ≤ sin2 (α + β), where α > 0, β > 0 and α + β ≤ π2 . √ √ √ 2 2 2 √ a+b−c) +( b+c−a) a+b−c+ b+c−a 20. ≤(  b. 2 2 √ 21. a3 + c2 a ≥ 2 a3 · c2 a  2a2 c. abc abc c 22. a3 +babc 3 +abc  (a+b)(a 2 −ab+b2 )+abc ≤ (a+b)ab+abc  a+b+c . 2 23. (uv + vw + wu) ≥ 3(uv · vw + vw · wu + wu · uv)  3uvw. 2  1 . 24. ai1·aj ≥ 4 · ai +a j 25. Let min(p, q, r)  r, then

7(pq + qr + rp) − 9pqr  (7 − 9r)pq + 7r(p + q) ≤ ≤ (7 − 9r)

1 (p + q)2 + 7r(p + q)  2 − (3r − 1)2 (r + 1). 4 4

√  x2 + y2 + z 2 ≥ 33 (x + y + z). √  27. x2 + 2y2 ≥ 33 (x + 2y).  

2  √ 2 2 2 28. x + 2y + 23 y . x + xy + y    3 3 3 3  −b |a−b|·|b−c|·|c−a| −c3 −a3  + bb+c + c c+a 29.  aa+b   (a+b)(b+c)(c+a) (ab + bc + ca).

26.

Let a ≤ b ≤ c, then

|c−a| c+a

< 1,

ab+bc+ca (a+b)(b+c)

< 1 and

  6|a − b| · |b − c|  2|a − b| · |b − c| + 4|a − b| · |b − c| ≤ (a − b)2 + (b − c)2 + ((b − a) + (c − b))2 .

30. 31. 32. 33.

(1 − x)(1 − y)(1 − z) ≥ 0, and therefore (1 − xk )(1 − yk )(1 − z k ) ≥ 0. x2 − x3 ≥ y4 − y3 ≥ y3 − y2 and (x + y)(x3 + y3 ) ≥ (x2 + y2 )2 ≥ (x3 + y3 )2 . loga (loga b) > logb (loga b). · cot α2 + 21 sin(nα). sin α + · · · + sin(nα)  sin2 nα 2

Hints

231 (n+1)α · cot α2 − 21 sin((n + 1)α). 2 3 xi and 3xi5 − 5xi3 + 2  x15 +x25 +x35 +x45 +x55 +x65 +4

34. sin α + · · · + sin(nα)  sin2 35. 36.

37. 38. 39.

40. 41.

xi3 x15 +x25 +x35 +x45 +x55 +x65 −xi5 +5

≤

6xi2 + 4xi + 2). Set a  x+α, b  y+α, c  z+α, d  t +α, X  xy+xz+xt +yz+yt +zt, Y  xyz + xyt + xzt + yzt. In this case, the given inequality can be rewritten as

2

X − 6α 2 + 12 ≥ 6 Y + 2αX − 8α 3 , where x + y + z + t  −4α. Let us choose the number α such that 36α 4 + 12  −48α 3 , for example α  −1. We need to prove that X 2 ≥ 6Y , where x + y + z + t  4. If Y ≤ 0, then X 2 ≥ 0 ≥ 6Y . And if Y > 0, then X 2  2Y (x + y + z + t) + 2 + (yz)2 + (yt)2 ≥ 8Y > 6Y . Therefore X 2 > 6Y . (xy − zt)2 + (xz)2 + (xt)  5 3 . y + x ≥ 4y + x ≥ 5 xy4√ √ √ √ √ √ 3 4 5 3 4 5 (a+3b)(b+4c)(c+2a) ≥ 4 ab3 ·5 bc4 ·3 ca2 ≥ 60 ab3 · bc4 · c3/ 5 b2/ 5 a2 . (a + b + 3c)(a + 3b + c)(3a + b + c) + 8(ab + bc + ac + abc) ≥ ( 31 + 2c)( 13 + 2b)( 13 + √ 2a) + 8(ab + bc + ac) + 8abc ≥ 19 + 28 (ab + bc + ac) + 16abc ≥ 19 + 28 ( ab · 3 3 √ √ √ √ √ √ √ bc + bc · ca + ca · ab) + 16abc  19 + 28 abc + 16abc ≥ 349 abc. 3 27 1 p p p q q p q q k+l . k + l ≥ p + q  k + · · · + k + l + · · · + l ≥ (k + l) · k · · · k · l · · · l 1  n−1    a+xi a+xi a+xi a+xi + · · · + a+x − 1 ≥ (n − 1)a a+x · · · a+x . xi ≥ a a+x 1 n 1 n

