313

Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 4, Issue 3, Article 63, 2003

A SURVEY ON CAUCHY-BUNYAKOVSKY-SCHWARZ TYPE DISCRETE INEQUALITIES S.S. DRAGOMIR S CHOOL OF C OMPUTER S CIENCE AND M ATHEMATICS V ICTORIA U NIVERSITY, PO B OX 14428, MCMC 8001, M ELBOURNE , V ICTORIA , AUSTRALIA . [email protected]

Received 22 January, 2003; accepted 14 May, 2003 Communicated by P.S. Bullen

A BSTRACT. The main purpose of this survey is to identify and highlight the discrete inequalities that are connected with (CBS)− inequality and provide refinements and reverse results as well as to study some functional properties of certain mappings that can be naturally associated with this inequality such as superadditivity, supermultiplicity, the strong versions of these and the corresponding monotonicity properties. Many companion, reverse and related results both for real and complex numbers are also presented. Key words and phrases: Cauchy-Bunyakovsky-Schwarz inequality, Discrete inequalities. 2000 Mathematics Subject Classification. 26D15, 26D10.

C ONTENTS 1. Introduction 2. (CBS) – Type Inequalities 2.1. (CBS) −Inequality for Real Numbers 2.2. (CBS) −Inequality for Complex Numbers 2.3. An Additive Generalisation 2.4. A Related Additive Inequality 2.5. A Parameter Additive Inequality 2.6. A Generalisation Provided by Young’s Inequality 2.7. Further Generalisations via Young’s Inequality 2.8. A Generalisation Involving J−Convex Functions 2.9. A Functional Generalisation 2.10. A Generalisation for Power Series 2.11. A Generalisation of Callebaut’s Inequality 2.12. Wagner’s Inequality for Real Numbers 2.13. Wagner’s inequality for Complex Numbers ISSN (electronic): 1443-5756 c 2003 Victoria University. All rights reserved.

010-03

4 5 5 6 7 8 10 11 12 16 18 19 21 22 24

2

S.S. D RAGOMIR

References 3. Refinements of the (CBS) −Inequality 3.1. A Refinement in Terms of Moduli 3.2. A Refinement for a Sequence Whose Norm is One 3.3. A Second Refinement in Terms of Moduli 3.4. A Refinement for a Sequence Less than the Weights 3.5. A Conditional Inequality Providing a Refinement 3.6. A Refinement for Non-Constant Sequences 3.7. De Bruijn’s Inequality 3.8. McLaughlin’s Inequality 3.9. A Refinement due to Daykin-Eliezer-Carlitz 3.10. A Refinement via Dunkl-Williams’ Inequality 3.11. Some Refinements due to Alzer and Zheng References 4. Functional Properties 4.1. A Monotonicity Property 4.2. A Superadditivity Property in Terms of Weights 4.3. The Superadditivity as an Index Set Mapping 4.4. Strong Superadditivity in Terms of Weights 4.5. Strong Superadditivity as an Index Set Mapping 4.6. Another Superadditivity Property 4.7. The Case of Index Set Mapping 4.8. Supermultiplicity in Terms of Weights 4.9. Supermultiplicity as an Index Set Mapping References 5. Reverse Inequalities 5.1. The Cassels’ Inequality 5.2. The Pólya-Szegö Inequality 5.3. The Greub-Rheinboldt Inequality 5.4. A Cassels’ Type Inequality for Complex Numbers 5.5. A Reverse Inequality for Real Numbers 5.6. A Reverse Inequality for Complex Numbers 5.7. Shisha-Mond Type Inequalities 5.8. Zagier Type Inequalities 5.9. A Reverse Inequality in Terms of the sup −Norm 5.10. A Reverse Inequality in Terms of the 1−Norm 5.11. A Reverse Inequality in Terms of the p−Norm 5.12. A Reverse Inequality Via an Andrica-Badea Result 5.13. A Refinement of Cassels’ Inequality 5.14. Two Reverse Results Via Diaz-Metcalf Results ˇ 5.15. Some Reverse Results Via the Cebyšev Functional 5.16. Another Reverse Result via a Grüss Type Result References 6. Related Inequalities 6.1. Ostrowski’s Inequality for Real Sequences 6.2. Ostrowski’s Inequality for Complex Sequences 6.3. Another Ostrowski’s Inequality 6.4. Fan and Todd Inequalities 6.5. Some Results for Asynchronous Sequences 6.6. An Inequality via A − G − H Mean Inequality 6.7. A Related Result via Jensen’s Inequality for Power Functions 6.8. Inequalities Derived from the Double Sums Case

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

26 28 28 29 31 33 35 37 40 41 42 43 45 50 51 51 52 54 55 57 59 61 63 67 71 72 72 73 75 75 77 79 82 83 85 87 90 92 94 97 99 105 107 109 109 110 111 113 114 115 116 117

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

6.9. A Functional Generalisation for Double Sums 6.10. A (CBS) −Type Result for Lipschitzian Functions 6.11. An Inequality via Jensen’s Discrete Inequality 6.12. An Inequality via Lah-Ribari´c Inequality 6.13. An Inequality via Dragomir-Ionescu Inequality 6.14. An Inequality via a Refinement of Jensen’s Inequality 6.15. Another Refinement via Jensen’s Inequality 6.16. An Inequality via Slater’s Result 6.17. An Inequality via an Andrica-Ra¸sa Result 6.18. An Inequality via Jensen’s Result for Double Sums ˇ 6.19. Some Inequalities for the Cebyšev Functional ˇ 6.20. Other Inequalities for the Cebyšev Functional ˇ 6.21. Bounds for the Cebyšev Functional References Index

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

3

118 119 121 122 123 124 127 129 131 132 134 136 137 140 141

http://jipam.vu.edu.au/

4

S.S. D RAGOMIR

1. I NTRODUCTION The Cauchy-Bunyakovsky-Schwarz inequality, or for short, the (CBS)− inequality, plays an important role in different branches of Modern Mathematics including Hilbert Spaces Theory, Probability & Statistics, Classical Real and Complex Analysis, Numerical Analysis, Qualitative Theory of Differential Equations and their applications. The main purpose of this survey is to identify and highlight the discrete inequalities that are connected with (CBS)− inequality and provide refinements and reverse results as well as to study some functional properties of certain mappings that can be naturally associated with this inequality such as superadditivity, supermultiplicity, the strong versions of these and the corresponding monotonicity properties. Many companions and related results both for real and complex numbers are also presented. The first section is devoted to a number of (CBS)− type inequalities that provides not only natural generalizations but also several extensions for different classes of analytic functions of a real variable. A generalization of the Wagner inequality for complex numbers is obtained. Several results discovered by the author in the late eighties and published in different journals of lesser circulation are also surveyed. The second section contains different refinements of the (CBS)− inequality including de Bruijn’s inequality, McLaughlin’s inequality, the Daykin-Eliezer-Carlitz result in the version presented by Mitrinovi´c-Peˇcari´c and Fink as well as the refinements of a particular version obtained by Alzer and Zheng. A number of new results obtained by the author, which are connected with the above ones, are also presented. Section 4 is devoted to the study of functional properties of different mappings naturally associated to the (CBS)− inequality. Properties such as superadditivity, strong superadditivity, monotonicity and supermultiplicity and the corresponding inequalities are mentioned. In the next section, Section 5, reverse results for the (CBS)− inequality are surveyed. The results of Cassels, Pólya-Szegö, Greub-Rheinbold, Shisha-Mond and Zagier are presented with their original proofs. New results and versions for complex numbers are also obtained. Reverse results in terms of p−norms of the forward difference recently discovered by the author and some refinements of Cassels and Pólya-Szegö results obtained via Andrica-Badea inequality are mentioned. Some new facts derived from Grüss type inequalities are also pointed out. Section 6 is devoted to various inequalities related to the (CBS)− inequality. The two inequalities obtained by Ostrowski and Fan-Todd results are presented. New inequalities obtained ˇ via Jensen type inequality for convex functions are derived, some inequalities for the Ceby¸ sev functionals are pointed out. Versions for complex numbers that generalize Ostrowski results are also emphasised. It was one of the main aims of the survey to provide complete proofs for the results considered. We also note that in most cases only the original references are mentioned. Each section concludes with a list of the references utilized and thus may be read independently. Being self contained, the survey may be used by both postgraduate students and researchers interested in Theory of Inequalities & Applications as well as by Mathematicians and other Scientists dealing with numerical computations, bounds and estimates where the (CBS)− inequality may be used as a powerful tool. The author intends to continue this survey with another one devoted to the functional and integral versions of the (CBS)− inequality. The corresponding results holding in inner-product and normed spaces will be considered as well.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

5

2. (CBS) – T YPE I NEQUALITIES 2.1. (CBS) −Inequality for Real Numbers. The following inequality is known in the literature as Cauchy’s or Cauchy-Schwarz’s or Cauchy-Bunyakovsky-Schwarz’s inequality. For simplicity, we shall refer to it throughout this work as the (CBS) −inequality. ¯ = (b1 , . . . , bn ) are sequences of real numbers, then Theorem 2.1. If ¯ a = (a1 , . . . , an ) and b !2 n n n X X X 2 b2k (2.1) ak b k ≤ ak k=1

k=1

k=1

¯ are proportional, i.e., there is a r ∈ R such with equality if and only if the sequences ¯ a and b that ak = rbk for each k ∈ {1, . . . , n} . Proof.

(1) Consider the quadratic polynomial P : R → R, P (t) =

(2.2)

n X

(ak t − bk )2 .

k=1

It is obvious that P (t) =

n X

! a2k

2

t −2

k=1

n X

! ak b k

k=1

t+

n X

b2k

k=1

for any t ∈ R. Since P (t) ≥ 0 for any t ∈ R it follows that the discriminant ∆ of P is negative, i.e., !2 n n n X X X 1 2 0≥ ∆= ak b k − ak b2k 4 k=1 k=1 k=1 and the inequality (2.1) is proved. (2) If we use Lagrange’s identity (2.3)

n X i=1

a2i

n X

b2i −

i=1

n X i=1

!2 ai b i

n 1X (ai bj − aj bi )2 2 i,j=1 X = (ai bj − aj bi )2

=

1≤i<j≤n

then (2.1) obviously holds. The equality holds in (2.1) iff (ai bj − aj bi )2 = 0 ¯ are proportional. for any i, j ∈ {1, . . . , n} which is equivalent with the fact that ¯ a and b



Remark 2.2. The inequality (2.1) apparently was firstly mentioned in the work [2] of A.L. Cauchy in 1821. The integral form was obtained in 1859 by V.Y. Bunyakovsky [1]. The corresponding version for inner-product spaces obtained by H.A. Schwartz is mainly known as Schwarz’s inequality. For a short history of this inequality see [3]. In what follows we use the spelling adopted in the paper [3]. For other spellings of Bunyakovsky’s name, see MathSciNet.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

6

S.S. D RAGOMIR

2.2. (CBS) −Inequality for Complex Numbers. The following version of the (CBS) −inequality for complex numbers holds [4, p. 84]. ¯ = (b1 , . . . , bn ) are sequences of complex numbers, Theorem 2.3. If ¯ a = (a1 , . . . , an ) and b then 2 n n n X X X 2 (2.4) ak b k ≤ |ak | |bk |2 , k=1

k=1

k=1

with equality if and only if there is a complex number c ∈ C such that ak = c¯bk for any k ∈ {1, . . . , n} . Proof. (2.5)

(1) For any complex number λ ∈ C one has the equality n n X X

¯ k ak − λ¯bk 2 = ak − λ¯bk a ¯k − λb k=1

=

k=1 n X

2

2

|ak | + |λ|

k=1

n X

¯ |bk | − 2 Re λ 2

k=1

n X

! ak b k

.

k=1

If in (2.5) we choose λ0 ∈ C, Pn ak b k λ0 := Pnk=1 2, |b | k k=1

¯ 6= 0 b

then we get the identity (2.6)

P n n 2 X 2 X | nk=1 ak bk | 2 ¯ 0≤ ak − λ 0 b k = |ak | − Pn 2 , k=1 |bk | k=1 k=1

which proves (2.4). By virtue of (2.6), we conclude that equality holds in (2.4) if and only if ak = λ0¯bk for any k ∈ {1, . . . , n} . (2) Using Binet-Cauchy’s identity for complex numbers (2.7)

n X i=1

xi yi

n X i=1

zi ti −

n X i=1

xi ti

n X

zi yi

i=1

n 1X = (xi zj − xj zi ) (yi tj − yj ti ) 2 i,j=1 X = (xi zj − xj zi ) (yi tj − yj ti ) 1≤i<j≤n

(2.8)

for the choices xi = a ¯i , zi = bi , yi = ai , ti = ¯bi , i = {1, . . . , n}, we get 2 n n n n X X X 1X |ai |2 |bi |2 − ai b i = |¯ ai b j − a ¯j bi |2 2 i=1 i=1 i=1 i,j=1 X = |¯ ai b j − a ¯j bi |2 . 1≤i<j≤n

Now the inequality (2.4) is a simple consequence of (2.8). The case of equality is obvious by the identity (2.8) as well. 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

7

Remark 2.4. By the (CBS) −inequality for real numbers and the generalised triangle inequality for complex numbers n n X X |zi | ≥ zi , zi ∈ C, i ∈ {1, . . . , n} i=1

we also have

i=1

n 2 X ak b k ≤ k=1

n X

!2 |ak bk |

k=1

≤

n X

2

|ak |

n X

|bk |2 .

k=1

k=1

Remark 2.5. The Lagrange identity for complex numbers stated in [4, p. 85] is wrong. It should be corrected as in (2.8). 2.3. An Additive Generalisation. The following generalisation of the (CBS) −inequality was obtained in [5, p. 5]. ¯ = (b1 , . . . , bn ) , ¯ ¯ = (d1 , . . . , dn ) Theorem 2.6. If ¯ a = (a1 , . . . , an ) , b c = (c1 , . . . , cn ) and d are sequences of real numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) are nonnegative, then n n n n n n X X X X X X 2 2 2 2 (2.9) p i ai qi b i + pi ci qi di ≥ 2 p i ai c i qi bi di . i=1

i=1

i=1

i=1

i=1

i=1

If p ¯ and q ¯ are sequences of positive numbers, then the equality holds in (2.9) iff ai bj = ci dj for any i, j ∈ {1, . . . , n} . Proof. We will follow the proof from [5]. From the elementary inequality a2 + b2 ≥ 2ab for any a, b ∈ R

(2.10) with equality iff a = b, we have (2.11)

a2i b2j + c2i d2j ≥ 2ai ci bj dj for any i, j ∈ {1, . . . , n} .

Multiplying (2.11) by pi qj ≥ 0, i, j ∈ {1, . . . , n} and summing over i and j from 1 to n, we deduce (2.9). If pi , qj > 0 (i = 1, . . . , n) , then the equality holds in (2.9) iff ai bj = ci dj for any i, j ∈ {1, . . . , n} .  Remark 2.7. The condition ai bj = ci dj for ci 6= 0, bj 6= 0 (i, j = 1, . . . , n) is equivalent with d ai ¯ d ¯ are proportional with the same constant k. = bjj (i, j = 1, . . . , n) , i.e., ¯ a, ¯ c and b, ci Remark 2.8. If in (2.9) we choose pi = qi = 1 (i = 1, . . . , n) , ci = bi , and di = ai (i = 1, . . . , n) , then we recapture the (CBS) −inequality. The following corollary holds [5, p. 6]. ¯ ¯ ¯ are nonnegative, then Corollary 2.9. If ¯ a, b, c and d " n # n n n n n X X X X 1 X 3 X 3 3 3 2 2 (2.12) a ci b di + c i ai di bi ≥ ai c i b2i d2i , 2 i=1 i i=1 i i=1 i=1 i=1 i=1 (2.13)

" n # n n n X X X 1 X 2 a bi di · b2i ai ci + c2i bi di · d2i ai ci ≥ 2 i=1 i i=1 i=1 i=1

n X

!2 ai bi ci di

.

i=1

Another result is embodied in the following corollary [5, p. 6].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

8

S.S. D RAGOMIR

¯ ¯ ¯ are sequences of positive and real numbers, then: Corollary 2.10. If ¯ a, b, c and d # " n n n n n n X X X 1 X a3i X b3i X 2 (2.14) bi di ≥ ai b2i , + ai c i 2 i=1 ci i=1 di i=1 i=1 i=1 i=1 " n # n n n X 1 X a2i bi X b2i ai X + bi c i ai di ≥ 2 i=1 ci i=1 di i=1 i=1

(2.15)

n X

!2 ai b i

.

i=1

Finally, we also have [5, p. 6]. ¯ are positive, then Corollary 2.11. If ¯ a, and b  !2  n n n n n 3 X 3 X X X X 1 ai bi 2  − ai b i ≥ ai b2i − 2 i=1 bi i=1 ai i=1 i=1 i=1

n X

!2 ai b i

≥ 0.

i=1

The following version for complex numbers also holds. ¯ = (b1 , . . . , bn ) , ¯ ¯ = (d1 , . . . , dn ) Theorem 2.12. Let ¯ a = (a1 , . . . , an ) , b c = (c1 , . . . , cn ) and d be sequences of complex numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) are nonnegative. Then one has the inequality " n # n n n n n X X X X X X (2.16) pi |ai |2 qi |bi |2 + pi |ci |2 qi |di |2 ≥ 2 Re pi ai c¯i qi bi d¯i . i=1

i=1

i=1

i=1

i=1

i=1

The case of equality for p ¯, q ¯ positive holds iff ai bj = ci dj for any i, j ∈ {1, . . . , n} . Proof. From the elementary inequality for complex numbers   |a|2 + |b|2 ≥ 2 Re a¯b , a, b ∈ C, with equality iff a = b, we have   ¯ |ai |2 |bj |2 + |ci |2 |dj |2 ≥ 2 Re ai c¯i bj dj

(2.17)

for any i, j ∈ {1, . . . , n} . Multiplying (2.17) by pi qj ≥ 0 and summing over i and j from 1 to n, we deduce (2.16). The case of equality is obvious and we omit the details.  Remark 2.13. Similar particular cases may be stated but we omit the details. 2.4. A Related Additive Inequality. The following inequality was obtained in [5, Theorem 1.1]. ¯ = (b1 , . . . , bn ) are sequences of real numbers and Theorem 2.14. If ¯ a = (a1 , . . . , an ) , b ¯ ¯ c = (c1 , . . . , cn ) , d = (d1 , . . . , dn ) are nonnegative, then (2.18)

n X i=1

di

n X

ci a2i +

i=1

n X i=1

ci

n X i=1

di b2i ≥ 2

n X

c i ai

i=1

n X

di bi .

i=1

¯ = k ¯ where If ci and di (i = 1, . . . , n) are positive, then equality holds in (2.18) iff ¯ a = b ¯ k = (k, k, . . . , k) is a constant sequence. Proof. We will follow the proof from [5]. From the elementary inequality (2.19)

a2 + b2 ≥ 2ab for any a, b ∈ R

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

9

with equality iff a = b; we have a2i + b2j ≥ 2ai bj for any i, j ∈ {1, . . . , n} .