42. Let tan αi  xi , then 43.

(xi − 1)2 (3xi3 +

xi n−1+xi



xi (n−1)x1α ···xnα +xi

1 1+x1



+ ··· +

≤ 1.

1 1+xn

xi1−α x1α ···xnα (n−1) xα i

+xi1−α

≥

xi1−α , where (n−1)α x11−α +···+xn1−α

 1−α.

 4 3  4 3  4 3 4 4 4 8 4 4 8 8 4 4 8 8 4 a 3 + b 3 + c 3 + 3a 3 · b 3 · c 3 ≥ a 3 b 3 + a 3 b 3 + b 3 c 3 + b 3 c 3 + c 3 a 3 +  8 4  8 4  8 4 4 8 4 8 4 8 4 8 c3 a3 ≥ 2 a3 b3 · a3 b3 + 2 b3 c3 · b3 c3 + 2 c3 a3 · c3 a3 . 4 45. abc(a + b + c) ≥ 3(abc) 3 . a2 +b a+b 46. b+c  b+c − a. (a+1)3 47. a2 + bc ≥ 2a(d . +1) 44.

48. Let a + b + c  p, ab + bc + ca  p −q , where q ≥ 0. Then if q < p ≤ 2q, 3 2 p−2q then min(abc)  0, and if p > 2q, then min(abc)  p+q · 3 . 3 49. Let x + y + z  3u, xy + yz + zx  3v2 , xyz  w3 . Then one needs to prove that 4(27u3 − 27uv2 + 4w3 )2 ≥ (9u2 − 6v2 + u2 )3 . 50. 1.5(x2 − x + 1)(y2 − y + 1) ≥ ((xy)2 − xy + 1). 51. Let a ≥ b ≥ c. If a > c, then without loss of generality one can assume   that (a − c)2 + (b − c)2  1. Therefore a  c + cos α, b  c + sin α, α ∈ 0, π4 . √ √ √ x · 1 + 1 · y ≤ (x + 1)(1 + y), where x ≥ 0, y ≥ 0. 52.

2 √

2 √ √ 2 53. p(x2 )  a0 · xn + a1 · xn−1 + · · · + an . 54. Let yi  xi − i, then 0 ≤ y1 ≤ · · · ≤ yn . 2

2

232

14 Miscellaneous Inequalities

55. Let

56.

1 (1+a)2

+

1 (1+b)2

+

1 (1+c)2

+

 R2 , then

1 (1+d )2

1 1  R cos α cos β,  R cos α sin β, 1+a 1+b  π 1 1 .  R sin α cos γ ,  R sin α sin γ , α, β, γ ∈ 0, 1+c 1+d 2 n  xi x2  x1 +  √ √ 1 + x + · · · + x · x + · · · + x 0 i−1 i n 1 − x2 i1 1

xn

+ ··· +  > x1 + x2 + · · · + xn  1. 1 − (x1 + · · · + xn−1 )2

 Let x1 + · · · + xi  cos αi , αi ∈ 0, π2 , i  1, 2, . . . , n. Then xi √ < sin(αi−1 − αi ), i  2, 3, . . . , n. 1−(x1 +···+xi−1 )2  √ 2 2 57. πn4 − n < n2 − 12 + · · · + n2 − (n − 1)2 < πn4 . 2 a2 b2 c2 58. (a+b)(a+c) + (b+c)(b+a) + (c+a)(c+b) ≥ a2 +b2 +c(a+b+c) 2 +3ab+3bc+3ca . 59. 60. 61. 62. 63. 64.