(2.20)

Multiplying (2.20) by ci dj ≥ 0, i, j ∈ {1, . . . , n} and summing over i from 1 to n and over j from 1 to n, we deduce (2.18). If ci , dj > 0 (i = 1, . . . , n) , then the equality holds in (2.18) iff ai = bj for any i, j ∈ {1, . . . , n} which is equivalent with the fact that ai = bi = k for any i ∈ {1, . . . , n} .  The following corollary holds [5, p. 4]. ¯ are nonnegative sequences, then Corollary 2.15. If ¯ a and b " n # n n n n n X X X X X X 1 3 2 3 (2.21) ai bi + ai bi ≥ ai b2i ; 2 i=1 i=1 i=1 i=1 i=1 i=1 " n # n n n X X 1 X X 2 ai a bi + bi b2i ai ≥ 2 i=1 i=1 i i=1 i=1

(2.22)

n X

!2 ai b i

.

i=1

Another corollary that may be obtained is [5, p. 4 – 5]. ¯ are sequences of positive real numbers, then Corollary 2.16. If ¯ a and b Pn 1 Pn 1 n X a2i + b2i i=1 ai i=1 Pn 1 bi , (2.23) ≥ 2a b i i i=1 ai bi i=1 n X

(2.24)

i=1

n n n X 1 X 1 X + bi ≥ 2n2 , ai b a i i i=1 i=1 i=1

and n

(2.25)

n X a2 + b 2 i

i=1

i

2a2i b2i

n n X 1 X1 ≥ . a b i=1 i i=1 i

The following version for complex numbers also holds. ¯ = (b1 , . . . , bn ) are sequences of complex numbers, then Theorem 2.17. If ¯ a = (a1 , . . . , an ) , b for p ¯ = (p1 , . . . , pn ) and q ¯ = (q1 , . . . , qn ) two sequences of nonnegative real numbers, one has the inequality " n # n n n n n X X X X X X (2.26) qi pi |ai |2 + pi qi |bi |2 ≥ 2 Re p i ai qi¯bi . i=1

i=1

i=1

i=1

i=1

i=1

¯=k ¯ = (k, . . . , k) . For p ¯, q ¯ positive sequences, the equality holds in (2.26) iff ¯ a=b The proof goes in a similar way with the one in Theorem 2.14 on making use of the following elementary inequality holding for complex numbers   (2.27) |a|2 + |b|2 ≥ 2 Re a¯b , a, b ∈ C; with equality iff a = b.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

10

S.S. D RAGOMIR

2.5. A Parameter Additive Inequality. The following inequality was obtained in [5, Theorem 4.1]. ¯ = (b1 , . . . , bn ) be sequences of real numbers and Theorem 2.18. Let ¯ a = (a1 , . . . , an ) , b ¯ ¯ c = (c1 , . . . , cn ) , d = (d1 , . . . , dn ) be nonnegative. If α, β > 0 and γ ∈ R such that γ 2 ≤ αβ, then n n n n n n X X X X X X 2 2 (2.28) α di ai c i + β ci bi di ≥ 2γ c i ai di bi . i=1

i=1

i=1

i=1

i=1

i=1

Proof. We will follow the proof from [5]. Since α, β > 0 and γ 2 ≤ αβ, it follows that for any x, y ∈ R one has αx2 + βy 2 ≥ 2γxy.

(2.29)

Choosing in (2.29) x = ai , y = bj (i, j = 1, . . . , n) , we get αa2i + βb2j ≥ 2γai bj for any i, j ∈ {1, . . . , n} .

(2.30)

If we multiply (2.30) by ci dj ≥ 0 and sum over i and j from 1 to n, we deduce the desired inequality (2.28).  The following corollary holds. ¯ are nonnegative sequences and α, β, γ are as in Theorem 2.18, then Corollary 2.19. If ¯ a and b α

(2.31)

n X

bi

i=1

(2.32)

α

n X i=1

ai

n X

a3i + β

n X

i=1 n X

a2i bi

+β

ai

n X

i=1

i=1

n X

n X

i=1

i=1

bi

b3i ≥ 2γ

n X

a2i

n X

i=1

b2i ai

i=1 n X

≥ 2γ

i=1

b2i , !2

ai b i

.

i=1

The following particular case is important [5, p. 8]. ¯ be sequences of real numbers. If p Theorem 2.20.PLet ¯ a, b ¯ is a sequence of nonnegative real n numbers with i=1 pi > 0, then: Pn Pn Pn n n X X 2 2 i=1 pi bi i=1 pi ai bi Pi=1 pi ai . (2.33) p i ai p i bi ≥ n i=1 pi i=1 i=1 In particular, (2.34)

n X

a2i

i=1

n X i=1

n

b2i

n

n

X X 1X ≥ ai b i ai bi . n i=1 i=1 i=1

Proof. We will follow the proof from [5, p. 8]. Pn 2 we choose inPTheorem 2.18, ci = di = pi (i = 1, . . . , n) and α = i=1 pi bi , β = PIf n n 2 i=1 pi ai , γ = i=1 pi ai bi , we observe, by the (CBS) −inequality with the weights pi (i = 1, . . . , n) one has γ 2 ≤ αβ, and then by (2.28) we deduce (2.33).  ¯ are asynchronous, i.e., Remark 2.21. If we assume that ¯ a and b (ai − aj ) (bi − bj ) ≤ 0 for any i, j ∈ {1, . . . , n} , ˇ then by Cebyšev’s inequality (2.35)

n X

p i ai

i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

n X i=1

pi bi ≥

n X i=1

pi

n X

p i ai b i

i=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

11

respectively n X

(2.36)

ai

i=1

n X

bi ≥ n

i=1

n X

ai b i ,

i=1

we have the following refinements of the (CBS) −inequality Pn Pn Pn n n X X 2 2 i=1 pi ai bi Pi=1 pi ai i=1 pi bi p i ai p i bi ≥ (2.37) n i=1 pi i=1 i=1 ! 2 n X ≥ p i ai b i i=1

provided

Pn

i=1

pi ai bi ≥ 0, respectively n X

(2.38)

a2i

i=1

provided

Pn

i=1

n X i=1

n

n

n

X X 1X b2i ≥ ai b i ai bi ≥ n i=1 i=1 i=1

n X

!2 ai b i

i=1

ai bi ≥ 0.

2.6. A Generalisation Provided by Young’s Inequality. The following result was obtained in [5, Theorem 5.1]. ¯ = (b1 , . . . , bn ) , p Theorem 2.22. Let ¯ a = (a1 , . . . , an ) , b ¯ = (p1 , . . . , pn ) and q ¯ = (q1 , . . . , qn ) 1 1 be sequences of nonnegative real numbers and α, β > 1 with α + β = 1. Then one has the inequality (2.39)

α

n X i=1

qi

n X

pi bβi

+β

i=1

n X i=1

pi

n X i=1

qi aαi

≥ αβ

n X i=1

p i bi

n X

q i ai .

i=1

If p ¯ and q ¯ are sequences of positive real numbers, then the equality holds in (2.39) iff there exists a constant k ≥ 0 such that aαi = bβi = k for each i ∈ {1, . . . , n} . Proof. It is, by the Arithmetic-Geometric inequality [6, p. 15], well known that (2.40)

1 1 1 1 1 1 x + y ≥ x α y β for x, y ≥ 0, + = 1, α, β > 1 α β α β

with equality iff x = y. Applying (2.40) for x = aαi , y = bβj (i, j = 1, . . . , n) we have (2.41)

αbβj + βaαi ≥ αβai bj for any i, j ∈ {1, . . . , n}

with equality iff aαi = bβj for any i, j ∈ {1, . . . , n} . If we multiply (2.41) by qi pj ≥ 0 (i, j ∈ {1, . . . , n}) and sum over i and j from 1 to n we deduce (2.39). The case of equality is obvious by the above considerations.  The following corollary is a natural consequence of the above theorem. ¯ α and β be as in Theorem 2.22. Then Corollary 2.23. Let ¯ a, b, (2.42)

n n n n n n 1 X X α+1 1 X X β+1 X 2 X 2 bi a + ai b ≥ ai bi ; α i=1 i=1 i β i=1 i=1 i i=1 i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

12

S.S. D RAGOMIR

n n n n 1X X α 1X X β b i ai + ai b i ≥ ai bi α i=1 i=1 β i=1 i=1

(2.43)

n X

!2 ai b i

.

i=1

The following result which provides a generalisation of the (CBS) −inequality may be obtained by Theorem 2.22 as well [5, Theorem 5.2]. Theorem 2.24. Let x ¯ and y ¯ be sequences of positive real numbers. If α, β are as above, then ! n !2 n n n X X 1 X α 2−α 1 X β 2−β (2.44) x y + x y · yi2 ≥ xi yi . α i=1 i i β i=1 i i i=1 i=1 The equality holds iff x ¯ and y ¯ are proportional. Proof. Follows by Theorem 2.22 on choosing pi = qi = yi2 , ai =

xi , yi

bi =

xi , yi

i ∈ {1, . . . , n} . 

Remark 2.25. For α = β = 2, we recapture the (CBS) −inequality. Remark 2.26. For ai = |zi | , bi = |wi | , with zi , wi ∈ C; i = 1, . . . , n, we may obtain similar inequalities for complex numbers. We omit the details. 2.7. Further Generalisations via Young’s Inequality. The following inequality is known in the literature as Young’s inequality 1 1 (2.45) pxq + qy p ≥ pqxy, x, y ≥ 0 and + = 1, p > 1 p q q p with equality iff x = y . The following result generalising the (CBS) −inequality was obtained in [7, Theorem 2.1] (see also [8, Theorem 1]). Theorem 2.27. Let x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) be sequences of complex numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) be two sequences of nonnegative real numbers. If p > 1, 1 + 1q = 1, then p (2.46)

n n n n n n X X X X 1X 1X pk |xk |p qk |yk |p + qk |xk |q pk |yk |q ≥ pk |xk yk | qk |xk yk | . p k=1 q k=1 k=1 k=1 k=1 k=1

Proof. We shall follow the proof in [7]. Choosing x = |xj | |yi | , y = |xi | |yj | , i, j ∈ {1, . . . , n} , we get from (2.45) q |xi |p |yj |p + p |xj |q |yi |q ≥ pq |xi yi | |xj yj |

(2.47)

for any i, j ∈ {1, . . . , n} . Multiplying with pi qj ≥ 0 and summing over i and j from 1 to n, we deduce the desired result (2.46).  The following corollary is a natural consequence of the above theorem [7, Corollary 2.2] (see also [8, p. 105]). Corollary 2.28. If x ¯ and y ¯ are as in Theorem 2.27 and m ¯ = (m1 , . . . , mn ) is a sequence of nonnegative real numbers, then !2 n n n n n X X X X X 1 1 (2.48) mk |xk |p mk |yk |p + mk |xk |q mk |yk |q ≥ mk |xk yk | , p k=1 q k=1 k=1 k=1 k=1 where p > 1,

1 p

+

1 q

= 1.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

13

Remark 2.29. If in (2.48) we assume that mk = 1, k ∈ {1, . . . , n} , then we obtain [7, p. 7] (see also [8, p. 105]) !2 n n n n n X X X X 1 1X (2.49) |xk |p |yk |p + |xk |q |yk |q ≥ |xk yk | , p k=1 q k=1 k=1 k=1 k=1 which, in the particular case p = q = 2 will provide the (CBS) −inequality. The second generalisation of the (CBS) −inequality via Young’s inequality is incorporated in the following theorem [7, Theorem 2.4] (see also [8, Theorem 2]). Theorem 2.30. Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 2.27. Then one has the inequality n n n n X X 1X 1X p q q pk |xk | qk |yk | + qk |xk | pk |yk |p (2.50) p k=1 q k=1 k=1 k=1

≥

n X

p−1

pk |xk | |yk |

k=1

n X

qk |xk | |yk |q−1 .

k=1

Proof. We shall follow the proof in [7]. |x | i| Choosing in (2.45), x = |yjj | , y = |x , we get |yi |  q  p |xi | |xi | |xj | |xj | +q ≥ pq (2.51) p |yj | |yi | |yi | |yj | for any yi 6= 0, i, j ∈ {1, . . . , n} . It is easy to see that (2.51) is equivalent to (2.52)

q |xi |p |yj |q + p |yi |p |xj |q ≥ pq |xi | |yi |p−1 |xj | |yj |q−1

for any i, j ∈ {1, . . . , n} . Multiplying (2.52) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequlality (2.50).  The following corollary holds [7, Corollary 2.5] (see also [8, p. 106]). Corollary 2.31. Let x ¯, y ¯, m ¯ and p ¯, q ¯ be as in Corollary 2.28. Then n n n n X X 1X 1X p q q (2.53) mk |xk | mk |yk | + mk |xk | mk |yk |p p k=1 q k=1 k=1 k=1

≥

n X k=1

p−1

mk |xk | |yk |

n X

mk |xk | |yk |q−1 .

k=1

Remark 2.32. If in (2.53) we assume that mk = 1, k ∈ {1, . . . , n} , then we obtain [7, p. 8] (see also [8, p. 106]) (2.54)

n n n n n n X X X X 1X 1X p q q p p−1 |xk | |yk | + |xk | |yk | ≥ |xk | |yk | |xk | |yk |q−1 , p k=1 q k=1 k=1 k=1 k=1 k=1

which, in the particular case p = q = 2 will provide the (CBS) −inequality. The third result is embodied in the following theorem [7, Theorem 2.7] (see also [8, Theorem 3]).

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

14

S.S. D RAGOMIR

Theorem 2.33. Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 2.27. Then one has the inequality n n n n X X 1X 1X q p p (2.55) qk |yk | + pk |yk |q pk |xk | qk |xk | p k=1 q k=1 k=1 k=1

≥

n X

pk |xk yk |

k=1

n X

pk |xk |p−1 |yk |q−1 .

k=1

Proof. We shall follow the proof in [7]. |yi | |xi | If we choose x = |y and y = |x in (2.45) we get j| j|   q p |xi | |xi | |yi | |yi | p +q ≥ pq , |yj | |xj | |xj | |yj | for any xi , yj 6= 0, i, j ∈ {1, . . . , n} , giving q |xi |p |yj |q + p |yi |q |xj |p ≥ pq |xi yi | |xj |p−1 |yj |q−1

(2.56)

for any i, j ∈ {1, . . . , n} . Multiplying (2.56) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (2.55).  The following corollary is a natural consequence of the above theorem [8, p. 106]. Corollary 2.34. Let x ¯, y ¯, m ¯ and p ¯, q ¯ be as in Corollary 2.28. Then one has the inequality: n n n n X X X X mk |yk |q ≥ mk |xk yk | mk |xk |p−1 |yk |q−1 . (2.57) mk |xk |p k=1

k=1

k=1

k=1

Remark 2.35. If in (2.57) we assume that mk = 1, k = {1, . . . , n} , then we obtain [7, p. 8] (see also [8, p. 10]) n n n n X X X X p q (2.58) |xk | |yk | ≥ |xk yk | |xk |p−1 |yk |q−1 , k=1

k=1

k=1

k=1

which, in the particular case p = q = 2 will provide the (CBS) −inequality. The fourth generalisation of the (CBS) −inequality is embodied in the following theorem [7, Theorem 2.9] (see also [8, Theorem 4]). Theorem 2.36. Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 2.27. Then one has the inequality n n n n X X 1X 1X 2 q 2 (2.59) pk |xk | qk |yk | + pk |yk | qk |xk |p q k=1 p k=1 k=1 k=1

≥

n X

qk |xk yk |

k=1

n X

2

2

pk |xk | q |yk | p .

k=1

Proof. We shall follow the proof in [7]. 2 2 Choosing in (2.45), x = |xi | q |yj | , y = |xj | |yi | p , we get (2.60)

2

2

p |xi |2 |yj |q + q |xj |p |yi |2 ≥ pq |xi | q |yi | p |xj yj |

for any i, j ∈ {1, . . . , n} . Multiply (2.60) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (2.60). 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

15

The following corollary holds [7, Corollary 2.10] (see also [8, p. 107]). Corollary 2.37. Let x ¯, y ¯, m ¯ and p, q be as in Corollary 2.28. Then one has the inequality: n n n n X X 1X 1X 2 q 2 mk |xk | mk |yk | + mk |yk | mk |xk |p (2.61) q k=1 p k=1 k=1 k=1

≥

n X

mk |xk yk |

k=1

n X

2

2

mk |xk | q |yk | p .

k=1

Remark 2.38. If in (2.61) we take mk = 1, k ∈ {1, . . . , n} , then we get (2.62)

n n n n n n X X X X 2 2 1X 1X |xk |2 |yk |q + |yk |2 |xk |p ≥ |xk yk | |xk | q |yk | p , q k=1 p k=1 k=1 k=1 k=1 k=1

which, in the particular case p = q = 2 will provide the (CBS) −inequality. The fifth result generalising the (CBS) −inequality is embodied in the following theorem [7, Theorem 2.12] (see also [8, Theorem 5]). Theorem 2.39. Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 2.27. Then one has the inequality n n n n X X 1X 1X 2 q 2 (2.63) pk |xk | qk |yk | + pk |yk | qk |xk |p p k=1 q k=1 k=1 k=1

≥

n X

2 p

pk |xk | |yk |

k=1

2 q

n X

qk |xk |p−1 |yk |q−1 .

k=1

Proof. We will follow the proof in [7]. 2

Choosing in (2.45), x = p

|yi | q |yj |

2 p

, y = |x|xij| | , yi , xj 6= 0, i, j ∈ {1, . . . , n} , we may write !q !p 2 2 2 2 q p |yi | |xi | |yi | q |xi | p +q ≥ pq , |yj | |xj | |xj | |yj |

from where results (2.64)

2

2

p |yi |2 |xj |p + q |xi |2 |yj |q ≥ pq |xi | p |yi | q |xj |p−1 |yj |q−1

for any i, j ∈ {1, . . . , n} . Multiplying (2.64) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (2.63).  The following corollary holds [7, Corollary 2.13] (see also [8, p. 108]). Corollary 2.40. Let x ¯, y ¯, m ¯ and p, q be as in Corollary 2.28. Then one has the inequality: n n n n X X 1X 1X 2 q 2 (2.65) mk |xk | mk |yk | + mk |yk | mk |xk |p p k=1 q k=1 k=1 k=1

≥

n X k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

2 p

mk |xk | |yk |

2 q

n X

mk |xk |p−1 |yk |q−1 .

k=1

http://jipam.vu.edu.au/

16

S.S. D RAGOMIR

Remark 2.41. If in (2.46) we choose mk = 1, k ∈ {1, . . . , n} , then we get [7, p. 10] (see also [8, p. 108]) (2.66)

n n n n n n X X X 2 2 X 1X 1X 2 q 2 p |xk | |yk | + |yk | |xk | ≥ |xk | p |yk | q |xk |p−1 |yk |q−1 , p k=1 q k=1 k=1 k=1 k=1 k=1

which in the particular case p = q = 2 will provide the (CBS) −inequality. Finally, the following result generalising the (CBS) −inequality holds [7, Theorem 2.15] (see also [8, Theorem 6]). Theorem 2.42. Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 2.27. Then one has the inequality: (2.67)

n n n n X X 1X 1X pk |xk |2 qk |yk |p + qk |yk |2 pk |xk |q p k=1 q k=1 k=1 k=1

≥

n X

2

pk |xk | p |yk |

k=1

n X

2

qk |xk | q |yk | .

k=1

Proof. We shall follow the proof in [7]. From (2.45) one has the inequality  p  q 2 2 2 2 (2.68) q |xi | p |yj | + p |xj | q |yi | ≥ pq |xi | p |yi | |xj | q |yj | for any i, j ∈ {1, . . . , n} . Multiplying (2.68) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (2.67).  The following corollary also holds [7, Corollary 2.16] (see also [8, p. 108]). Corollary 2.43. With the assumptions in Corollary 2.28, one has the inequality   X n n n n X X X 2 2 1 1 2 p q (2.69) mk |xk | mk |yk | + |yk | ≥ mk |xk | p |yk | mk |xk | q |yk | . p q k=1 k=1 k=1 k=1 Remark 2.44. If in (2.69) we choose mk = 1 (k ∈ {1, . . . , n}) , then we get  X n n  n n X X X 2 2 1 1 2 p q (2.70) |xk | |yk | + |yk | ≥ |xk | p |yk | |xk | q |yk | , p q k=1 k=1 k=1 k=1 which, in the particular case p = q = 2, provides the (CBS) −inequality. 2.8. A Generalisation Involving J−Convex Functions. For a > 1, we denote by expa the function (2.71)

expa : R → (0, ∞) , expa (x) = ax .

Definition 2.1. A function f : I ⊆ R → R is said to be J−convex on an interval I if   f (x) + f (y) x+y (2.72) f ≤ for any x, y ∈ I. 2 2 It is obvious that any convex function on I is a J convex function on I, but the converse does not generally hold. The following lemma holds (see [7, Lemma 4.3]).

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

17

Lemma 2.45. Let f : I ⊆ R → R be a J−convex function on I, a > 1 and x, y ∈ R\ {0} with loga x2 , loga y 2 ∈ I. Then loga |xy| ∈ I and 
 
 (2.73) {expb [f (loga |xy|)]}2 ≤ expb f loga x2 expb f loga y 2 for any b > 1. Proof. I, being an interval, is a convex set in R and thus loga |xy| =

 1 loga x2 + loga y 2 ∈ I. 2

Since f is J−convex, one has 
1 2 2 f (loga |xy|) = f loga x + loga y 2 f (loga x2 ) + f (loga y 2 ) ≤ . 2 

(2.74)

Taking the expb in both parts, we deduce  f (loga x2 ) + f (loga y 2 ) expb [f (loga |xy|)] ≤ expb 2  
 
 21 = expb f loga x2 expb f loga y 2 , 



which is equivalent to (2.73).

The following generalisation of the (CBS) −inequality in terms of a J− convex function holds [7, Theorem 4.4]. Theorem 2.46. Let f : I ⊆ R → R be a J−convex function on I, a, b > 1 and ¯ a = ¯ = (b1 , . . . , bn ) sequences of nonzero real numbers. If loga a2 , loga b2 ∈ I (a1 , . . . , an ) , b k k for all k ∈ {1, . . . , n} , then one has the inequality: ( n )2 n n X X 
 X 
 2 (2.75) expb [f (loga |ak bk |)] ≤ expb f loga ak expb f loga b2k . k=1

k=1

k=1

Proof. Using Lemma 2.45 and the (CBS) −inequality one has n X

expb [f (loga |ak bk |)]

k=1 n X
 1  
  ≤ expb f loga a2k expb f loga b2k 2 k=1

≤

n n n n X  
 12 o2 X  
 1 o2 2 expb f loga ak expb f loga b2k 2 k=1

which is clearly equivalent to (2.75).

! 12

k=1



Remark 2.47. If in (2.75) we choose a = b > 1 and f (x) = x, x ∈ R, then we recapture the (CBS) −inequality.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

18

S.S. D RAGOMIR

2.9. A Functional Generalisation. The following result was proved in [10, Theorem 2]. Theorem 2.48. Let A be a subset of real numbers R, f : A → R and ¯ a = (a1 , . . . , an ) , ¯ b = (b1 , . . . , bn ) sequences of real numbers with the properties that (i) ai bi , a2i , b2i ∈ A for any i ∈ {1, . . . , n} , (ii) f (a2i ) , f (b2i ) ≥ 0 for any i ∈ {1, . . . , n} , (iii) f 2 (ai bi ) ≤ f (a2i ) f (b2i ) for any i ∈ {1, . . . , n} . Then one has the inequality: " n #2 n n X X
X
2 (2.76) f (ai bi ) ≤ f ai f b2i . i=1

i=1

i=1

Proof. We give here a simpler proof than that found in [10]. We have n n X X f (ai bi ) ≤ |f (ai bi )| i=1

i=1

n X
 1 
 1  ≤ f a2i 2 f b2i 2 i=1

"

n  n  X 
 12 2 X 
 1 2 2 f b2i 2 ≤ f ai i=1

" =

n X

# 12 (by the (CBS)-inequality)

i=1 n
X 2

f ai

i=1

# 21
2

f bi

i=1



and the inequality (2.76) is proved.

Remark 2.49. It is obvious that for A = R and f (x) = x, we recapture the (CBS) −inequality. Assume that ϕ : N → N is Euler’s indicator. In 1940, T. Popoviciu [11] proved the following inequality for ϕ

(2.77) [ϕ (ab)]2 ≤ ϕ a2 ϕ b2 for any natural number a, b; with equality iff a and b have the same prime factors. A simple proof of this fact may be done by using the representation     1 1 ϕ (n) = n 1 − ··· 1 − , p1 pk where n = pα1 1 pα2 2 · · · pαk k [9, p. 109]. The following generalisation of Popoviciu’s result holds [10, Theorem 1]. Theorem 2.50. Let ai , bi ∈ N (i = 1, . . . , n) . Then one has the inequality " n #2 n n X X
X
2 (2.78) ϕ (ai bi ) ≤ ϕ ai ϕ b2i . i=1

i=1

i=1

Proof. Follows by Theorem 2.48 on taking into account that, by (2.77),

[ϕ (ai bi )]2 ≤ ϕ a2i ϕ b2i for any i ∈ {1, . . . , n} . 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

19

Further, let us denote by s (n) the sum of all relatively prime numbers with n and less than n. Then the following result also holds [10, Theorem 1]. Theorem 2.51. Let ai , bi ∈ N (i = 1, . . . , n) . Then one has the inequality #2 " n n n X X

X 2 (2.79) s (ai bi ) ≤ s ai s b2i . i=1

i=1

i=1

Proof. It is known (see for example [9, p. 109]) that for any n ∈ N one has 1 s (n) = nϕ (n) . 2

(2.80) Thus





1 1 [s (ai bi )]2 = a2i b2i ϕ2 (ai bi ) ≤ a2i b2i ϕ a2i ϕ b2i = s a2i s b2i 4 4 for each i ∈ {1, . . . , n} . Using Theorem 2.48 we then deduce the desired inequality (2.79).

(2.81)



The following corollaries of Theorem 2.48 are also natural to be considered [10, p. 126]. Corollary 2.52. Let ai , bi ∈ R (i = 1, . . . , n) and a > 1. Denote expa x = ax , x ∈ R. Then one has the inequality #2 " n n n X X
X
2 (2.82) expa (ai bi ) ≤ expa ai expa b2i . i=1

i=1

i=1

Corollary 2.53. Let ai , bi ∈ (−1, 1) (i = 1, . . . , n) and m > 0. Then one has the inequality: " n #2 n n X X X 1 1 1 (2.83) ≤ . m 2 m 2 m (1 − a b ) (1 − a ) (1 − b ) i i i i i=1 i=1 i=1 2.10. A Generalisation for Power Series. The following result holds [12, Remark 2]. P∞ k Theorem 2.54. Let F : (−r, r) → R, F (x) = a = k=0 αk x with αk ≥ 0, k ∈ N. If ¯ ¯ = (b1 , . . . , bn ) are sequences of real numbers such that (a1 , . . . , an ) , b (2.84)

ai bi , a2i , b2i ∈ (−r, r) for any i ∈ {1, . . . , n} ,

then one has the inequality: (2.85)

n X

n
X 2

F ai

i=1

"
2

F bi ≥

i=1

n X

#2 F (ai bi )

.

i=1

Proof. Firstly, let us observe that if x, y ∈ R such that xy, x2 , y 2 ∈ (−r, r) , then one has the inequality

(2.86) [F (xy)]2 ≤ F x2 F y 2 . Indeed, by the (CBS) −inequality, we have #2 " n n n X X X k k 2k αk y 2k , (2.87) αk x y ≤ αk x k=0

k=0

n ≥ 0.

k=0

Taking the limit as n → ∞ in (2.87), we deduce (2.86).

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

20

S.S. D RAGOMIR

Using the (CBS) −inequality and (2.86) we have n n X X F (ai bi ) ≤ |F (ai bi )| i=1

i=1 n X

≤


 1 
 1 F a2i 2 F b2i 2

i=1

)1 ( n  n   2 X  1 2 X 1 2

 ≤ F a2i 2 F b2i 2 i=1

" =

i=1

n X

n
X 2

F ai

i=1

# 12
2

F bi

,

i=1



which is clearly equivalent to (2.85). The following particular inequalities of (CBS) −type hold [12, p. 164]. ¯ are sequences of real numbers, then one has the inequality (1) If ¯ a, b n X

(2.88)

n
X 2

exp ak

(2.89)


2

sinh ak

k=1 n X

(2.90)

exp bk ≥

k=1

k=1 n X

"
2

n X

cosh

n
X


2

sinh bk ≥

exp (ak bk )

;

n X

#2 sinh (ak bk )

;

k=1

" cosh

b2k




≥

k=1

k=1

#2

k=1

"

k=1

a2k

n X

n X

#2 cosh (ak bk )

.

k=1

¯ are such that ai , bi ∈ (−1, 1) , i ∈ {1, . . . , n} , then one has the inequalities (2) If ¯ a, b " n #2 n n X X
X
2 2 (2.91) tan ak tan bk ≥ tan (ak bk ) ; k=1 n X

(2.92)

k=1 n
X 2

arcsin ak

k=1

" (2.93)

 n  Y 1 + a2 k=1

" (2.94)

ln

n  Y k=1

(2.95)

n X k=1


2

arcsin bk ≥

k=1

#

k

ln

k=1

"

1 − a2k 1 1 − a2k

"

 n  Y 1 + b2 k=1

#

" ln

#2 arcsin (ak bk )

;

k=1

#

k

ln

n X

n  Y k=1

1 − b2k 1 1 − b2k

( " ≥

ln

 n  Y 1 + ak b k k=1

#

( " ≥

ln

n  Y k=1

#)2

1 − ak b k

1 1 − ak b k

;

#)2 ;

" n #2 n X X 1 1 1 , m > 0. m m ≥ (1 − ak bk )m (1 − a2k ) k=1 (1 − b2k ) k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

21

2.11. A Generalisation of Callebaut’s Inequality. The following result holds (see also [12, Theorem 2] for a generalisation for positive linear functionals). P∞ k Theorem 2.55. Let F : (−r, r) → R, F (x) = a = k=0 αk x with αk ≥ 0, k ∈ N. If ¯ ¯ (a1 , . . . , an ) , b = (b1 , . . . , bn ) are sequences of nonnegative real numbers such that (2.96)

ai bi , aαi bi2−α , a2−α bαi ∈ (0, r) for any i ∈ {1, . . . , n} ; α ∈ [0, 2] , i

then one has the inequality " (2.97)

n X

#2 ≤

F (ai bi )

i=1

n X

F

aαi b2−α i

n
X


F a2−α bαi . i

i=1

i=1

Proof. Firstly, we note that for any x, y > 0 such that xy, xα y 2−α , x2−α y α ∈ (0, r) one has

[F (xy)]2 ≤ F xα y 2−α F x2−α y α .

(2.98)

Indeed, using Callebaut’s inequality, i.e., we recall it [4] m X

(2.99)

!2 ≤

αi xi yi

m X

i=1

αi xαi yi2−α

i=1

m X

αi x2−α yiα , i

i=1

we may write, for m ≥ 0, that m X

(2.100)

!2 αi xi y i

≤

i=0

m X

αi xα y 2−α

i=0

m
i X


i αi x2−α y α .

i=0

Taking the limit as m → ∞, we deduce (2.98). Using the (CBS) −inequality and (2.98) we may write: n n X X F (ai bi ) ≤ |F (ai bi )| i=1

≤

i=1 n X


 1 
 21  F a2−α bαi 2 F aαi b2−α i i

i=1

( n )1  n   2 X  1 2 X 1 2

 2 ≤ F aαi b2−α F a2−α bαi 2 i i i=1

" =

n X

i=1 n
X 2−α

F aαi bi

i=1

# 21 F a2−α bi i


α

i=1

which is clearly equivalent to (2.97).



The following particular inequalities also hold [12, pp. 165-166].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

22

S.S. D RAGOMIR

¯ be sequences of nonnegative real numbers. Then one has the inequalities (1) Let ¯ a and b " (2.101)

n X

#2 ≤

exp (ak bk )

k=1

" (2.102)

n X

(2.103)

n X

exp aαk b2−α k

n
X

≤

sinh (ak bk )

n X

sinh

aαk b2−α k

n
X

≤

cosh (ak bk )

k=1


sinh ak2−α bαk ;

k=1

k=1

#2


exp ak2−α bαk ;

k=1

k=1

#2

k=1

"

n X

n X

cosh

aαk bk2−α

n
X

k=1


bαk . cosh a2−α k

k=1

¯ be such that ak , bk ∈ (0, 1) for any k ∈ {1, . . . , n} . Then one has the (2) Let ¯ a and b inequalities: " (2.104)

n X

#2 ≤

tan (ak bk )

k=1

" (2.105)

n X

n X

(2.106)

ln

#2 ≤

arcsin (ak bk )

 n  Y 1 + ak b k

( " (2.107)

ln

n  Y k=1

1 1 − ak b k


tan ak2−α bαk ;

k=1

n X

arcsin

aαk bk2−α

n
X

k=1

k=1

"

#

#)2

 n  Y 1 + aα b2−α k k

≤ ln

1 − ak b k

k=1

n
X

k=1

k=1

( "

tan aαk b2−α k

k=1

#)2

"

n  Y

≤ ln

k=1

"

 n  Y 1 + a2−α bα k

ln

1 − aαk b2−α k

1 1 − aαk b2−α k


arcsin ak2−α bαk ;

" ln

n  Y k=1

k

1 − a2−α bαk k

k=1

#

#

1 1 − a2−α bαk k

;

# .

2.12. Wagner’s Inequality for Real Numbers. The following generalisation of the (CBS) − inequality for sequences of real numbers is known in the literature as Wagner’s inequality [15], or [14] (see also [4, p. 85]). ¯ = (b1 , . . . , bn ) be sequences of real numbers. If Theorem 2.56. Let ¯ a = (a1 , . . . , an ) and b 0 ≤ x ≤ 1, then one has the inequality

(2.108)

n X k=1

!2 X

ak b k + x

ai b j

1≤i6=j≤n

" ≤

n X k=1

#" a2k + 2x

X

ai aj

1≤i<j≤n

n X k=1

# b2k + 2x

X

bi bj .

1≤i<j≤n

Proof. We shall follow the proof in [13] (see also [4, p. 85]).

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

23

For any x ∈ [0, 1] , consider the quadratic polynomial in y P (y) := (1 − x)

n X

" 2

(ak y − bk ) + x

= (1 − x) y 2

(ak y − bk )

k=1

k=1

"

#2

n X

n X

n X

a2k − 2y

k=1

ak b k +

k=1



n X

+ x y 2

n X

− 2y

ak

= (1 − x)

n X

n X

a2k + x

 a2k + x 

k=1

! bk

n X

n X

" − 2y

n X

n X

bk



n X

ak b k + x

n X k=1

ak

n X

# bk

k=1

!2

n X

b2k + x

!2 

k=1

k=1

bk

k=1

!2 −

ak

k=1

  2 ak y2 

n X

n X

n X

ak b k + x

k=1 n X

+

k=1

"

 y 2 − 2y (1 − x)

ak

k=1

+

n X

ak

!2 

k=1



!

k=1

+ (1 − x)

=

b2k

k=1

k=1

 n X

#

k=1

!2

k=1



n X



n X

b2k + x 

k=1

ak

k=1

k=1

!2

n X

−

bk

bk −

n X

!# ak b k

k=1

 b2k  .

k=1

k=1

Since, it is obvious that: n X

n X

!2

k=1 n X

ak

k=1

−

ak

n X

ak b k =

k=1

ai aj ,

1≤i<j≤n

k=1 n X

bk −

X

a2k = 2

k=1

X

ai b j

1≤i6=j≤n

and n X

!2 −

bk

k=1

n X

b2k = 2

k=1

X

bi bj ,

1≤i<j≤n

we get P (y) =

n X k=1

! a2k + 2x

X

ai aj

y2

1≤i<j≤n

− 2y

n X k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

! ak b k + x

X 1≤i6=j≤n

ai b j

+

n X k=1

b2k + 2x

X

bi bj .

1≤i<j≤n

http://jipam.vu.edu.au/

24

S.S. D RAGOMIR

Taking into consideration, by the definition of P, that P (y) ≥ 0 for any y ∈ R, it follows that the discriminant ∆ ≤ 0, i.e., 1 0≥ ∆= 4

n X

!2 X

ak b k + x

k=1

ai b j

1≤i6=j≤n n X

−

! X

a2k + 2x

ai aj

1≤i<j≤n

k=1

n X

! X

b2k + 2x

bi bj

1≤i<j≤n

k=1



and the inequality (2.108) is proved.

Remark 2.57. If x = 0, then from (2.108) we recapture the (CBS) −inequality for real numbers. 2.13. Wagner’s inequality for Complex Numbers. The following inequality which provides a version for complex numbers of Wagner’s result holds [16]. ¯ = (b1 , . . . , bn ) be sequences of complex numbers. Theorem 2.58. Let ¯ a = (a1 , . . . , an ) and b Then for any x ∈ [0, 1] one has the inequality " (2.109)

n X

#2 X

Re ak¯bk + x


k=1

Re ai¯bj




1≤i6=j≤n

" ≤

n X

#" X

|ak |2 + 2x

Re (ai a ¯j )

1≤i<j≤n

k=1

n X

# X

|bk |2 + 2x

Re bi¯bj




n X

#

.

1≤i<j≤n

k=1

Proof. Start with the function f : R → R,

(2.110)

f (t) = (1 − x)

n X k=1

n 2 X |tak − bk | + x (tak − bk ) . 2

k=1

We have (2.111)

f (t) = (1 − x)

n X

(tak − bk ) t¯ ak − ¯bk




k=1

+x t " 2

= (1 − x) t

n X

ak −

k=1 n X

n X

! bk

k=1 n X

2

|ak | − t

k=1

" + x t2

t

k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

2

|ak | − t

a ¯k −

k=1 n X

bk a ¯k − t

k=1 n X

n X

!

n X

¯bk

k=1

ak¯bk +

n X

k=1 n X k=1

bk

n X k=1

a ¯k − t

# 2

|bk |

k=1 n X k=1

ak

n X k=1

¯bk +

2

|bk |

k=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

 = (1 − x)

n X k=1

25

n  X 2 |ak |2 + x ak  t2 k=1

" + 2 (1 − x)

n X

" Re ak¯bk + x Re


k=1

+ (1 − x)

n X k=1

n X

ak

k=1

n X

## ¯bk

t

k=1

2 n X |bk |2 + x bk . k=1

Observe that 2 n n X X ak = ai a ¯j

(2.112)

k=1

i,j=1 n X

X

i=1

1≤i6=j≤n

|ai |2 +

=

n X

=

i=1 n X

=

ai a ¯j

X

|ai |2 +

1≤i<j≤n

i=1

ai a ¯j

1≤j
X

|ai |2 + 2

X

ai a ¯j +

Re (ai a ¯j )

1≤i<j≤n

and, similarly, n n X 2 X X
bk = |bi |2 + 2 Re bi¯bj .

(2.113)

i=1

k=1

1≤i<j≤n

Also n X

ak

k=1

n X

¯bk =

n X i=1

k=1

X

ai¯bi +

ai¯bj

1≤i6=j≤n

and thus n X

Re

(2.114)

k=1

ak

n X

! ¯bk

=

n X i=1

k=1

X


Re ai¯bi +


Re ai¯bj .