1 1 1 32 + 1+bc + 1+ca ≥ 3+ab+bc+ca . 1+ab 2 a a  ab+2ac . b+2c n n n    ai xi2 xi2  − xi2 . 1−ai 1−ai i1 i1 i1 2 2 2 (a+b+c)2 . a3 + b3 + c3  1a a + 1b b + 1c c ≥ 1 a+1 / 2 / 2 / 2 / / b+1/ c a b c 2 2 2 a x + b y + c z  1 x + 1 y + 1 z. / 2 / √ /2 √ √ 2 x−1) y−1) z−1 ( ( + +( z ) . 1 x y x1k−1 ·1 k−1 1 x +x1

+ ··· + 65. n − 1  1   1 1 ≤ 0. k−1 − 1 k−1 1 +x +x xi

i

a2 +ab+b2  a+b

xj

xnk−1 ·1 k−1 1 xn +xn

xk−1 +···+xk−1 ·n ≤ 1(+x1 k−1 +···+ n1 +x) k−1 , as x1

1

xn

n



xik−1 k−1 1 x +xi

−

i

j

b2 +bc+c2 b+c

2

2

2

2

2

+4ab+4bc+4ac + c +ca+a ≤ 5a +5b +5c . c+a 2(a+b+c)  √ m 67. m m+1 (m + 1) . . . n n < 2, where 2 ≤ m ≤ n.    √ 68. m + m + 1 + · · · + n < m + 1, where 1 ≤ m ≤ n. 69. If n ≥ 5, then

66.

+



70.

2n+1 2n+2

>

√ n √ n+1

   n  7 7n + 9 n 7n + 16 7n + 16 n+1 · 1+ :  16 16 16 7n + 9   7n n(n − 1) 49 7n + 16 · 1+ + · ≥ 16 7n + 9 2 (7n + 9)2   49n(n − 1) 1 · 7n + 16 + 7n + > n + 1. > 16 14n + 18 and

1 2

· 43 · · · 2n−1 ≤ 2n

√1 . 3n+1

xjk−1 1 xj

+xjk−1

 ·

Hints

233

k  k+2  √k √ √ √ √ √ k+1 k+1 k+1 · k+2 bk bk+1 · k+2 bk b √2 √ k 71. If 2 ≤ k ≤ n − 2, then k+2√bbk+1  k+2b√k+1 ≥  . k+1 k+2 bk+2 ·bk bk+1 k+2 bk+1      72. (k + 1) a + a2 + · · · + ak+1  (a + a ) + a + a2 + a 1 1 k 1 k−1 + · · · 2 k + 1 2   ak a2 + ··· + + a1 + ak+1 + a1 + 2 k ≥ (a1 + ak ) + (a2 + ak−1 ) + · · · + (ak + a1 ) + ak+1 ≥ (k + 1)ak+1 . 73. (a2 + a1 )(f (a2 ) − f (a1 )) + · · · + (an + an−1 )(f (an ) − f (an−1 )) ≥ (an + a1 )(f (an ) − f (a1 )), where f (x)  x4 is convex in R. 74. 2un ≤ un+1 < 3un , n  1, 2, . . . . / A and n ∈ / A, then from k ∈ A 75. Let A  {i|ai  min(a1 , . . . , an )}. Then if 1 ∈ it follows that k − 1 ∈ A. 76. Let A  {i|ai ≥ 0} and |A|  m. If max(a1 , a2 , . . . , an ) < 2, then 4m + (2m − n)2 > a12 + · · · + an2 ≥ n2 , and therefore m  n. 77. Let aai+1i  bi , i  0, 1, . . . , n − 1, S  a1 + · · · + an−1 . Then b0 ≤ b1 ≤ . . . ≤ bn−1 , and therefore,

a0 an−i  b0 · · · bi−1 ≤ bn−i · · · bn−1  , 1 ≤ i ≤ n, S ≥ (n − 1)2 a0 an . ai an n 2 n+1 n−1 78. 1−x · 1−x > 1−x , n  2, 3, . . . n+1 n−1 n 79. sin x > π2 · x, where 0 < x < π2 . 80. Let 0 < x < 1, 0 < y < 1. Then xy 