1≤i6=j≤n

Utilising (2.112) – (2.114), by (2.111), we deduce " (2.115) f (t) =

n X

# |ak |2 + 2x

+2

n X

Re (ai a ¯j ) t2

1≤i<j≤n

k=1

"

X

# Re ak¯bk + x


k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

X 1≤i6=j≤n

Re ai¯bj




t+

n X k=1

|bk |2 + 2x

X


Re bi¯bj .

1≤i<j≤n

http://jipam.vu.edu.au/

26

S.S. D RAGOMIR

Since, by (2.110), f (t) ≥ 0 for any t ∈ R, it follows that the discriminant of the quadratic function given by (2.115) is negative, i.e., 1 0≥ ∆ 4 " n #2 X X

= Re ak¯bk + x Re ai¯bj 1≤i6=j≤n

k=1

" −

n X

#" X

2

|ak | + 2x

Re (ai a ¯j )

1≤i<j≤n

k=1

n X

# 2

|bk | + 2x

X

Re bi¯bj




1≤i<j≤n

k=1



and the inequality (2.109) is proved. Remark 2.59. If x = 0, then we get the (CBS) −inequality #2 " n n n X X X
2 ¯ |bk |2 . (2.116) Re ak bk ≤ |ak | k=1

k=1

k=1

R EFERENCES [1] V.Y. BUNIAKOWSKI, Sur quelques inégalités concernant les intégrales aux differences finies, Mem. Acad. St. Petersburg, (7) 1 (1859), No. 9, 1-18. [2] A.L. CAUCHY, Cours d’Analyse de l’École Royale Polytechnique, I re Partie, Analyse Algébrique, Paris, 1821. [3] P. SCHREIDER, The Cauchy-Bunyakovsky-Schwarz inequality, Hermann Graßmann (Lieschow, 1994), 64–70. ´ J.E. PECARI ˇ [4] D.S. MITRINOVIC, C´ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993. [5] S.S. DRAGOMIR, On some inequalities (Romanian), “Caiete Metodico Stiin¸ ¸ tifice”, No. 13, 1984, pp. 20. Faculty of Mathematics, Timi¸soara University, Romania. [6] E.F. BECKENBACH Heidelberg, 1961.

AND

R. BELLMAN, Inequalities, Springer-Verlag, Berlin-Göttingen-

[7] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s Inequality for Real Numbers (Romanian), “Caiete Metodico-Stiin¸ ¸ tifice”, No. 57, pp. 24, 1989. Faculty of Mathematics, Timi¸soara University, Romania. [8] S.S. DRAGOMIR AND J. SÁNDOR, Some generalisations of Cauchy-Buniakowski-Schwartz’s inequality (Romanian), Gaz. Mat. Metod. (Bucharest), 11 (1990), 104–109. [9] I. CUCUREZEANU, Problems on Number Theory (Romanian), Ed. Technicˇa, Bucharest, 1976. [10] S.S. DRAGOMIR, On an inequality of Tiberiu Popoviciu (Romanian), Gaz. Mat. Metod., (Bucharest) 8 (1987), 124–128. ZBL No. 712:110A. [11] T. POPOVICIU, Gazeta Matematicˇa, 16 (1940), p. 334. [12] S.S. DRAGOMIR, Inequalities of Cauchy-Buniakowski-Schwartz’s type for positive linear functionals (Romanian), Gaz. Mat. Metod. (Bucharest), 9 (1988), 162–164. [13] T. ANDRESCU, D. ANDRICA AND M.O. DRÎMBE, The trinomial principle in obtaining inequalities (Romanian), Gaz. Mat. (Bucharest), 90 (1985), 332–338. [14] P. FLOR, Über eine Unglichung von S.S. Wagner, Elemente Math., 20 (1965), 136.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

27

[15] S.S. WAGNER, Amer. Math. Soc., Notices, 12 (1965), 220. [16] S.S. DRAGOMIR, A version of Wagner’s inequality for complex numbers, submitted.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

28

S.S. D RAGOMIR

3. R EFINEMENTS OF THE (CBS) −I NEQUALITY 3.1. A Refinement in Terms of Moduli. The following result was proved in [1]. ¯ = (b1 , . . . , bn ) be sequences of real numbers. Then Theorem 3.1. Let ¯ a = (a1 , . . . , an ) and b one has the inequality !2 n n n n n n n X X X X X X X 2 2 bk − ak b k ≥ ak |ak | bk |bk | − ak |bk | |ak | bk ≥ 0. (3.1) ak k=1

k=1

k=1

k=1

k=1

k=1

k=1

Proof. We will follow the proof from [1]. For any i, j ∈ {1, . . . , n} the next elementary inequality is true: |ai bj − aj bi | ≥ ||ai bj | − |aj bi || .

(3.2)

By multiplying this inequality with |ai bj − aj bi | ≥ 0 we get (3.3)

(ai bj − aj bi )2 ≥ |(ai bj − aj bi ) (|ai | |bj | − |aj | |bi |)| = |ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi || .

Summing (3.3) over i and j from 1 to n, we deduce n X

(ai bj − aj bi )2

i,j=1

n X a |a | b |b | + b |b | a |a | − |a | b a |b | − a b |a | |b | ≥ i i j j i i j j i i j j i j j i i,j=1 n X ≥ (ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi |) , i,j=1



giving the desired inequality (3.1). The following corollary is a natural consequence of (3.1) [1, Corollary 4]. Corollary 3.2. Let ¯ a be a sequence of real numbers. Then !2 n n n n n X X X X X 1 1 1 1 1 2 (3.4) ak − ak ≥ ak |ak | − ak · |ak | ≥ 0. n k=1 n k=1 n k=1 n k=1 n k=1

There are some particular inequalities that may also be deduced from the above Theorem 3.1 (see [1, p. 80]). ¯ sequences of real numbers, one has sgn (ak ) = sgn (bk ) = (1) Suppose that for ¯ a and b ek ∈ {−1, 1} . Then one has the inequality !2 n !2 n n n n n X X X X X X (3.5) a2k b2k − ak b k ≥ ek a2k ek b2k − ek ak bk ≥ 0. k=1 k=1 k=1 k=1 k=1 k=1 (2) If ¯ a = (a1 , . . . , a2n ) , then we have the inequality " 2n #2 2n 2n 2n X X X X k k 2 (3.6) 2n ak − (−1) ak ≥ ak (−1) |ak | ≥ 0. k=1

k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

k=1

k=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

29

(3) If ¯ a = (a1 , . . . , a2n+1 ) , then we have the inequality !2 2n+1 2n+1 2n+1 2n+1 X X X X (3.7) (2n + 1) a2k − (−1)k ak ≥ ak (−1)k |ak | ≥ 0. k=1

k=1

k=1

k=1

The following version for complex numbers is valid as well. ¯ = (b1 , . . . , bn ) be sequences of complex numbers. Theorem 3.3. Let ¯ a = (a1 , . . . , an ) and b Then one has the inequality n 2 n n n n n n X X X X X X X (3.8) |ai |2 |bi |2 − ai b i ≥ |ai | a ¯i |bi | bi − |ai | bi |bi | a ¯i ≥ 0. i=1

i=1

i=1

i=1

i=1

i=1

i=1

Proof. We have for any i, j ∈ {1, . . . , n} that |¯ ai b j − a ¯j bi | ≥ ||ai | |bj | − |aj | |bi || . Multiplying by |¯ ai b j − a ¯j bi | ≥ 0, we get |¯ ai b j − a ¯j bi |2 ≥ ||ai | a ¯i |bj | bj + |aj | a ¯j |bi | bi − |ai | bi |bj | a ¯j − |bi | a ¯i |aj | bj | . Summing over i and j from 1 to n and using the Lagrange’s identity for complex numbers: 2 n n n n X X X 1X 2 2 |¯ ai b j − a ¯j bi |2 |ai | |bi | − ai b i = 2 i,j=1 i=1 i=1 i=1 

we deduce the desired inequality (3.8). Remark 3.4. Similar particular inequalities may be stated, but we omit the details.

3.2. A Refinement for a Sequence Whose Norm is One. The following result holds [1, Theorem 6]. ¯ Theorem 3.5. Let ¯ a = (a e= P1 ,n. . . ,2an ), b = (b1 , . . . , bn ) be sequences of real numbers and ¯ (e1 , . . . , en ) be such that i=1 ei = 1. Then the following inequality holds #2 " n n n n n n n X X X X X X X 2 2 ai bi ≥ ak b k − ek ak ek bk + ek ak (3.9) ek bk i=1 i=1 k=1 k=1 k=1 k=1 k=1 !2 n X ≥ ak b k . k=1

Proof. We will follow the proof from [1]. From the (CBS) −inequality, one has (3.10)

n X k=1

" ak −

n X

!

#2

ei ai ek

i=1

n X

" bk −

n X

≥

k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

ei bi ek

i=1

k=1

( n " X

#2

!

ak −

n X i=1

! ei ai ek

#" bk −

n X

! ei bi ek

#)2 .

i=1

http://jipam.vu.edu.au/

30

S.S. D RAGOMIR

Since

Pn

2 k=1 ek

= 1, a simple calculation shows that ! #2 " n n n X X X ak − ei ai ek = a2k − i=1

k=1 n X

"

n X

bk −

#2

! ei bi ek

=

i=1

k=1

n X

k=1

k=1

n X

n X

b2k −

k=1

!2 ek ak

,

!2 ek bk

,

k=1

and n X

" ak −

n X

!

#"

n X

bk −

ei ai ek

i=1

k=1

!

ei bi ek =

i=1

and then the inequality (3.10) becomes  !2   n n n X X X (3.11)  a2k − ek ak   b2k − k=1

#

k=1

k=1

n X

ak bk −

n X

k=1

n X

ek ak

k=1

n X

ek bk

k=1

!2  ek bk



k=1

≥

n X

ak b k −

k=1

Using the elementary inequality

m2 − l2 p2 − q 2 ≤ (mp − lq)2 ,

n X

ek ak

k=1

n X

!2 ek bk

≥ 0.

k=1

m, l, p, q ∈ R

for the choices m=

n X

! 12 a2k

,

n X l= ek ak ,

p=

k=1

k=1 n X and q = ek bk

n X

! 12 b2k

k=1

k=1

the above inequality (3.11) provides the following result  (3.12)

n X

 k=1

! 21 a2k

n X

! 21 b2k

k=1

n 2 n X X − ek ak ek bk  k=1 k=1 n 2 n n X X X ≥ ak b k − ek ak ek bk . k=1

k=1

k=1

Since n X

! 21 a2k

k=1

n X k=1

! 12 b2k

n n X X ≥ ek ak ek bk k=1

k=1

then, by taking the square root in (3.12) we deduce the first part of (3.9). The second part is obvious, and the theorem is proved.



The following corollary is a natural consequence of the above theorem [1, Corollary 7].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

¯ ¯ Corollary 3.6. Let ¯ a, b, e be as in Theorem 3.5. If n X

(3.13)

a2k

n X

b2k ≥ 4

k=1

k=1

n X

Pn

k=1

!2 ek ak

k=1

31

ak bk = 0, then one has the inequality: n X

!2 ek bk

.

k=1

The following inequalities are interesting as well [1, p. 81]. ¯ one has the inequality (1) For any ¯ a, b n #2 " n n n n n n X 1 X X X X X X 1 b2k ≥ ak b k − (3.14) a2k ak bk + ak bk n k=1 k=1 n k=1 k=1 k=1 k=1 k=1 !2 n X ≥ ak b k . k=1

(2) If

Pn

k=1

ak bk = 0, then n X

n X

4 a2k b2k ≥ 2 n k=1 k=1

(3.15)

n X k=1

!2 ak

n X

!2 bk

.

k=1

In a similar manner, we may state and prove the following result for complex numbers. ¯ = (b1 , . . . , bn ) be sequences of complex numbers and Theorem 3.7. Let ¯ a = (a1 , . . . , an ), b P ¯ e = (e1 , . . . , en ) a sequence of complex numbers satisfying the condition ni=1 |ei |2 = 1. Then the following refinement of the (CBS) −inequality holds n #2 " n n n n n n X X X X X X X 2 2 ¯ ¯ ¯ (3.16) |ai | |bi | ≥ ak bk − ak e¯k · ek bk + ak e¯k · ek bk i=1 i=1 k=1 k=1 k=1 k=1 k=1 2 n X ≥ ak¯bk . k=1

The proof is similar to the one in Theorem 3.5 on using the corresponding (CBS) −inequality for complex numbers. Remark 3.8. Similar particular inequalities may be stated, but we omit the details. 3.3. A Second Refinement in Terms of Moduli. The following lemma holds. Lemma 3.9. Let ¯ a = (a1 , . . . , an ) beP a sequence of real numbers and p ¯ = (p1 , . . . , pn ) a sequence of positive real numbers with ni=1 pi = 1. Then one has the inequality: !2 n n n n n X X X X X 2 (3.17) p i ai − p i ai ≥ pi |ai | ai − pi |ai | p i ai . i=1

i=1

i=1

i=1

i=1

Proof. By the properties of moduli we have (ai − aj )2 = |(ai − aj ) (ai − aj )| ≥ |(|ai | − |aj |) (ai − aj )| for any i, j ∈ {1, . . . , n} . This is equivalent to (3.18)

a2i − 2ai aj + a2j ≥ ||ai | ai + |aj | aj − |ai | aj − |aj | ai |

for any i, j ∈ {1, . . . , n} .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

32

S.S. D RAGOMIR

If we multiply (3.18) by pi pj ≥ 0 and sum over i and j from 1 to n we deduce n n n n n n X X X X X X p i ai p j aj + pi pj a2j pj pi a2i − 2 j=1

i=1

i=1

≥

n X

j=1

i=1

j=1

pi pj |ai | ai + |aj | aj − |ai | aj − |aj | ai

i,j=1

n X pi pj (|ai | ai + |aj | aj − |ai | aj − |aj | ai ) , ≥ i,j=1



which is clearly equivalent to (3.17).

Using the above lemma, we may prove the following refinement of the (CBS) -inequality. ¯ = (b1 , . . . , bn ) be two sequences of real numbers. Theorem 3.10. Let ¯ a = (a1 , . . . , an ) and b Then one has the inequality !2 n n n n n n n X X X X X X X (3.19) a2i b2i − ai b i ≥ a2i · sgn (ai ) |bi | bi − |ai bi | ai bi ≥ 0. i=1

i=1

i=1

i=1

i=1

i=1

i=1

Proof. If we choose (for ai 6= 0, i ∈ {1, . . . , n}) in (3.17), that a2 pi := Pn i

2 k=1 ak

, xi =

bi , i ∈ {1, . . . , n} , ai

we get n X i=1

 2 bi Pn · − 2 ai k=1 ak a2i

!2 bi Pn · 2 k=1 ak ai i=1 n n X bi bi X a2i a2i Pn P ≥ · − n 2 2 ai k=1 ak ai k=1 ak i=1 i=1 n X

a2i

n 2 bi X a b i i Pn · 2 ai k=1 ak ai i=1

from where we get n X

b2 Pn i k=1 i=1

Pn |a | Pn Pn Pn i 2 |b | b i i ( i=1 ai bi ) i=1 ai i=1 |ai bi | i=1 ai bi P − Pn − P n 2 ≥ 2 2 a2k ( k=1 a2k ) ( nk=1 a2k ) k=1 ak 

which is clearly equivalent to (3.19). The case for complex numbers is as follows.

Lemma 3.11. Let ¯ z = (z1 , . . . , zn ) be aPsequence of complex numbers and p ¯ = (p1 , . . . , pn ) a n sequence of positive real numbers with i=1 pi = 1. Then one has the inequality: 2 n n n n n X X X X X 2 (3.20) pi |zi | − pi zi ≥ pi |zi | zi − pi |zi | pi zi . i=1

i=1

i=1

i=1

i=1

Proof. By the properties of moduli for complex numbers we have |zi − zj |2 ≥ |(|zi | − |zj |) (zi − zj )| for any i, j ∈ {1, . . . , n} , which is clearly equivalent to |zi |2 − 2 Re (zi z¯j ) + |zj |2 ≥ ||zi | zi + |zj | zj − zi |zj | − |zi | zj | for any i, j ∈ {1, . . . , n} .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

33

If we multiply with pi pj ≥ 0 and sum over i and j from 1 to n, we deduce the desired inequality (3.20).  Now, in a similar manner to the one in Theorem 3.10, we may state the following result for complex numbers. ¯ = (b1 , . . . , bn ) be two seTheorem 3.12. Let ¯ a = (a1 , . . . , an ) (ai 6= 0, i = 1, . . . , n) and b quences of complex numbers. Then one has the inequality: (3.21)

n X i=1

n 2 n n n n X X |a | X X X i |bi | bi − |ai | bi a ¯i bi ≥ 0. |ai |2 |bi |2 − a ¯ i bi ≥ ai i=1

i=1

i=1

i=1

i=1

3.4. A Refinement for a Sequence Less than the Weights. The following result was obtained in [1, Theorem 9] (see also [2, Theorem 3.10]). ¯ = (b1 , . . . , bn ) be sequences of real numbers and Theorem 3.13. Let ¯ a = (a1 , . . . , an ), b p ¯ = (p1 , . . . , pn ), q ¯ = (q1 , . . . , qn ) be sequences of nonnegative real numbers such that pk ≥ qk for any k ∈ {1, . . . , n} . Then we have the inequality

(3.22)

 ! 21 2 n n n X X X pk a2k pk b2k ≥  (pk − qk ) ak bk + qk a2k qk b2k  k=1 k=1 k=1 k=1 k=1 n #2 " n X X ≥ (pk − qk ) ak bk + q k ak b k k=1 k=1 !2 n X ≥ p k ak b k .

n X

n X

k=1

Proof. We shall follow the proof in [1]. Since pk − qk ≥ 0, then the (CBS) −inequality for the weights rk := pk − qk will give (3.23)

n X

pk a2k −

k=1

n X

! qk a2k

k=1

n X

pk b2k −

k=1

n X

! qk b2k

" ≥

n X

#2 (pk − qk ) ak bk

.

k=1

k=1

Using the elementary inequality (ac − bd)2 ≥ a2 − b2





c2 − d2 ,

a, b, c, d ∈ R

for the choices a=

n X

! 21 pk a2k

, b=

k=1

n X

! 12 qk a2k

k=1

, c=

n X

! 12 pk b2k

k=1

and d=

n X

! 12 qk b2k

k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

34

S.S. D RAGOMIR

we deduce by (3.23) that  (3.24)

n X



! 21

n X

pk a2k

k=1

! 12 pk b2k

−

n X

! 12

n X

qk a2k

k=1

k=1

! 12 2 qk b2k



k=1

" ≥

n X

#2 (pk − qk ) ak bk

.

k=1

Since, obviously, n X

! 21

n X

pk a2k

k=1

! 12 pk b2k

≥

k=1

n X

! 12 qk a2k

k=1

! 12

n X

qk b2k

k=1

then, by (3.24), on taking the square root, we would get ! 12 ! 12 ! 12 ! 21 n n n n n X X X X X pk a2k pk b2k ≥ qk a2k qk b2k + (pk − qk ) ak bk , k=1

k=1

k=1

k=1

k=1

which provides the first inequality in (3.22). The other inequalities are obvious and we omit the details.