1 1 x . > x+y y > 1+y( 1x −1) (1+ 1x −1) 2n 81. c2 > c3 > · · · > cn > · · · and cn ≥ n−√n−1 , n  2, 3, . . . , where cn  2

n−2 2

n−3 2

) +2(t ) +···+n−1) . min (t+1) t(t((tn−2 +2t n−3 +···+n−1) (0,1]

82. Let f (x)  x arctan x. Then f  (x)  (1+x2 2 )2 , and therefore, x1 arctan x1 + · · · + n n xn arctan xn ≥ n · x1 +···+x arctan x1 +···+x . n n       2 n n n 83. p a   p i bi ≥ ai2 · b i i1 i1 i1 ⎛ ⎞ 2p ⎞2 ⎛ n  n p a n ⎜  ⎜ i1 i ⎟ ⎟ ⎜ ⎟ ⎟ ⎟  n2−p ≥⎜ n⎜ ai . ⎝ i1 ⎝ n ⎠ ⎠ i1 84. Let f (x)  x ln k + ln(ln x) − k ln x − ln(ln k). If x ≥ k ≥ 3, then f  (x)  ln k + ln1x · 1x − kx > ln k − 1 > 0, and if x ≥ k  2, then

234

14 Miscellaneous Inequalities

  1 1 2 1 1 f (x)  ln 2 + · −  x ln 2 + −2 ln x x x x ln x     1 1 1 √ 2 2 ln 2 − 2 > 0. ≥ −2 > 2 x ln 2 · x ln x x 

x

85. 86. 87. 88.

k

Hence if x ≥ k ≥ 2, k ∈ N, then f (x) ≥ f (k), and therefore xk ≥ k x . Let a2  x and f (x)  2−1√x . If 0 ≤ x ≤ 3, then f (x) ≥ f (1) + f  (1)(x − 1).

1

2 1 1 Let 4+a + f  15 ·  x and f (x)  17xx−4x 2 −8x+1 . If 0 < x ≤ 4 , then f (x) ≤ f 5

x − 15 . See  11.25(c).  n Problem     ak bk   |(a1 − a2 )b1 + (a2 − a3 )(b1 + b2 )    i1

+ · · · + (an−1 − an )(b1 + · · · + bn−1 ) + an (b1 + · · · + bn )| ≤ |a1 − a2 |B + |a2 − a3 |B + · · · + |an−1 − an |B + |an |B   |a1 − an |B + |an |B ≤ B(|a1 | + 2|an |).  n  89.    ak bk   |(a1 − a2 )b1 + (a2 − a3 )(b1 + b2 )    i1

+ . . . + (an−1 − an )(b1 + · · · + bn−1 ) + an (b1 + · · · + bn )| ≤ |a1 − a2 |B + |a2 − a3 |B + · · · + |an−1 − an |B + |an |B  Ba1 . 90. Without loss of generality one can assume that a1 ≥ a2 ≥ · · · ≥ an . n 

ai2  (a1 − a2 )a1 + (a2 − a3 )(a1 + a2 )

i1

+ · · · + (an−1 − an )(a1 + · · · + an−1 ) + an (a1 + · · · + an ) √ √ √ √ > (a1 − a2 ) 1 + (a2 − a3 ) 2 + · · · + (an−1 − an ) n − 1 + an n         1 2 1 1 1 1 2 1 a1 · √ + · · · + an · √ > · + ··· + √ > , √ 2 4 n n 1 1 √ 91. since 2 k > 1 + · · · +

√1 , k

k  1, . . . , n.