The following corollary is a natural consequence of the above theorem [2, Corollary 3.11]. ¯ be sequences of real numbers and ¯ Corollary 3.14. Let ¯ a, b s = (s1 , . . . , sn ) be such that 0 ≤ sk ≤ 1 for any k ∈ {1, . . . , n} . Then one has the inequalities  ! 12 2 n n n n n X X X X X b2k ≥  (1 − sk ) ak bk + sk b2k  a2k sk a2k (3.25) k=1 k=1 k=1 k=1 k=1 n #2 " n X X ≥ (1 − sk ) ak bk + s k ak b k k=1 k=1 !2 n X ≥ ak b k . k=1

¯ and ¯ Remark 3.15. Assume that ¯ a, b s are as in Corollary 3.14. The following inequalities hold (see [2, p. 15]). P a) If nk=1 ak bk = 0, then !2 n n n X X X (3.26) a2k b2k ≥ 4 s k ak b k . k=1

b) If (3.27)

Pn

k=1

k=1

k=1

sk ak bk = 0, then n X k=1

a2k

n X k=1

 n X b2k ≥  ak b k +

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

k=1

n X k=1

αk a2k

n X

! 21 2 αk b2k

 .

k=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

35

In particular, we may obtain the following particular inequalities involving trigonometric functions (see [2, p. 15])  ! 12 2 n n n n n X X X X X (3.28) a2k b2k ≥  ak bk cos2 αk + a2k sin 2 αk b2k sin2 αk  k=1 k=1 k=1 k=1 k=1 #2 " n n X X ak bk sin2 αk ≥ ak bk cos2 αk + k=1 k=1 !2 n X ≥ ak b k , k=1

where ak , bk , αk ∈ R, k = 1,P . . . , n. If one would assume that nk=1 ak bk = 0, then n X

(3.29)

k=1

If

Pn

k=1

(3.30)

a2k

n X

b2k ≥ 4

k=1

!2 ak bk sin2 αk

.

k=1

k=1

ak bk sin2 αk = 0, then  n n n X X X 2 2  ak bk ≥ ak b k + k=1

n X

n X

k=1

a2k sin2 αk

k=1

n X

! 21 2 b2k sin2 αk

 .

k=1

3.5. A Conditional Inequality Providing a Refinement. The following lemma holds [2, Lemma 4.1]. Lemma 3.16. Consider the sequences of real numbers x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) and ¯ z = (z1 , . . . , zn ) . If yk2 ≤ |xk zk | for any k ∈ {1, . . . , n} ,

(3.31) then one has the inequality:

n X

(3.32)

!2 |yk |

≤

n X

k=1

|xk |

n X

k=1

|zk | .

k=1

Proof. We will follow the proof in [2]. Using the condition (3.31) and the (CBS) −inequality, we have n n X X 1 1 |yk | ≤ |xk | 2 |zk | 2 k=1

k=1

" ≤

n  X

|xk |

1 2

n  2 X

k=1

=

n X

|zk |

1 2

2

# 12

k=1

|xk |

k=1

which is clearly equivalent to (3.32).

n X

! 12 |zk |

k=1



The following result holds [2, Theorem 4.6].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

36

S.S. D RAGOMIR

¯ = (b1 , . . . , bn ) and ¯ Theorem 3.17. Let ¯ a = (a1 , . . . , an ), b c = (c1 , . . . , cn ) be sequences of real numbers such that (i) |bk | + |ck | = 6 0 (k ∈ {1, . . . , n}); 2|bk ck | (ii) |ak | ≤ |bk |+|ck | for any k ∈ {1, . . . , n} . Then one has the inequality n X

(3.33)

|ak | ≤

k=1

2

Pn P |bk | nk=1 |ck | k=1 Pn . k=1 (|bk | + |ck |)

Proof. We will follow the proof in [2]. By (ii) we observe that   2 |bk | 2 |bk ck | for any k ∈ {1, . . . , n} |ak | ≤ ≤ |bk | + |ck |  2 |c | k

and thus (3.34)

xk := 2 |bk | − |ak | ≥ 0

and

zk := 2 |ck | − |ak | ≥ 0

for any k ∈ {1, . . . , n} .

A simple calculation also shows that the relation (ii) is equivalent to (3.35)

a2k ≤ (2 |bk | − |ak |) (2 |ck | − |ak |) for any k ∈ {1, . . . , n} .

If we consider yk := ak and take xk , zk (k = 1, . . . , n) as defined by (3.34), then we get yk2 ≤ xk zk (with xk , zk ≥ 0) for any k ∈ {1, . . . , n}. Applying Lemma 3.16 we deduce

(3.36)

n X

!2 |ak |

≤

k=1

2

n X k=1

|bk | −

n X

! |ak |

k=1

2

n X k=1

|ck | −

n X

! |ak |

k=1



which is clearly equivalent to (3.33). The following corollary is a natural consequence of the above theorem [2, Corollary 4.7].

Corollary 3.18. For any sequence x ¯ and y ¯ of real numbers, with |xk |+|yk | = 6 0 (k = 1, . . . , n) , one has: P P n X 2 nk=1 |xk | nk=1 |yk | |xk yk | (3.37) ≤ Pn . |x k | + |yk | k=1 (|xk | + |yk |) k=1 For two positive real numbers, let us recall the following means a+b A (a, b) := (the arithmetic mean) 2 √ G (a, b) := ab (the geometric mean) and 2 H (a, b) := 1 1 (the harmonic mean). +b a ¯ = (b1 , . . . , bn ) are sequences of real numbers, then We remark that if ¯ a = (a1 , . . . , an ), b obviously ! n n n X X X (3.38) A (ai , bi ) = A ai , bi , i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

i=1

i=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

37

and, by the (CBS) −inequality, n X

(3.39)

G (ai , bi ) ≤ G

i=1

n X

ai ,

i=1

n X

! bi .

i=1

The following similar result for harmonic means also holds [2, p. 19]. ¯ we have the property: Theorem 3.19. For any two sequences of positive real numbers ¯ a and b ! n n n X X X (3.40) H (ai , bi ) ≤ H ai , bi . i=1

i=1

i=1

Proof. Follows by Corollary 3.18 on choosing xk = ak , yk = bk and multiplying the inequality (3.37) with 2.  The following refinement of the (CBS) −inequality holds [2, Corollary 4.9]. This result is known in the literature as Milne’s inequality [8]. Theorem 3.20. For any two sequences of real numbers p ¯ = (p1 , . . . , pn ), q ¯ = (q1 , . . . , qn ) with |pk | + |qk | = 6 0 (k = 1, . . . , n) , one has the inequality: !2 n n n n n X X X X
X p2k qk2 2 2 2 p k + qk ≤ pk qk2 . (3.41) p k qk ≤ 2 2 p + q k k=1 k k=1 k=1 k=1 k=1 Proof. We shall follow the proof in [2]. The first inequality is obvious by Lemma 3.16 on p2 q 2 choosing yk = pk qk , xk = p2k + qk2 and zk = p2k+qk2 (k = 1, . . . , n) . k k The second inequality follows by Corollary 3.18 on choosing xk = p2k and y = qk2 (k = 1, . . . , n).  Remark 3.21. The following particular inequality is obvious by (3.41) !2 n n X X (3.42) sin αk cos αk ≤n sin2 αk cos2 αk i=1

i=1

≤

n X

2

sin αk

i=1

n X

cos2 αk ;

i=1

for any αk ∈ R, k ∈ {1, . . . , n} . 3.6. A Refinement for Non-Constant Sequences. The following result was proved in [3, Theorem 1]. ¯ = (bi ) , p Theorem 3.22. Let ¯ a = (ai )i∈N , b i∈N ¯ = (pi )i∈N be sequences of real numbers such that (i) ai 6= aj and bi 6= bj for i 6= j, i, j ∈ N; (ii) pi > 0 for all i ∈ N. Then for any H a finite part of N one has the inequality: !2 X X X (3.43) pi a2i pi b2i − p i ai b i ≥ max {A, B} ≥ 0, i∈H

i∈H

i∈H

where hP (3.44)

A := max

i∈H

p i ai b i

P

j∈J

p j aj −

PJ

P

i∈H

pi a2i −

J⊆H J6=∅

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

2 i∈H pi ai

P

P

i∈J

p i ai

P

j∈J

p j bj

i2


2

http://jipam.vu.edu.au/

38

S.S. D RAGOMIR

and hP

and PJ :=

i∈H

B := max

(3.45)

p i ai b i

P

j∈J

p j bj −

PJ

P

i∈H

pi b2i −

J⊆H J6=∅

P

j∈J

2 i∈H pi bi

P

P

i∈J

P

j∈J

p i bi

p j aj

i2


2

pj .

Proof. We shall follow the proof in [3]. Let J be a part of H. Define the mapping fJ : R → R given by    2 X X X X X fJ (t) = pi a2i  pi b2i + pi (bi + t)2  −  p i ai b i + pi ai (bi + t) . i∈H

i∈J

i∈H\J

i∈J

i∈H\J

Then by the (CBS) −inequality we have that fJ (t) ≥ 0 for all t ∈ R. On the other hand we have " # " #2 X X X X X 2 2 2 fJ (t) = p i ai pi bi + 2t p i b i + t PJ − p i ai b i + t p i ai i∈H

i∈H

i∈H

i∈H



i∈J

!2  X

= t2 PJ

X

pi a2i −

i∈H

p i ai



i∈J

#

" + 2t

X

pi a2i

i∈H

X

p i bi −

i∈J

i∈H

pi a2i −

X

p i ai b i

i∈H

 X X + pi a2i pi b2i − i∈H

X

X

p i ai

i∈J

!2  X

p i ai b i



i∈H

for all t ∈ R. Since !2 PJ

X i∈H

p i ai

!2 ≥ PJ

X

pi a2i −

i∈J

i∈J

X

p i ai

>0

i∈J

as ai 6= aj for all i, j ∈ {1, . . . , n} with i 6= j, then, by the inequality fJ (t) ≥ 0 for any t ∈ R we get that " #2 X X X X 1 0≥ ∆= p i ai b i p j aj − pi a2i p j bj 4 i∈H j∈J i∈H j∈J  !2   X X X X − PJ pi a2i − p i ai   pi a2i pi b2i − i∈H

i∈J

i∈H

i∈H

!2  X

p i ai b i



i∈H

from where results the inequality !2 X

pi a2i

i∈H

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

X i∈H

pi b2i −

X

p i ai b i

≥ A.

i∈H

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

39

The second part of the proof goes likewise for the mapping gJ : R → R given by    2 X X X X X gJ (t) =  pi a2i + pi (ai + t)2  pi b2i −  p i ai b i + pi bi (ai + t) i∈J

i∈H\J

i∈H

i∈J

i∈H\J



and we omit the details. The following corollary also holds [3, Corollary 1]. Corollary 3.23. With the assumptions of Theorem 3.22 and if P 2 P P P 2 i∈H pi ai bi i∈H pi ai − i∈H pi ai i∈H pi bi , (3.46) C :=
2 P P PH i∈H pi a2i − p a i i i∈H P

i∈H

D :=

(3.47)

p i ai b i

P

PH

P

i∈H

p i bi −

P

2 i∈H pi bi −

i∈H

pi b2i

P

P

i∈H

i∈H pi bi

p i ai

2


2

,

then one has the inequality !2 X

(3.48)

i∈H

pi a2i

X

X

pi b2i −

i∈H

≥ max {C, D} ≥ 0.

p i ai b i

i∈H

The following corollary also holds [3, Corollary 2]. Corollary 3.24. If ai , bi 6= 0 for i ∈ N and H is a finite part of N, then one has the inequality !2 X X X p i ai b i (3.49) pi a2i pi b2i − i∈H

i∈H

i∈H

1 ≥ max card (H) − 1

(P

) P 2 pj c2j j∈H pj dj P ,P ≥ 0, 2 2 i∈H pi ai i∈H pi bi j∈H

where cj := aj

(3.50)

X

p i ai b i − b j

i∈H

X

pi a2i , j ∈ H

i∈H

and dj := aj

(3.51)

X

pi b2i − bj

i∈H

X

pi ai bi , j ∈ H.

i∈H

Proof. Choosing in Theorem 3.22, J = {j} , we get the inequality !2 X X X p2j c2j 2 2 P p i ai p i bi − p i ai b i ≥ , j∈H pj i∈H pi a2i − p2j a2j i∈H i∈H i∈H from where we obtain X

! X X pi a2i − pj a2j  pi a2i pi b2i −

X

i∈H

i∈H

i∈H

i∈H

!2  p i ai b i

Summing these inequalities over j ∈ H, we get  X X X [card (H) − 1] pi a2i  pi a2i pi b2i − i∈H

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

i∈H

i∈H

 ≥ pj c2j for any j ∈ H.

!2  X i∈H

p i ai b i

≥

X

pj c2j

j∈H

http://jipam.vu.edu.au/

40

S.S. D RAGOMIR

from where we get the first part of (3.49). The second part goes likewise and we omit the details.



Remark 3.25. The following particular inequalities provide refinement for the (CBS) −inequality [3, p. 60 – p. 61]. (1) Assume that ¯ a = (a1 , . . . , an ), b = (b1 , . . . , bn ) are nonconstant sequences of real numbers. Then !2 n n n X X X (3.52) a2i b2i − ai b i i=1

i=1

i=1

    [Pn a2 Pn b − Pn a Pn a b ]2 i=1 i i=1 i i=1 i i=1 i i ≥ max , n P P  2 n 2  n ai − ( i=1 ai )  i=1

P P P P [ ni=1 bi ni=1 ai bi − ni=1 ai ni=1 n P P 2 n b2i − ( ni=1 bi ) i=1

  2 b2 ]  i

.

  

¯ are sequences of real numbers with not all elements equal to zero, (2) Assume that ¯ a and b then !2 n n n X X X b2i − ai b i (3.53) a2i i=1

i=1

i=1

≥

  2 n n n P P P  2  aj ai b i − b j ai  

1 j=1 max  n−1   

i=1

i=1

n P

,

a2i

i=1 n P



j=1

aj

n P

a2i − bj

i=1

n P i=1

n P i=1

b2i

2   ai b i   

.

   

3.7. De Bruijn’s Inequality. The following refinement of the (CBS) −inequality was proved by N.G. de Bruijn in 1960, [4] (see also [5, p. 89]). Theorem 3.26. If ¯ a = (a1 , . . . , an ) is a sequence of real numbers and ¯ z = (z1 , . . . , zn ) is a sequence of complex numbers, then 2 n # " n n n X X X X 1 a2k |zk |2 + zk2 . (3.54) ak zk ≤ 2 k=1

k=1

k=1

k=1

Equality holds in (3.54) only if for k ∈ {1, . . . , n} , ak = Re (λzk ) , where λ is a complex Pn if 2and 2 number such that k=1 λ zk is a nonnegative real number. Proof. We shall follow the proof in [5, p. 89 – p. 90].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

41

By a simultaneous rotation of all the zk ’s about the origin, we get n X ak zk ≥ 0. k=1

This rotation does not affect the moduli n n X X ak zk , zk2 and |zk | for k ∈ {1, . . . , n} . k=1 k=1 P Hence, it is sufficient to prove inequality (3.54) for the case where nk=1 ak zk ≥ 0. If we put zk = xk + iyk (k ∈ {1, . . . , n}) , then, by the (CBS) −inequality for real numbers, we have n 2 !2 n n n X X X X 2 (3.55) ak zk = ak zk ≤ ak x2k . k=1

k=1

k=1

k=1

Since 2x2k = |zk |2 + Re zk2 for any k ∈ {1, . . . , n} we obtain, by (3.55), that 2 " n # n n n X X 1X 2 X ak |zk |2 + Re zk2 . (3.56) ak zk ≤ 2 k=1

k=1

As

n X

k=1

Re zk2 = Re

k=1

n X

k=1

! zk2

k=1

n X ≤ zk2 , k=1



then by (3.56) we deduce the desired inequality (3.54).

3.8. McLaughlin’s Inequality. The following refinement of the (CBS) −inequality for sequences of real numbers was obtained in 1966 by H.W. McLaughlin [7, p. 66]. ¯ = (b1 , . . . , b2n ) are sequences of real numbers, then Theorem 3.27. If ¯ a = (a1 , . . . , a2n ), b !2 " n #2 2n 2n 2n X X X X 2 (3.57) ai b i + (ai bn+i − an+i bi ) ≤ ai b2i i=1

i=1

i=1

i=1

with equality if and only if for any i, j ∈ {1, . . . , n} ai bj − aj bi − an+i bn+j + an+j bn+i = 0

(3.58) and

ai bn+j − aj bn+i + an+i bj − an+j bi = 0.

(3.59)

Proof. We shall follow the proof in [6] by M.O. Drîmbe. The following identity may be obtained by direct computation !2 " n #2 2n 2n 2n X X X X (3.60) a2i b2i − ai b i − (ai bn+i − an+i bi ) i=1

i=1

i=1

=

X

i=1

(ai bj − aj bi − an+i bn+j + an+j bn+i )2

1≤i<j≤n

+

X

(ai bn+j − aj bn+i + an+i bj − an+j bi )2 .

1≤i<j≤n

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

42

S.S. D RAGOMIR

It is obvious that (3.57) is a simple consequence of the identity (3.60). The case of equality is also obvious.  Remark 3.28. For other similar (CBS) −type inequalties see the survey paper [7]. An anal¯ having 4n terms each may be found in [7, p. ogous inequality to (3.57) for sequences ¯ a and b 70]. 3.9. A Refinement due to Daykin-Eliezer-Carlitz. We will present now the version due to Mitrinovi´c, Peˇcari´c and Fink [5, p. 87] of Daykin-Eliezer-Carlitz’s refinement of the discrete (CBS) −inequality [8]. ¯ = (b1 , . . . , bn ) be two sequences of positive numTheorem 3.29. Let ¯ a = (a1 , . . . , an ) and b bers. The ineuality !2 n n n n n X X X X X 2 ai bi ≤ f (ai bi ) g (ai bi ) ≤ ai b2i (3.61) i=1

i=1

i=1

i=1

i=1

holds if and only if f (a, b) g (a, b) = a2 b2 ,

(3.62)

f (ka, kb) = k 2 f (a, b) ,

(3.63)

bf (a, 1) af (b, 1) a b + ≤ + af (b, 1) bf (a, 1) b a

(3.64) for any a, b, k > 0.

Proof. We shall follow the proof in [5, p. 88 – p. 89]. Necessity. Indeed, for n = 1, the inequality (3.61) becomes (ab)2 ≤ f (a, b) g (a, b) ≤ a2 b2 ,

a, b > 0

which gives the condition (3.62). For n = 2 in (3.61), using (3.62), we get 2a1 b1 a2 b2 ≤ f (a1 , b1 ) g (a2 , b2 ) + f (a2 , b2 ) g (a1 , b1 ) ≤ a21 b22 + a22 b21 . By eliminating g, we get (3.65)

2≤

f (a1 , b1 ) a2 b2 f (a2 , b2 ) a1 b1 a1 b 2 a2 b 1 · + · ≤ + . f (a2 , b2 ) a1 b1 f (a1 , b1 ) a2 b2 a2 b 1 a1 b 2

By substituting in (3.65) a, b for a1 , b1 and ka, kb for a2 , b2 (k > 0), we get 2≤

f (a, b) 2 f (ka, kb) −2 k + k ≤2 f (ka, kb) f (a, b)

and this is valid only if k 2 f (a, b) (f (ka, kb)) = 1, i.e., the condition (3.63) holds. Using (3.65), for a1 = a, b1 = b, a2 = b, b2 = 1, we have (3.66)

2≤

f (a,1) a f (b,1) b

+

f (b,1) b f (a,1) a

≤

a b + . b a

The first inequality in (3.66) is always satisfied while the second inequality is equivalent to (3.64).

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

43

Sufficiency. Suppose that (3.62) holds. Then inequality (3.61) can be written in the form 2

X

X

ai b i aj b j ≤

1≤i<j≤n

[f (ai , bi ) g (aj , bj ) + f (aj , bj ) g (ai , bi )]

1≤i<j≤n

X

≤


a2i b2j + a2j b2i .

1≤i<j≤n

Therefore, it is enough to prove (3.67)

2ai bi aj bj ≤ f (ai , bi ) g (aj , bj ) + f (aj , bj ) g (ai , bi ) ≤ a2i b2j + a2j b2i .

Suppose that (3.64) holds. Then (3.66) holds and putting a = (3.63), we get 2≤

ai , bi

b=

aj bj

in (3.66) and using

f (ai , bi ) aj bj f (aj , bj ) ai bi ai b j aj b i · + · ≤ + . f (aj , bj ) ai bi f (ai , bi ) aj bj aj b i ai b j

Multiplying the last inequality by ai bi aj bj and using (3.62), we obtain (3.67).



Remark 3.30. In [8] (see [5, p. 89]) the condition (3.64) is given as (3.68)

f (b, 1) ≤ f (a, 1) ,

f (a, 1) f (b, 1) ≤ for a ≥ b > 0. 2 a b2

Remark 3.31. O.E. Daykin, C.J. Eliezer and C. Carlitz [8] stated that examples for f, g satisfying (3.62) – (3.64) were obtained in the literature. The choice f (x, y) = x2 + y 2 , g (x, y) = x2 y 2 will give the Milne’s inequality x2 +y 2

(3.69)

n X

!2 ≤

ai b i

i=1

n X i=1

a2i

+

b2i

n n n X X
X a2i b2i 2 · ≤ ai · b2i . 2 2 a + b i i=1 i i=1 i=1

For a different proof of this fact, see Section 3.5. The choice f (x, y) = x1+α y 1−α , g (x, y) = x1−α y 1+α (α ∈ [0, 1]) will give the Callebaut inequality

(3.70)

n X i=1

!2 ai b i

≤

n X i=1

a1+α b1−α i i

n X

a1−α b1+α ≤ i i

i=1

n X i=1

a2i ·

n X

b2i .

i=1

3.10. A Refinement via Dunkl-Williams’ Inequality. We will use the following version of Dunkl-Williams’ inequality established in 1964 in inner product spaces [9]. Lemma 3.32. Let a, b be two non-null complex numbers. Then (3.71)

a 1 b |a − b| ≥ (|a| + |b|) − . 2 |a| |b|

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

44

S.S. D RAGOMIR

Proof. We start with the identity (see also [5, pp. 515 – 516]) 2   ¯b  a a a ¯ b b − = − − |a| |b| |a| |b| |a| |b|   a ¯b · = 2 − 2 Re |a| |b|

1 2 |a| |b| − 2 Re a · ¯b = |a| |b|
 1  = 2 |a| |b| − |a|2 + |b|2 − |a − b|2 |a| |b|  1  = |a − b|2 − (|a| − |b|)2 . |a| |b| Hence 2 2 a  1 b (|a| − |b|)2  |a − b| − (|a| + |b|) − = (|a| + |b|)2 − |a − b|2 2 |a| |b| 4 |a| |b| 2





and (3.71) is proved.

Using the above result, we may prove the following refinement of the (CBS) −inequality for complex numbers. ¯ = (b1 , . . . , bn ) are two sequences of nonzero complex Theorem 3.33. If ¯ a = (a1 , . . . , an ), b numbers, then 2 n n n X X X 2 2 |bk | − ak b k (3.72) |ak | k=1 k=1 k=1 2 n |bi | |aj | |ai | |bj | 1 X ≥ a ¯ i bj − a ¯ j bi + a ¯i · bj − bi · a ¯j ≥ 0. 8 i,j=1 |ai | |bj | |bi | |aj | Proof. The inequality (3.71) is clearly equivalent to 2 1 |b| |a| 2 (3.73) |a − b| ≥ a − b + ·a− ·b 4 |a| |b| for any a, b ∈ C, a, b 6= 0. We know the Lagrange’s identity for sequences of complex numbers n 2 n n n X X X 1X 2 2 (3.74) |ak | |bk | − ak b k = |¯ ai bj − a ¯j bi |2 . 2 i,j=1 k=1 k=1 k=1 By (3.73), we have 2 1 |aj | |bi | |ai | |bj | |¯ ai b j − a ¯ j bi | ≥ a ¯ i bj − a ¯ j bi + a ¯ i bj − a ¯j bi . 4 |ai | |bj | |aj | |bi | 2

Summing over i, j from 1 to n and using the (CBS) −inequality for double sums, we deduce (3.72). 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

45

3.11. Some Refinements due to Alzer and Zheng. In 1992, H. Alzer [10] presented the following refinement of the Cauchy-Schwarz inequality written in the form !2 n n n X X X (3.75) xk yk ≤ yk x2k yk . k=1

k=1

k=1

Theorem 3.34. Let xk and yk (k = 1, . . . , n) be real numbers satisfying 0 = x0 < x1 ≤ · · · ≤ xnn and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 . Then !2  n n n  X X X 1 2 (3.76) xk yk ≤ yk xk − xk−1 xk yk , 4 k=1 k=1 k=1

x2 2

≤

with equality holding if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn . In 1998, Liu Zheng [11] pointed out an error in the proof given in [10], which can be corrected as shown in [11]. Moreover, Liu Zheng established the following result which sharpens (3.76). Theorem 3.35. Let xk and yk (k = 1, . . . , n) be real numbers satisfying 0 < x1 ≤ xn and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 . Then n !2 n n n X X X (3.77) xk yk ≤ yk δk yk , k=1

k=1

x2 2

≤ ··· ≤

k=1

with (3.78)

δ1 = x21 and δk =

7k + 1 2 k xk − x2 8k 8 (k − 1) k−1

(k ≥ 2) .

Equality holds in (3.77) if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn . In 1999, H. Alzer improved the above results as follows. To present his results, we will follow [12]. In order to prove the main result, we need some technical lemmas. Lemma 3.36. Let xk (k = 1, . . . , n) be real numbers such that x2 xn 0 < x1 ≤ ≤ ··· ≤ . 2 n Then n X (3.79) 2 xk ≤ (n + 1) xn , k=1

with equality holding if and only if xk = kx1 (k = 1, . . . , n) . A proof of Lemma 3.36 is given in [10]. Lemma 3.37. Let xk (k = 1, . . . , n) be real numbers such that x2 xn 0 < x1 ≤ ≤ ··· ≤ . 2 n Then !2 n n X X 3k + 1 2 (3.80) xk ≤n xk , 4k k=1 k=1 with equality holding if and only if xk = kx1 (k = 1, . . . , n) .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

46

S.S. D RAGOMIR

Proof. Let Sn = Sn (x1 , . . . , xn ) = n

n X 3k + 1 k=1

4k

n X

x2k −

!2 .

xk

k=1

Then we have for n ≥ 2 : (3.81)

Sn − Sn−1 =

n−1 X 3k + 1 k=1

4k

x2k

− 2xn

n−1 X

xk +

k=1

3 (n − 1) 2 xn 4

= f (xn ) , say. We differentiate with respect to xn and use (3.79) and xn ≥

n x . n−1 n−1

This yields

n−1 X 3 (n − 1) n f (xn ) = xn − 2 xk ≥ xn−1 > 0 2 2 k=1 0

and  (3.82)

f (xn ) ≥ f =

n xn−1 n−1

n X 3k + 1 k=1

4k

x2k



n−1 X 2n 3n2 − xn−1 xk + x2n−1 n−1 4 (n − 1) k=1

= Tn−1 (x1 , . . . , xn−1 ) , say. We use induction on n to establish that Tn−1 (x1 , . . . , xn−1 ) ≥ 0 for n ≥ 2. We have T1 (x1 ) = 0. Let n ≥ 3; applying (3.79) we obtain n−1

∂ 3n + 2 2n X Tn−1 (x1 , . . . , xn−1 ) = xn−1 − xk ∂xn−1 2 n − 1 k=1 ≥

(n − 2) (n + 1) xn−1 > 0 2 (n − 1)

and  (3.83)

Tn−1 (x1 , . . . , xn−1 ) ≥ Tn−1

 n−1 x1 , . . . , xn−2 , xn−2 . n−2

Using the induction hypothesis Tn−2 (x1 , . . . , xn−2 ) ≥ 0 and (3.79), we get " #   n−2 X n−1 xn−2 (3.84) Tn−1 x1 , . . . , xn−2 , xn−2 ≥ (n − 1) xn−2 − 2 xk . n−2 n−2 k=1 From (3.83) and (3.84) we conclude Tn−1 (x1 , . . . , xn−1 ) ≥ 0 for n ≥ 2, so that (3.81) and (3.82) imply (3.85)

Sn ≥ Sn−1 ≥ · · · ≥ S2 ≥ S1 = 0.

This proves inequality (3.80). We discuss the cases of equality. A simple calculation reveals that Sn (x1 , 2x1 , . . . , nx1 ) = 0. We use induction on n to prove the implication (3.86)

Sn (x1 , . . . , xn ) = 0 =⇒ xk = kx1 for k = 1, . . . , n.

If n = 1, then (3.86) is obviously true. Next, we assume that (3.86) holds with n − 1 instead of n. Let n ≥ 2 and Sn (x1 , . . . , xn ) = 0. Then (3.85) leads to Sn−1 (x1 , . . . , xn−1 ) = 0 which implies xk = kx1 for k = 1, . . . , n − 1. Thus, we have Sn (x1 , 2x1 , . . . , (n − 1) x1 , xn ) = 0 which is equivalent to (xn − nx1 ) (3xn − nx1 ) = 0. Since 3xn > nx1 , we get xn = nx1 . 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

47

Lemma 3.38. Let xk (k = 1, . . . , n) be real numbers such that 0 < x1 ≤

x2 xn ≤ ··· ≤ . 2 n

If the natural numbers n and q satisfy n ≥ q + 1, then !2 q q X X (3n + 1) q 2 2 xk − 2qxn xk + (3.87) 0< xn . 4n k=1 k=1 Proof. We denote the expression on the right-hand  side of (3.87) by u (xn ) . Then we differentiate with respect to xn and apply (3.79), xn ≥ nq xq and n ≥ q + 1. This yields q X 1 0 (3n + 1) q u (xn ) = xn − 2q xk q 2n k=1

(3n + 1) q xn − (q + 1) xq 2n 3n − 2q − 1 ≥ xq > 0. 2 ≥

Hence, we get  (3.88)

u (xn ) ≥ u

n xq q



q

X (3n + 1) n 2 = xq − 2nxq xk + 4 k=1

q X

!2 xk

.

k=1

Let q X (3t + 1) t v (t) = xq − 2t xk and t ≥ q + 1; 4 k=1

from (3.79) we conclude that v 0 (t) =

q X 6t + 1 2q + 3 xq − 2 xk ≥ xq > 0. 4 4 k=1

This implies that the expression on the right-hand side of (3.88) is increasing on [q + 1, ∞) with respect to n. Since n ≥ q + 1, we get from (3.88): !2 q q X X (3q + 4) (q + 1) 2 (3.89) u (xn ) ≥ xq − 2 (q + 1) xq xk + xk 4 k=1 k=1 = Pq (x1 , . . . , xq ) , say. We use induction on q to show that Pq (x1 , . . . , xq ) > 0 for q ≥ 1. We have P1 (x1 ) = 21 x21 . If Pq−1 (x1 , . . . , xq−1 ) > 0, then we obtain for q ≥ 2 : (3.90) Pq (x1 , . . . , xq ) > 2q (xq−1 − xq )

q−1 X k=1

xk −

(3q + 1) q 2 xq−1 4 +

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

q (3q − 1) 2 xq = w (xq ) , say. 4

http://jipam.vu.edu.au/

48

S.S. D RAGOMIR

  q xq−1 . Then we get We differentiate with respect to xq and use (3.79) and xq ≥ q−1 " # q−1 X 3q − 1 w0 (xq ) = q xq − 2 xk 2 k=1 q 2 (q + 1) ≥ xq−1 2 (q − 1) >0 and 

(3.91)

 q w (xq ) ≥ w xq−1 q−1 " # q−1 X 4q 2 − q − 1 q = xq−1 xq−1 − 2 xk q−1 4 (q − 1) k=1 (3q − 1) q 2 x 4 (q − 1) q−1 > 0. ≥

From (3.89), (3.90) and (3.91), we obtain u (xn ) > 0.



We are now in a position to prove the following companion of inequalities (3.76) and (3.77) (see [12]). Theorem 3.39. The inequality (3.92)

n X

!2 xk yk

≤

k=1

n X k=1

yk

n  X k=1

β α+ k

 yk ,

holds for all natural numbers n and for all real numbers xk and yk (k = 1, . . . , n) with x2 xn (3.93) 0 < x1 ≤ ≤ ··· ≤ and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 , 2 n if and only if 3 α≥ and β ≥ 1 − α. 4 Proof. First, we assume that (3.92) is valid for all n ≥ 1 and for all real numbers xk and yk (k = 1, . . . , n) which satisfy (3.93). We set xk = k and yk = 1 (k = 1, . . . , n). Then (3.92) leads to   3 3 (3.94) 0≤ α− 2n + α + 3β − (n ≥ 1) . 4 2 This implies α ≥ 34 . And, (3.94) with n = 1 yields α + β ≥ 1. Now, we suppose that α ≥ 34 and β ≥ 1 − α. Then we obtain for k ≥ 1 : β 1−α 3 1 ≥α+ ≥ + , k k 4 4k 3 so that is suffices to show that (3.92) holds with α = 4 and β = 14 . Let α+

F (y1 , . . . , yn ) =

n X k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

yk

n X 3k + 1 k=1

4k

x2k yk −

n X

!2 xk yk

k=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

49

and (1 ≤ q ≤ n − 1) .

Fq (y) = F (y, . . . , y, yq+1 , . . . , yn )

We shall prove that Fq is strictly increasing on [yq+1 , ∞). Since yq+1 ≤ yq , we obtain Fq (yq ) ≥ Fq (yq+1 ) = Fq+1 (yq+1 )

(3.95)

(1 ≤ q ≤ n − 1) ,

and Lemma 3.37 imply F (y1 , . . . , yn ) = F1 (y1 ) ≥ F1 (y2 ) = F2 (y2 ) ≥ F2 (y3 ) ≥ · · · ≥ Fn−1 (yn−1 ) ≥ Fn−1 (y)  !2  n n X X 3k + 1 2 xk − xk  ≥ 0. = yn2 n 4k k=1 k=1 If F (y1 , . . . , yn ) = 0, then we conclude from the strict monotonicity of Fq and from Lemma 3.37 that y1 = · · · = yn and xk = kx1 (k = 1, . . . , n) . It remains to show that Fq is strictly increasing on [yq+1 , ∞). Let y ≥ yq+1 ; we differentiate Fq and apply Lemma 3.37. This yields  !2  q q n X X X 3k + 1 2 3k + 1 2 0   xk − xk +q xk yk Fq (y) = 2y q 4k 4k k=1 k=q+1 k=1 n X

+

yk

q X 3k + 1

k=q+1

k=1

q X 1 00 3k + 1 2 Fq (y) = q xk − 2 4k k=1

q X

and

4k

x2k

−2

n X k=q+1

xk yk

q X

xk

k=1

!2 xk

≥ 0.

k=1

Hence, we have (3.96)

Fq0 (y) ≥ Fq0 (yq+1 ) =

! q !2  q X 3k + 1 X 1 2qyq+1 + yk  x2k − xk  4k q k=q+1 k=1 k=1  !2  q q n   2 X X X 1 (3k + 1) q 2 + yk xk − 2qxk xi + xi .  q k=q+1  4k i=1 i=1 n X

From Lemma 3.37 and Lemma 3.38 we obtain Fq0 (yq+1 ) > 0, so that (3.96) implies Fq0 (y) > 0 for y ≥ yq+1 . This completes the proof of the theorem.  Remark 3.40. The proof of the theorem reveals that the sign of equality holds in (3.92) (with α = 34 and β = 14 ) if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn . Remark 3.41. If δk is defined by (3.78), then we have for k ≥ 2 : "    2 #   2 3 1 k (k − 1) xk xk−1 δk − + x2k = − , 4 4k 8 k k−1 which implies that inequality (3.92) (with α =

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

3 4

and β = 14 ) sharpens (3.77).

http://jipam.vu.edu.au/

50

S.S. D RAGOMIR

Remark 3.42. It is
shown in [10] that if a sequence (xk ) satisfies x0 = 0 and 2xk ≤ xk−1 +xk+1 (k ≥ 1) , then xkk is increasing. Hence, inequality (3.92) is valid for all sequences (xk ) which are positive and convex. R EFERENCES [1] S.S. DRAGOMIR AND N.M. IONESCU, Some refinements of Cauchy-Buniakowski-Schwartz inequality for sequences, Proc. of the Third Symp. of Math. and Its Appl., 3-4 Nov. 1989, Polytechnical Institute of Timi¸soara, Romania, 78–82. [2] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s inequality for real numbers (Romanian), “Caiete Metodico Stiin¸ ¸ tifice”, No. 57, 1989, pp. 24. Faculty of Mathematics, Timi¸soara University, Romania. ´ On Cauchy-Buniakowski[3] S.S. DRAGOMIR, Š.Z. ARSLANAGIC´ AND D.M. MILOŠEVIC, Schwartz’s inequality for real numbers, Mat. Bilten, 18 (1994), 57–62. [4] N.G. de BRUIJN, Problem 12, Wisk. Opgaven, 21 (1960), 12–14. ˘ ´ J.E. PECARI [5] D.S. MITRINOVIC, C´ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993. [6] M.O. DRÎMBE, A generalisation of Lagrange’s identity (Romanian), Gaz. Mat. (Bucharest), 89 (1984), 139-141. [7] H.W. McLAUGHLIN, Inequalities complementary to the Cauchy-Schwartz inequality for finite sums of quaternions, refinements of the Cauchy-Schwartz inequality for finite sums of real numbers; inequalities concerning ratios and differences of generalised means of different order, Maryland University, Technical Note BN-454, (1966), 129 pp. [8] D.E. DAYKIN, C.J. ELIEZER AND C. CARLITZ, Problem 5563, Amer. Math. Monthly, 75 (1968), p. 198 and 76 (1969), 98–100. [9] C.F. DUNKL AND K.S. WILLIAMS, A simple inequality, Amer. Math. Monthly, 71 (1964), 53–54. [10] H. ALZER, A refinement of the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 168 (1992), 596–604. [11] L. ZHENG, Remark on a refinement of the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 218 (1998), 13–21. [12] H. ALZER, On the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 234 (1999), 6–14.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

51

4. F UNCTIONAL P ROPERTIES 4.1. A Monotonicity Property. The following result was obtained in [1, Theorem]. ¯ = (b1 , . . . , bn ) be sequences of real numbers and p Theorem 4.1. Let ¯ a = (a1 , . . . , an ), b ¯= (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) be sequences of nonnegative real numbers such that pk ≥ qk for any k ∈ {1, . . . , n} . Then one has the inequality