1 1 · (x1 − y1 ) + · · · + · (xn − yn ) x1 y1 xn yn     1 1 1 1 · (x1 − y1 ) + · · · + − −  x1 y1 x2 y2 xn−1 yn−1 xn yn 1 · (x1 − y1 + · · · + xn − yn ) ≥ 0. · (x1 − y1 + · · · + xn−1 − yn−1 ) + xn yn

Hints

235

 n−1    n−1 n−1  sin(kx)  92.  cos(n − k)x  cos(n + k)x     −   2|sin(nx)|      k k k  k1 k1 k1  n−1   sin(kx)    ≤ 2  < 6.  k  k1 n−1    cos(n−k)x n−1 √  cos(n+k)x    0 < 4 2. If 0 < x ≤ π , then 93. If x  0, then  −  k k 2 k1 k1  x 1 1 there exists a positive integer m such that √2(m+1) < sin 2  ≤ √2m . n−1 n−1     sin(kx)    cos(n−k)x n−1  cos(n+k)x      ≤ For n ≤ m + 1 we have  −  ≤ 2 k k k  k1 k1 k1 √ 2m|sin x| < 2 2.   √ 1 < + . . . + sin((n−1)x) Prove that if n ≥ m + 2, then  sin(m+1)x < 2. m+1 n (m+1)|sin 2x | 94. Without loss of generality one can prove that a1 ≤ a2 ≤ · · · ≤ an . Let Ai   1 k−1  1 k−1  1 k−2 1 + bik · aik + · · · + aik . Then Ai ≤ Ai+1 , i  1, . . . , n − 1, bik and  1 1 1 1 b1k + · · · + bnk − a1k + . . . + ank   bn − an 1 1 b1 − a1 + ··· +  (b1 − a1 ) −  A1 An A1 A2   1 1 + · · · + (b1 − a1 + · · · + bn−1 − an−1 ) − An−1 An 1 + (b1 − a1 + · · · + bn − an ) ≥ 0. An 95. Let x1 ≤ · · · ≤ xn . Then one can prove that x11 ≤ u11 , . . . , x11 + · · · + 1 1 + · · · + un−1 . u1 1 1 If x1 + . . . + xn > u11 + . . . + u1n , then     1 1 1 1 1 <1− ≤1− + ... + + ... + x1 · . . . · xn x1 xn u1 un 1 1  .  un+1 − 1 u1 · . . . · un

1 xn−1

≤

96. (a) A(b − c)2 + B(c − a)2 + C(a − b)2 ≥ A(b − c)2 + B(a − b)2 + B(b − c)2 + C(a − b)2 ≥ (B + A)(b − c)2 + (B + C)(a − b)2 ≥ 0.

(b) A(b − c)2 + B(c − a)2 + C(a − b)2 ≥ min(A, C) (b − c)2 + (a − b)2 + 2 B(c − a)2 ≥ min(A, C) · (a−c) + B(c − a)2  21 (min(A, C) + 2B)(a − c)2 ≥ 0. 2 2 2 2 2 2 (c) A(b − c) +B(c − a) + C(a − b)  ≥ A(b − c) + B(c − a)  (b − 2 2 c)2 A + (c−a) B ≥ (b − c)2 A + ab2 B ≥ 0. (b−c)2

236

14 Miscellaneous Inequalities

(d) Let A ≥ B ≥ C. Then A + B ≥ 23 (A + B + C) ≥ 0. If A + B > 0, then A(b − c)2 + B(c − a)2 + C(a − b)2  (A + B)c2 − 2c(bA + aB) + (A + C)b2 + (B + C)a2 − 2Cab ≥ 0, since D  −(AB + BC + CA)(a − b)2 ≤ 0. If A + B  0, then AB + BC + CA  AB ≥ 0, and therefore, A  B  C  0. (e) A(b − c)2 + B(c − a)2 + C(a − b)2 ≥ B(c − a)2 + C(a − b)2  (a − 2 (c−a)2 b)2 B (a−b) ≥ (a − b)2 B bc2 + C ≥ 0. 2 + C 97. Let x + y  a, y + z  b, z + x  c and a ≥ b ≥ c. Then one needs to prove that A(b − c)2 + B(c − a)2 + C(a − b)2 ≥ 0, where a, b, c are the sides 2 2 2 − a12 , B  ac − b12 , C  ab − c12 . We have of some triangle and A  bc 2 2 2 1 A  bc − a12 ≥ a22 − a12 > 0, B  ac − b12 ≥ ac − bc > 0, Bb2 + Cc2 