(4.1)

n X

! 21

n X

pi a2i

i=1

! 12 pi b2i

i=1

n X − p i ai b i i=1

n X

≥

! 12

n X

qi a2i

! 12 qi b2i

i=1

i=1

n X − qi ai bi ≥ 0. i=1

Proof. We shall follow the proof in [1]. Since pk −qk ≥ 0, then the (CBS) −inequality for the weights rk = pk −qk (k ∈ {1, . . . , n}) will produce ! n ! " n #2 n n n X X X X X (4.2) pk a2k − qk a2k pk b2k − qk b2k ≥ (pk −qk ) ak bk . k=1

k=1

k=1

k=1

k=1

Using the elementary inequality (ac − bd)2 ≥ a2 − b2





c2 − d2 ,

a, b, c, d ∈ R

for the choices a=

n X

! 21 pk a2k

,

b=

d=

! 12 qk a2k

,

c=

k=1

k=1 n X

n X

n X

! 12 pk b2k

and

k=1

! 21 qk b2k

k=1

we deduce by (4.2), that n X k=1

! 12 pk a2k

n X k=1

! 12 pk b2k

−

n X

! 12 qk a2k

n X

! 21 qk b2k

k=1 k=1 n n n n X X X X ≥ p k ak b k − q k ak b k ≥ p k ak b k − q k ak b k k=1

k=1

k=1

k=1



proving the desired inequality (4.1). The following corollary holds [1, Corollary 1]. ¯ be as in Theorem 4.1. Denote Corollary 4.2. Let ¯ a and b Sn (1) := {¯ x = (x1 , . . . , xn ) |0 ≤ xi ≤ 1, i ∈ {1, . . . , n}} .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

52

S.S. D RAGOMIR

Then

(4.3)

n X

! 12 a2i

i=1

n X

! 12 b2i

i=1

n X − ai b i i=1  = sup 

n X

x ¯∈Sn (1)

! 12 xi a2i

i=1

n X

! 12 xi b2i

i=1

 n X − xi ai bi  ≥ 0. i=1

Remark 4.3. The following inequality is a natural particular case that may be obtained from (4.1) [1, p. 79]

(4.4)

n X i=1

! 12 a2i

n X

! 12 b2i

i=1

" ≥

n X

n X − ai b i i=1 # 12 "

a2i trig 2 (αi )

i=1

n X

# 12 b2i trig 2 (αi )

i=1

n X − ai bi trig 2 (αi ) ≥ 0, i=1

where trig (x) = sin x or cos x, x ∈ R and α ¯ = (α1 , . . . , αn ) is a sequence of real numbers. 4.2. A Superadditivity Property in Terms of Weights. Let Pf (N) be the family of finite parts of the set of natural numbers N, S (K) the linear space of real or complex numbers, i.e.,  S (K) := x|x = (xi )i∈N , xi ∈ K, i ∈ N and S+ (R) the family of nonnegative real sequences. Define the mapping ! 12 (4.5)

S (p, I, x, y) :=

X

X pi |xi |2 pi |yi |2

i∈I

i∈I

X − pi xi y¯i , i∈I

where p ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) . The following superadditivity property in terms of weights holds [2, p. 16]. Theorem 4.4. For any p, q ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) we have (4.6)

S (p + q, I, x, y) ≥ S (p, I, x, y) + S (q, I, x, y) ≥ 0.

Proof. Using the (CBS) −inequality for real numbers (4.7)

a2 + b2


12

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

c2 + d2


12

≥ ac + bd; a, b, c, d ≥ 0,

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

53

we have ! 12 S (p, I, x, y) =

X

pi |xi |2 +

i∈I

≥

X

! 12 X

qi |xi |2

i∈I

i∈I

X X − pi xi y¯i + qi xi y¯i i∈I i∈I ! 12 ! 12 X X 2 2 pi |yi | + pi |xi | i∈I

i∈I

pi |yi |2 +

X

qi |yi |2

i∈I

! 12 X

! 12 X

2

qi |xi |

i∈I

2

qi |yi |

i∈I

X X − pi xi y¯i − qi xi y¯i i∈I

i∈I

= S (p, I, x, y) + S (q, I, x, y) , 

and the inequality (4.6) is proved. The following corollary concerning the monotonicity of S (·, I, x, y) also holds [2, p. 16].

Corollary 4.5. For any p, q ∈ S+ (R) with p ≥ q and I ∈ Pf (N) , x, y ∈ S (K) one has the inequality: S (p, I, x, y) ≥ S (q, I, x, y) ≥ 0.

(4.8)

Proof. Using Theorem 4.4, we have S (p, I, x, y) = S ((p − q) + q, I, x, y) ≥ S (p − q, I, x, y) + S (q, I, x, y) giving S (p, I, x, y) − S (q, I, x, y) ≥ S (p − q, I, x, y) ≥ 0 

and the inequality (4.8) is proved. Remark 4.6. The following inequalities follow by the above results [2, p. 17].

(1) Let αi ∈ R (i ∈ {1, . . . , n}) and xi , yi ∈ K (i ∈ {1, . . . , n}) . Then one has the inequality:

(4.9)

n X

n X 2 |xi | |yi |2

i=1

i=1

≥

! 12

n X − xi y¯i i=1

n X i=1

! 12

n X 2 2 2 − xi y¯i sin αi |xi | sin αi |yi | sin αi i=1 i=1 ! 12 n n n X X X 2 2 − xi y¯i cos2 αi ≥ 0. + |xi | cos2 αi |yi | cos2 αi n X

2

i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

2

i=1

i=1

http://jipam.vu.edu.au/

54

S.S. D RAGOMIR

(2) Denote Sn (1) := {p ∈ S+ (R) |pi ≤ 1 for all i ∈ {1, . . . , n}} . Then for all x, y ∈ S (K) one has the bound (see also Corollary 4.2): ! 12 n n n X X X 2 2 (4.10) 0≤ |xi | |yi | − xi y¯i i=1 i=1 i=1   ! 12 n n n  X  X X 2 2 pi |xi | − pi xi y¯i . = sup pi |yi |  p∈Sn (1)  i=1

i=1

i=1

4.3. The Superadditivity as an Index Set Mapping. We assume that we are under the hypothesis and notations in Section 4.2. Reconsider the functional S (·, ·, ·, ·) : S+ (R) × Pf (N) × S (K) × S (K) → R, ! 12 X X X (4.11) S (p, I, x, y) := pi |xi |2 pi |yi |2 − pi xi y¯i . i∈I

i∈I

i∈I

The following superadditivity property as an index set mapping holds [2]. Theorem 4.7. For any I, J ∈ Pf (N) \ {∅} with I ∩ J = ∅, one has the inequality (4.12)

S (p, I ∪ J, x, y) ≥ S (p, I, x, y) + S (p, J, x, y) ≥ 0.

Proof. Using the elementary inequality for real numbers
1
1 (4.13) a2 + b2 2 c2 + d2 2 ≥ ac + bd; a, b, c, d ≥ 0, we have ! 12 S (p, I ∪ J, x, y) =

X

pi |xi |2 +

i∈I

≥

X

pj |xj |2

j∈J

! 12 X i∈I

X X − pi xi y¯i + pj xj y¯j i∈I j∈J ! 12 ! 12 X X pi |yi |2 + pi |xi |2 i∈I

pi |yi |2 +

i∈I

X

pj |yj |2

j∈J

! 12 X

pj |xj |2

j∈J

! 12 X

pj |yj |2

j∈J

X X − pi xi y¯i − pj xj y¯j i∈I

j∈J

= S (p, I, x, y) + S (p, J, x, y) 

and the inequality (4.12) is proved.

The following corollary concerning the monotonicity of S (p, ·, x, y) as an index set mapping also holds [2, p. 16]. Corollary 4.8. For any I, J ∈ Pf (N) with I ⊇ J 6= ∅, one has (4.14)

S (p, I, x, y) ≥ S (p, J, x, y) ≥ 0.

Proof. Using Theorem 4.7, we may write S (p, I, x, y) = S (p, (I\J) ∪ J, x, y) ≥ S (p, I\J, x, y) + S (p, J, x, y)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

55

giving S (p, I, x, y) − S (p, J, x, y) ≥ S (p, I\J, x, y) ≥ 0 

which proves the desired inequality (4.14). Remark 4.9. The following inequalities follow by the above results [2, p. 17].

(1) Let pi ≥ 0 (i ∈ {1, . . . , 2n}) and xi , yi ∈ K (i ∈ {1, . . . , 2n}) . Then we have the inequality ! 21 2n 2n 2n X X X (4.15) pi |xi |2 pi |yi |2 − pi xi y¯i i=1 i=1 i=1 ! 12 n n n X X X ≥ p2i |x2i |2 p2i |y2i |2 − p2i xi y¯i i=1 i=1 i=1 ! 12 n n n X X X + p2i−1 |x2i−1 |2 p2i−1 |y2i−1 |2 − p2i−1 x2i−1 y¯2i−1 i=1

i=1

i=1

≥ 0. (2) We have the bound (4.16)

! 21

n X − pi xi y¯i i=1   ! 12 X X X = sup  pi |xi |2 pi |yi |2 − pi xi y¯i  ≥ 0. I⊆{1,...,n}

n X

n X pi |xi |2 pi |yi |2

i=1

i=1

i∈I

I6=∅

i∈I

i∈I

(3) Define the sequence (4.17)

Sn :=

n X

2

pi |xi |

i=1

n X

! 21 2

pi |yi |

i=1

n X − pi xi y¯i ≥ 0 i=1

where p = (pi )i∈N ∈ S+ (R) , x = (xi )i∈N , y = (yi )i∈N ∈ S (K) . Then Sn is monotontic nondecreasing and we have the following lower bound n
1
1 (4.18) Sn ≥ max pi |xi |2 + pj |xj |2 2 pi |yi |2 + pj |yj |2 2 1≤i,j≤n  − |pi xi y¯i + pj xj y¯j | ≥ 0. 4.4. Strong Superadditivity in Terms of Weights. With the notations in Section 4.2, define the mapping 2 X X X (4.19) S¯ (p, I, x, y) := pi |xi |2 pi |yi |2 − pi xi y¯i , i∈I

i∈I

i∈I

where p ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

56

S.S. D RAGOMIR

Denote also by k·k`,H the weighted Euclidean norm ! 12 X (4.20) kxk`,H := `i |xi |2 , ` ∈ S+ (R) , H ∈ Pf (N) . i∈H

The following strong superadditivity property in terms of weights holds [2, p. 18]. Theorem 4.10. For any p, q ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) we have S¯ (p + q, I, x, y) − S¯ (p, I, x, y) − S¯ (q, I, x, y) 2   kxkp,I kykp,I  ≥ 0. ≥ det  kxkq,I kykq,I

(4.21)

Proof. We have ! (4.22)

X

S¯ (p + q, I, x, y) =

2

pi |xi | +

i∈I

X

! X

2

qi |xi |

i∈I

2

pi |yi | +

i∈I

X

2

qi |yi |

i∈I

2 X X − pi xi y¯i + qi xi y¯i i∈I i∈I X X X X ≥ pi |xi |2 pi |yi |2 + qi |xi |2 qi |yi |2 i∈I

i∈I

i∈I

i∈I

!2 X X − pi xi y¯i + qi xi y¯i i∈I i∈I X X = S¯ (p, I, x, y) + S¯ (q, I, x, y) + pi |xi |2 qi |yi |2 i∈I

i∈I

X X X X + qi |xi |2 pi |yi |2 − 2 pi xi y¯i qi xi y¯i . i∈I

i∈I

i∈I

i∈I

By (CBS) −inequality, we have " # 12 X X X X X X pi xi y¯i qi xi y¯i ≤ pi |xi |2 pi |yi |2 qi |xi |2 qi |yi |2 i∈I

i∈I

i∈I

i∈I

i∈I

i∈I

and thus (4.23)

X i∈I

pi |xi |2

X i∈I

qi |yi |2 +

X i∈I

qi |xi |2

X

pi |yi |2

i∈I

 ! 21 X X  X 2 − 2 pi xi y¯i qi xi y¯i ≥ pi |xi | i∈I

i∈I

i∈I

! 12 X

2

qi |yi |

i∈I

! 12 2

! 12 −

X

qi |xi |2

i∈I

X

pi |yi |2

 .

i∈I



Utilising (4.22) and (4.23) we deduce the desired inequality (4.21). The following corollary concerning a strong monotonicity result also holds [2, p. 18].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

57

Corollary 4.11. For any p, q ∈ S+ (R) with p ≥ q one has the inequality: 



S¯ (p, I, x, y) − S¯ (q, I, x, y) ≥ det 

(4.24)

kxkq,I

kykq,I

2  ≥ 0.

kxkp−q,I

kykp−q,I

Remark 4.12. The following refinement of the (CBS) −inequality is a natural consequence of (4.21) [2, p. 19]

(4.25)

X i∈I

2 X X 2 2 |xi | |yi | − xi y¯i i∈I

i∈I

2 X |xi |2 sin2 αi |yi |2 sin2 αi − xi y¯i sin2 αi ≥ i∈I i∈I i∈I 2 X X X + |xi |2 cos 2 αi |yi |2 cos 2 αi − xi y¯i cos 2 αi i∈I i∈I i∈I     21   12 P P 2 2 2 2 |xi | sin αi |yi | sin αi     i∈I i∈I   + det      12   12   P P 2 2 |xi | cos 2 αi |yi | cos2 αi X

X

i∈I

2     ≥ 0.  

i∈I

where αi ∈ R, i ∈ I. 4.5. Strong Superadditivity as an Index Set Mapping. We assume that we are under the hypothesis and notations in Section 4.2. Reconsider the functional S¯ (·, ·, ·, ·) : S+ (R) × Pf (N) × S (K) × S (K) → R,

(4.26)

S¯ (p, I, x, y) :=

X i∈I

2

pi |xi |

X i∈I

2 X pi |yi | − pi xi y¯i . 2

i∈I

The following strong supperadditivity property as an index set mapping holds [2, p. 18]. Theorem 4.13. For any p ∈ S+ (R) , I, J ∈ Pf (N) \ {∅} with I ∩ J = ∅ and x, y ∈ S (K) , we have (4.27)

S¯ (p, I ∪ J, x, y) − S¯ (p, I, x, y) − S¯ (p, J, x, y)   2 kxkp,I kykp,I  ≥ 0. ≥ det  kxkp,J kykp,J

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

58

S.S. D RAGOMIR

Proof. We have S¯ (p, I ∪ J, x, y)

(4.28)

! =

X

2

pi |xi | +

i∈I

X

! X

2

pj |xj |

j∈J

2

pi |yi | +

i∈I

X

2

pj |yj |

j∈J

2 X X − pi xi y¯i + pj xj y¯j i∈I j∈I X X X X ≥ pi |xi |2 pi |yi |2 + pj |xj |2 pj |yj |2 i∈I

i∈I

+

X

j∈J 2

pi |xi |

i∈I

X

2

j∈J

pj |yj | +

j∈J

X

pi |yi |2

i∈I

X

pj |xj |2

j∈J

!2 X X − pi xi y¯i + pj xj y¯j i∈I j∈I X X pi |xi |2 pj |yj |2 = S¯ (p, I, x, y) + S¯ (p, J, x, y) + i∈I

j∈J

X X X X 2 2 + pi |yi | pj |xj | − 2 pi xi y¯i pj xj y¯j . i∈I

j∈J

i∈I

j∈I

By the (CBS) −inequality, we have " # 21 X X X X X X 2 2 2 2 pi xi y¯i pj xj y¯j ≤ pi |xi | pi |yi | pj |xj | pj |yj | i∈I

j∈I

i∈I

i∈I

j∈J

j∈J

and thus (4.29)

X i∈I

pi |xi |2

X j∈J

pj |yj |2 +

X i∈I

pi |yi |2

X

pj |xj |2

j∈J

 ! 21 X X X − 2 pi xi y¯i pj xj y¯j ≥  pi |xi |2 i∈I

j∈I

i∈I

! 12 X

pj |yj |2

j∈J

! 12 2

! 12 −

X i∈I

pi |yi |2

X

pj |xj |2

 .

j∈J

If we use now (4.28) and (4.29), we may deduce the desired inequality (4.27).



The following corollary concerning strong monotonicity also holds [2, p. 18]. Corollary 4.14. For any I, J ∈ Pf (N) \ {∅} with I ⊇ J one has the inequality   2 kxkp,J kykp,J  ≥ 0. (4.30) S¯ (p, I, x, y) − S¯ (p, J, x, y) ≥ det  kxkp,I\J kykp,I\J Remark 4.15. The following refinement of the (CBS) −inequality is a natural consequence of (4.27) [2, p. 19].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

Suppose pi ≥ 0, i ∈ {1, . . . , 2n} and xi , yi ∈ K, i ∈ {1, . . . , 2n} . Then inequality 2 2n 2n 2n X X X 2 2 pi |xi | pi |yi | − pi xi y¯i (4.31) i=1 i=1 i=1 2 n n n X X X p2i |x2i |2 p2i |y2i |2 − ≥ p2i x2i y¯2i i=1 i=1 i=1 n 2 n n X X X 2 2 p2i−1 |x2i−1 | p2i−1 |y2i−1 | − p2i−1 x2i−1 y¯2i−1 + i=1 i=1 i=1    n  12  21 n P P 2 2 p2i |y2i | p |x |     i=1 2i 2i i=1   + det      21 n  12 n   P P 2 2 p2i−1 |x2i−1 | p2i−1 |y2i−1 | i=1

59

we have the

2     ≥ 0.  

i=1

4.6. Another Superadditivity Property. Let Pf (N) be the family of finite parts of the set of natural numbers, S (R) the linear space of real sequences and S+ (R) the family of nonnegative real sequences. Consider the mapping C : S+ (R) × Pf (N) × S (R) × S (R) → R !2 X X X
(4.32) C p, I, a, b := pi a2i pi b2i − p i ai b i . i∈I

i∈I

i∈I

The following identity holds [3, p. 115]. Lemma 4.16. For any p, q ∈ S+ (R) one has


(4.33) C p + q, I, a, b = C p, I, a, b + C q, I, a, b +

X

pi qj (ai bj − aj bi )2 .

(i,j)∈I×I

Proof. Using the well-known Lagrange’s identity, we have
1 X (4.34) C p, I, a, b = pi pj (ai bj − aj bi )2 . 2 (i,j)∈I×I

Thus
C p + q, I, a, b 1 X = (pi + qi ) (pj + qj ) (ai bj − aj bi )2 2 (i,j)∈I×I

=

1 2

X

pi pj (ai bj − aj bi )2 +

(i,j)∈I×I

+

1 2

X

1 2

X

qi qj (ai bj − aj bi )2

(i,j)∈I×I

1 X pj qi (ai bj − aj bi )2 2 (i,j)∈I×I X pi qj (ai bj − aj bi )2

pi qj (ai bj − aj bi )2 +

(i,j)∈I×I



= C p, I, a, b + C q, I, a, b +

(i,j)∈I×I

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

60

S.S. D RAGOMIR

since, by symmetry, X

X

pi qj (ai bj − aj bi )2 =

pj qi (ai bj − aj bi )2 .

(i,j)∈I×I

(i,j)∈I×I

 Consider the following mapping: !2  12

 X X

 1 D p, I, a, b := C p, I, a, b 2 =  pi a2i pi b2i − i∈I

X

i∈I

p i ai b i

 .

i∈I

The following result has been obtained in [4, p. 88] as a particular case of a more general result holding in inner product spaces. Theorem 4.17. For any p, q ∈ S+ (R) , I ∈ Pf (N) and a, b ∈ S (R) , we have the superadditive property


(4.35) D p + q, I, a, b ≥ D p, I, a, b + D q, I, a, b ≥ 0. Proof. We will give here an elementary proof following the one in [3, p. 116 – p. 117]. By Lemma 4.16, we obviously have X


(4.36) D2 p + q, I, a, b = D2 p, I, a, b + D2 q, I, a, b + pi qj (ai bj − aj bi )2 . (i,j)∈I×I

We claim that X

(4.37)



pi qj (ai bj − aj bi )2 ≥ 2D p, I, a, b D q, I, a, b .