2(b3 + c3 ) − 2abc abc 2(b + c)bc − 2abc > 0. ≥ abc

 n  98.  1  x2i (x2i+1 − x2i−1 ) −    2 i1  n  n    x2i+1 + x2i−1    (x2i+1 − x2i−1 ) x2i (x2i+1 − x2i−1 ) −   2 i1 i1  n   x  2i+1 + x2i−1 − 2x2i    (x2i+1 − x2i−1 )   2 i1   n n    x2i+1 + x2i−1 − 2x2i  h h (x2i+1 − x2i−1 ) <  ≤ (x2i+1 − x2i−1 )  .   2 2 2 i1 i1 99. Without loss of generality one can assume that ai , bi ∈ [0, 1], i  1, . . . , n. Let fi (x)  1 if x ∈ [0, ai ] and fi (x)  0 if x ∈ (ai , 1], i  1, . . . , n, gi (x)  1 if x ∈ [0, bi ] and gi (x)  0 if x ∈ (bi , 1], i  1, . . . , n. Then   min(ai , aj ) + min(bi , bj ) 1≤i,j≤n

1≤i,j≤n

#1 

#1 (f1 (x) + · · · + fn (x)) dx +

(g1 (x) + · · · + gn (x))2 dx

2

0

0

#1 ≥

2(f1 (x) + · · · + fn (x))(g1 (x) + · · · + gn (x))dx  2 0

 1≤i,j≤n

min(ai , bj ).

Hints

237 n2 , i  1, . . . , n, t > 0. Then t ⎛ ⎞2 ⎛ ⎞  n n   1 2 ⎝ ⎝ ⎠ ai ≤ (ai ci ) ⎠ 2 c i i1 i1  ⎞ ⎛ n  2 a2 ⎞ ⎞ ⎛ ⎛ i ⎜ n n i ⎟  ⎟ ⎜  2 t ⎠ ⎟·⎝ ⎠t + i1 ⎝ a ⎜ ⎟ ⎜ i t t 2 + i2 ⎠ ⎝ i1 i1

100. Let ci2  t +

⎞  ⎛ n  ⎞ ⎛ ⎞ ⎛ i2 ai2 ⎟ ⎜  #i n n  ⎜ ⎟ 1 i1 ⎟ ⎜ ⎟·⎝ ⎝ ≤⎜ ai2 ⎠t + t dx⎠ ⎜ ⎟ t t 2 + x2 ⎝ i1 ⎠ i1 i−1

⎞  ⎛ n  2 a2 ⎞ ⎞ ⎛ n ⎛ i ⎜  # i ⎟ ⎜ n 2 ⎟ 1 i1 ⎜ ⎟ ⎠ ⎝ ⎝ ≤⎜ ai t + dx⎠ ⎟· t t t 2 + x2 ⎝ i1 ⎠ 0

 ⎞ ⎛ n  ⎞ ⎛ i2 ai2 ⎟ ⎜ n ⎜  2 ⎟ i1 ⎟ · arctan n ⎝ ⎜ ai ⎠t + ⎜ ⎟ t t ⎝ i1 ⎠ ⎛

⎞ ⎛ ⎜  n π ⎜ 2 ⎜ ⎝ ≤ ai ⎠t + · 2 ⎜ ⎝ i1

$  % n 2 2 % % i1 i ai if t  & n . i1

ai2



⎞ n  i2 ai2 ⎟ ⎟ i1 ⎟π ⎟ t ⎠

$⎛ ⎞⎛ ⎞ % n n %   %⎝ 2 2 2 ⎠ ⎝ ·& ai i ai ⎠, i1

i1

Appendix—Power Sums Triangle

Let n and k be positive integers. Sums of the forms 1k þ 2k þ    þ nk often arise in mathematics, particularly in algebra. These sums have been studied for hundreds of years. They have wide applications in mathematics and can also be used for proving algebraic Inequalities. In 1631, the Prussian mathematician Johann Faulhaber (1580–1635) published (Academiae Algebrae, 1631) the general formula for the sums of powers of the first n positive integers. In 1713, Nicolaus Bernoulli (also spelled Niklaus), the nephew of the Swiss mathematician Jacob Bernoulli, 7 years after Jacob Bernoulli’s death in 1705 published (the manuscript of Jacob Bernoulli called Summae Potestatum in the book Ars Conjectandi [18]) the general formula for the sums of kth powers of the first n integers as a (k + 1)th-degree polynomial in the variable n with coefficients involving numbers Bj, which are now known as Bernoulli numbers. Besides Bernoulli’s work there is a wide literature in which the interested reader can find this general formula (and its proof), which has the following form: n X i¼1

 k  1 X kþ1 Bj nk þ 1j ; i ¼ j k þ 1 j¼0 k

where Bj = 0 if j is odd (j > 1) and 1 1 1 1 1 5 B0 ¼ 0; B1 ¼ ; B2 ¼ ; B4 ¼  ; B6 ¼ ; B8 ¼  ; B10 ¼ ; . . . : 2 6 30 42 30 66 A more detailed list of Bernoulli numbers can be found in the literature. A full consideration and proof of this formula is outside the scope of this book. In studying the works of Jacob Bernoulli, Johann Faulhaber, and other authors, Hayk Sedrakyan noticed that for a high-school student to remember and apply this formula can be really challenging. So his idea was to use this result to create a simple and self-constructive Pascal-type triangle for sums of powers. © Springer International Publishing AG, part of Springer Nature 2018 H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books in Mathematics, https://doi.org/10.1007/978-3-319-77836-5

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Appendix—Power Sums Triangle

The “triangle.” Let us consider the following “triangle,” where k stands for the power of the considered sum and n; n2 ; n3 ; n4 ; n5 ;    ; nk þ 1 stand for the terms of the (k+ 1)th-degree polynomial in the variable n that the considered sum is equal to, and the number written at the intersection of each row and each column stands for the coefficient of the corresponding term of this polynomial.

Self-construction principle. Considering this “triangle” diagonally, we observe that the coefficients on the dashed diagonals remain unchanged, the denominator of each coefficient on the solid line increases by one for each next row (in other words, this coefficient is equal to k þ1 1), and the arrows indicate how each row can be recurrently constructed using the previous row (each time one needs to multiply the coefficient of the previous row by a fraction whose numerator is equal to the value of k written in that row and denominator is equal to the power of n written in that column). Note that Bk = 0 if k is odd (k > 1), so every odd row (starting from the third) starts with 0, and every even row k ¼ 2n; n 2 N; starts with B2n. Example (application). Calculate the sum 15 þ 25 þ    þ n5 . Using this triangle we obtain 5 3 4 5 1 1 15 þ 25 þ    þ n5 ¼ B4   n2 þ B2     n4 þ  n5 þ  n6 : 2 2 3 4 2 6 Thus, it follows that 15 þ 25 þ    þ n5 ¼ 

1 5n2 1 5n4 n5 n6  þ  þ þ : 30 2 6 2 2 6

Appendix—Power Sums Triangle

241

Therefore, we deduce that 15 þ 25 þ    þ n5 ¼ 

n2 5n4 n5 n6 þ þ þ : 12 12 2 6

Remark Note that neither this formula nor the general formula is easy to memorize for a high-school student. Using the “triangle,” however, one easily derives the formula even if it has been forgotten.

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