(i,j)∈I×I

Taking the square in both sides of (4.37), we must prove that " #2 X X X X X X (4.38) pi a2i qi b2i + qi a2i pi b2i −2 p i ai b i q i ai b i i∈I

i∈I

i∈I

i∈I

i∈I

 X X ≥ 4 pi a2i pi b2i − i∈I

i∈I

!2  X

i∈I

p i ai b i



i∈I

 X X × qi a2i qi b2i − i∈I

i∈I

!2  X

q i ai b i

.

i∈I

Let us denote ! 12

! 21 a :=

X

pi a2i

,

x :=

X

qi a2i

! 12 ,

b :=

i∈I

i∈I

X

pi b2i

,

i∈I

! 21 y :=

X

qi b2i

,

i∈I

c :=

X i∈I

p i ai b i ,

z :=

X

q i ai b i .

i∈I

With these notations (4.38) may be written in the following form
2

(4.39) a2 y 2 + b2 x2 − 2cz ≥ 4 a2 b2 − c2 x2 y 2 − z 2 . Using the elementary inequality

m2 − n2 p2 − q 2 ≤ (mp − nq)2 , m, n, p, q ∈ R

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

61

we may state that 4 (abxy − cz)2 ≥ 4 a2 b2 − c2

(4.40)





x2 y 2 − z 2 ≥ 0.

Since, by the (CBS) −inequality, we observe that abxy ≥ |cz| ≥ |cz| , we can state that a2 y 2 + b2 x2 − 2cz ≥ 2 (abxy − cz) ≥ 0 giving a2 y 2 + b2 x2 − 2cz

(4.41)


2

≥ 4 (abxy − cz)2 .

Utilizing (4.40) and (4.41) we deduce the inequality (4.39), and (4.37) is proved. Finally, by (4.36) and (4.37) we have


2 D2 p + q, I, a, b ≥ D p, I, a, b + D q, I, a, b , 

i.e., the superadditivity property (4.35). Remark 4.18. The following refinement of the (CBS) − inequality holds [4, p. 89]

(4.42)

 X X  a2i b2i − i∈I

i∈I

!2  21 X

ai b i



i∈I

 X X ≥ a2i sin2 αi b2i sin2 αi − i∈I

i∈I

!2  12 X

ai bi sin2 αi

i∈I

 X X + a2i cos 2 αi b2i cos2 αi − i∈I



i∈I

!2  12 X

ai bi cos2 αi

 ≥0

i∈I

for any αi ∈ R, i ∈ {1, . . . , n} . 4.7. The Case of Index Set Mapping. Assume that we are under the hypothesis and notations in Section 4.6. Reconsider the functional C : S+ (R) × Pf (N) × S (R) × S (R) → R given by !2 X X X
(4.43) C p, I, a, b := pi a2i pi b2i − p i ai b i . i∈I

i∈I

i∈I

The following identity holds. Lemma 4.19. For any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ one has the identity: X


(4.44) C p, I ∪ J, a, b = C p, I, a, b + C p, J, a, b + pi pj (ai bj − aj bi )2 . (i,j)∈I×J

Proof. Using Lagrange’s identity [5, p. 84], we may state (4.45)


1 C p, K, a, b = 2

X

pi pj (ai bj − aj bi )2 ,

K ∈ Pf (N) \ {∅} .

(i,j)∈K×K

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

62

S.S. D RAGOMIR

Thus C p, I ∪ J, a, b X 1 = 2


pi pj (ai bj − aj bi )2

(i,j)∈(I∪J)×(I∪J)

=

1 2

X

pi pj (ai bj − aj bi )2 +

(i,j)∈I×I

+

1 2

X

1 2

X

pi pj (ai bj − aj bi )2

(i,j)∈I×J

1 X pi pj (ai bj − aj bi )2 2 (i,j)∈J×J X pi pj (ai bj − aj bi )2

pi pj (ai bj − aj bi )2 +

(i,j)∈J×I



= C p, I, a, b + C p, J, a, b +

(i,j)∈I×J

since, by symmetry, X

X

pi pj (ai bj − aj bi )2 =

(i,j)∈I×J

pi pj (ai bj − aj bi )2 .

(i,j)∈J×I

 Now, if we consider the mapping !2  21

 X X

 D p, I, a, b := C p, I, a, b = pi a2i pi b2i − 1 2

i∈I

X

i∈I

p i ai b i

 ,

i∈I

then the following superadditivity property as an index set mapping holds: Theorem 4.20. For any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ one has


(4.46) D p, I ∪ J, a, b ≥ D p, I, a, b + D p, J, a, b ≥ 0. Proof. By Lemma 4.19, we have


(4.47) D2 p, I ∪ J, a, b = D2 p, I, a, b + D2 p, J, a, b +

X

pi pj (ai bj − aj bi )2

(i,j)∈I×J

To prove (4.46) it is sufficient to show that X

(4.48) pi pj (ai bj − aj bi )2 ≥ 2D p, I, a, b D p, J, a, b . (i,j)∈I×J

Taking the square in (4.48), we must demonstrate that " #2 X X X X X X pi a2i pj b2j + pj a2j pi b2i − 2 p i ai b i p j aj b j i∈I

j∈J

j∈J

i∈I

i∈I

j∈J

 X X ≥ 4 pi a2i pi b2i − i∈I

i∈I

!2  X

p i ai b i



i∈I

 X X × pj a2j pj b2j − j∈J

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

j∈J

X

!2  p j aj b j  .

j∈J

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

63

If we denote ! 12 a :=

X

! 12

pi a2i

,

X

x :=

i∈I

pj a2j

! 12 ,

X

b :=

j∈J

pi b2i

,

i∈I

! 21 y :=

X

pj b2j

,

c :=

j∈J

X

p i ai b i ,

z :=

i∈I

X

p j aj b j ,

j∈J

then we need to prove a2 y 2 + b2 x2 − 2cz

(4.49)


2

≥ 4 a2 b2 − c2





x2 y 2 − z 2 ,

which has been shown in Section 4.6. This completes the proof.



Remark 4.21. The following refinement of the (CBS) −inequality holds  2n 2n X X 2  p i ai pi b2i − i=1

i=1

2n X

!2  21 p i ai b i



i=1

 n n X X ≥ p2i a22i p2i b22i − i=1

i=1

n X

!2  12 p2i a2i b2i

i=1

 n n X X + p2i−1 a22i−1 p2i−1 b22i−1 − i=1



i=1

n X

!2  12 p2i−1 a2i−1 b2i−1

 ≥ 0.

i=1

4.8. Supermultiplicity in Terms of Weights. Denote by S+ (R) the set of nonnegative sequences. Assume that A : S+ (R) → R is additive on S+ (R) , i.e., (4.50)

A (p + q) = A (p) + A (q) ,

p, q ∈ S+ (R)

and L : S+ (R) → R is superadditive on S+ (R) , i.e., (4.51)

L (p + q) ≥ L (p) + L (q) , p, q ∈ S+ (R) .

Define the following associated functionals (4.52)

F (p) :=

L (p) and H (p) := [F (p)]A(p) . A (p)

The following result holds [3, Theorem 2.1]. Lemma 4.22. With the above assumptions, we have (4.53)

H (p + q) ≥ H (p) H (q) ;

for any p, q ∈ S+ (R) , i.e., H (·) is supermultiplicative on S+ (R) . Proof. We shall follow the proof in [3]. Using the well-known arithmetic mean-geometric mean inequality for real numbers (4.54)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

β α αx + βy ≥ x α+β y α+β α+β

http://jipam.vu.edu.au/

64

S.S. D RAGOMIR

for any x, y ≥ 0 and α, β ≥ 0 with α + β > 0, we have successively L (p + q) A (p + q) L (p + q) = A (p) + A (q) L (p) + L (q) ≥ A (p) + A (q)

F (p + q) =

(4.55)

L(q) L(p) A (p) A(p) + A (q) A(q)

=

A (p) + A (q) A (p) F (p) + A (q) F (q) = A (p) + A (q) A(p)

A(q)

≥ [F (p)] A(p)+A(q) · [F (q)] A(p)+A(q) for all p, q ∈ S+ (R) . However, A (p) + A (q) = A (p + q) , and thus (4.55) implies the desired inequality (4.53).  We are now able to point out the following inequality related to the (CBS) − inequality. The first result is incorporated in the following theorem [3, p. 115]. Theorem 4.23. For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality ( (4.56)

1 PI + QI

"

!2 PI +QI  2 2  (pi + qi ) ai (pi + qi ) bi − (pi + qi ) ai bi  i∈I i∈I i∈I   !2 PI 1 X  X X  ≥ pi a2i pi b2i − p i ai b i   PI  i∈I i∈I i∈I   !2 QI  1 X  X X 2 2   × q i ai qi b i − qi ai bi > 0,  QI 

X

X

X

i∈I

where PI :=

P

i∈I

pi > 0, QI :=

P

i∈I

i∈I

i∈I

qi > 0.

Proof. Consider the functionals A (p) :=

X

p i = PI ;

i∈I

!2 C (p) :=

X i∈I

pi a2i

X i∈I

pi b2i −

X

p i ai b i

.

i∈I

Then A (·) is additive and C (·) is superadditive (see for example Lemma 4.16) on S+ (R) . Applying Lemma 4.22 we deduce the desired inequality (4.56).  The following refinement of the (CBS) −inequality holds.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

65

Corollary 4.24. For any a, b, α ∈ S (R) , one has the inequality

(4.57)

n X i=1

a2i

n X

b2i −

n X

i=1

!2 ai b i

i=1

1

≥ 1 n

Pn


n1 Pni=1 sin2 αi

2

i=1 sin αi  n n X X 2 2  ai sin αi b2i sin2 αi − × i=1

1 n

Pn

n X

i=1

2 i=1 cos αi

!2  n1 ai bi sin2 αi

Pn

i=1

sin2 αi



i=1

 n n X X 2 2  × ai cos αi b2i cos2 αi − i=1


n1 Pni=1 cos 2 αi

i=1

n X

!2  n1 ai bi cos2 αi

Pn

i=1



cos2 αi

≥ 0.

i=1

The following result holds [3, p. 116]. Theorem 4.25. For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality

(4.58)

# 12 " # 12 X X 1 1 (pi + qi ) a2i (pi + qi ) b2i  PI + QI P + Q I I i∈I i∈I )PI +QI X 1 − (pi + qi ) ai bi PI + QI i∈I  PI ! 12 ! 12 X X X 1 1 1 2 2  ≥ p i ai p i bi − p i ai b i  PI PI i∈I PI i∈I i∈I " 

 ×

1 X 2 q i ai QI i∈I

! 12

1 X 2 qi b i QI i∈I

! 12

QI 1 X − qi ai bi  ≥ 0. QI i∈I

Proof. Follows by Lemma 4.22 on taking into account that the functional ! 12

! 12 B (p) :=

X

pi a2i

i∈I

X i∈I

pi b2i

X − p i ai b i

is superadditive on S+ (R) (see Section 4.2).

i∈I



The following refinement of the (CBS) −inequality holds.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

66

S.S. D RAGOMIR

Corollary 4.26. For any a, b, α ∈ S (R) , one has the inequality ! 12

n X

(4.59)

n X

a2i

! 12 b2i

i=1

i=1

1 n

×

i=1

1

≥ 

n X − ai b i

n X

Pn

2 i=1 sin αi ! 12


n1 Pni=1 sin2 αi ·


1 Pn 2 2 α n i=1 cos αi cos i i=1 P  n1 ni=1 sin2 αi ! 12 n n X X b2i sin2 αi − ai bi sin2 αi 

a2i sin2 αi

i=1

 ×

1 1 n

Pn

i=1

n X

! 12

i=1

n X

a2i cos2 αi

i=1

! 21 b2i cos2 αi

i=1

 n1 n X 2 − ai bi cos αi 

Pn

i=1

cos2 αi

≥ 0.

i=1

Finally, we may also state [3, p. 117]. Theorem 4.27. For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality " (4.60)

X X 1 1 (pi + qi ) a2i · (pi + qi ) b2i PI + QI i∈I PI + QI i∈I −

1 PI + QI

I !2  PI +Q 2 X (pi + qi ) ai bi 

i∈I

 1 X 2 1 X 2 ≥ p i ai · p i bi − PI i∈I PI i∈I

1 X p i ai b i PI i∈I

!2  P2I 

 ×

1 X 2 1 X 2 q i ai · qi b i − QI i∈I QI i∈I

1 X q i ai b i QI i∈I

!2  Q2I 

.

Proof. Follows by Lemma 4.22 on taking into account that the functional  n n X X D (p) :=  pi a2i pi b2i − i=1

i=1

is superadditive on S+ (R) (see Section 4.6).

n X

!2  12 p i ai b i 

i=1



The following corollary also holds.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

67

Corollary 4.28. For any a, b, α ∈ S (R) , one has the inequality

(4.61)

 n n X X 2  ai b2i − i=1

i=1

!2  12

n X

ai b i



i=1

1

≥ 1 n

Pn


n1 Pni=1 sin2 αi ·

2

i=1 sin αi  n n X X 2 2  × ai sin αi b2i sin2 αi − i=1

i=1

1 1 n

Pn

n X

i=1


n1 Pni=1 cos 2 αi

αi P !2  2n1 ni=1 sin2 αi

ai bi sin2 αi



i=1

 n n X X 2 2  × ai cos αi b2i cos2 αi − i=1

cos2

i=1

n X

!2  2n1 ai bi cos2 αi

Pn

i=1



cos2 αi

≥ 0.

i=1

4.9. Supermultiplicity as an Index Set Mapping. Denote by Pf (N) the set of all finite parts of the natural number set N and assume that B : Pf (N) → R is set-additive on Pf (N) , i.e., (4.62)

B (I ∪ J) = B (I) + B (J) for any I, J ∈ Pf (N) , I ∩ J 6= ∅,

and G : Pf (N) → R is set-superadditive on Pf (N) , i.e., (4.63)

G (I ∪ J) ≥ G (I) + G (J) for any I, J ∈ Pf (N) , I ∩ J 6= ∅.

We may define the following associated functionals

(4.64)

M (I) :=

G (I) and N (I) := [M (I)]A(I) . B (I)

With these notations we may prove the following lemma that is interesting in itself as well. Lemma 4.29. Under the above assumptions one has (4.65)

N (I ∪ J) ≥ N (I) N (J)

for any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅, i.e., N (·) is set-supermultiplicative on Pf (N) . Proof. Using the arithmetic mean – geometric mean inequality

(4.66)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

β α αx + βy ≥ x α+β y α+β α+β

http://jipam.vu.edu.au/

68

S.S. D RAGOMIR

for any x, y ≥ 0 and α, β ≥ 0 with α + β > 0, we have successively for I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ that G (I ∪ J) B (I ∪ J) G (I ∪ J) = B (I) + B (J) G (I) + G (J) ≥ B (I) + B (J)

M (I ∪ J) =

(4.67)

=

G(J) G(I) B (I) B(I) + B (J) B(J)

B (I) + B (J) B (I) M (I) + B (J) M (J) = B (I) + B (J) B(I)

B(J)

≥ (M (I)) B(I)+B(J) · (M (J)) B(I)+B(J) . Since B (I) + B (J) = B (I ∪ J) , we deduce by (4.67) the desired inequality (4.65).



Now, we are able to point out some set-superadditivity properties for some functionals associates to the (CBS) −inequality. The first result is embodied in the following theorem. Theorem 4.30. If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that I ∩ J 6= ∅, then one has the inequality

(4.68)

  !2 PI∪J   1 X X X  pk a2k pk b2k − p k ak b k    PI∪J k∈I∪J k∈I∪J k∈I∪J   !2 PI 1 X  X X  ≥ pi a2i pi b2i − p i ai b i   PI  i∈I i∈I i∈I   !2 PJ  1 X  X X 2 2   × p j aj pj bj − p j aj b j ,  PJ  j∈J j∈J j∈J

when PJ :=

P

j∈J

pj .

Proof. Consider the functionals B (I) :=

X

pi ;

i∈I

!2 G (I) :=

X i∈I

pi a2i

X

pi b2i −

i∈I

X

p i ai b i

.

i∈I

The functional B (·) is obviously set-additive and (see Section 4.7) the functional G (·) is setsuperadditive. Applying Lemma 4.29 we then deduce the desired inequality (4.68).  The following corollary is a natural application.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

69

Corollary 4.31. If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has the inequality

(4.69)

  !2 P2n 2n 2n 2n  1 X  X X 2 2   p i ai p i bi − p i ai b i  P2n  i=1 i=1 i=1   Pni=1 p2i ! 2 n n n   X X X 1  ≥ Pn p2i a22i p2i b22i − p2i a2i b2i   i=1 p2i  i=1 i=1 i=1 ( " n n X X 1 p2i−1 a22i−1 p2i−1 b22i−1 × Pn p i=1 2i−1 i=1 i=1 P !2  ni=1 p2i−1 n  X − p2i−1 a2i−1 b2i−1  .  i=1

The following result also holds. Theorem 4.32. If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that I ∩ J 6= ∅, then one has the inequality

(4.70)

" )PI∪J # 12 " # 12  1 X 1 X X 1 pk a2k pk b2k − p k ak b k  PI∪J P P I∪J I∪J k∈I∪J k∈I∪J k∈I∪J  PI ! 12 ! 12  1 X  X X 1 1 2 2 ≥ p i ai p i bi − p i ai b i  PI PI  PI i∈I i∈I i∈I  PJ ! 12 ! 12  1 X  X X 1 1 × pj a2j pj b2j − p j aj b j .  PJ  P P J J j∈J j∈J j∈J

Proof. Follows by Lemma 4.29 on taking into account that the functional ! 21 X

G (I) :=

! 12 X

pi a2i

pi b2i

i∈I

i∈I

X − p i ai b i i∈I

is set-superadditive on Pf (N) .



The following corollary is a natural application. Corollary 4.33. If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has the inequality  (4.71)

2n X  1 pi a2i P2n i=1

! 12

2n 1 X 2 p i bi P2n i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

! 12

P2n 2n 1 X − p i ai b i  P2n i=1

http://jipam.vu.edu.au/

70

S.S. D RAGOMIR

 ≥  Pn

n X

1

i=1 p2i 

! 12

n X

1

p2i a22i

Pn

i=1 p2i

i=1

! 12 p2i b22i

i=1

i=1

n X 1  Pn p2i−1 a22i−1 × p 2i−1 i=1 i=1

#Pni=1 p2i n X P 1 − n p2i a2i b2i i=1 p2i

! 12

n X 1 Pn p2i−1 b22i−1 p 2i−1 i=1 i=1

! 12

#Pni=1 p2i−1 n X P 1 − n p2i−1 a2i−1 b2i−1 . i=1 p2i−1 i=1

Finally, we may also state: Theorem 4.34. If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that I ∩ J 6= ∅, then one has the inequality  !2  PI∪J 2 X X X 1 1 1 2 2   (4.72) p k ak · p k bk − p k ak b k PI∪J k∈I∪J PI∪J k∈I∪J PI∪J k∈I∪J  ≥

1 X 2 1 X 2 p i ai · p i bi − PI i∈I PI i∈I

1 X p i ai b i PI i∈I

!2  P2I 

 ×

1 X 1 X 2 pj a2j · p j bj − PJ j∈J PJ j∈J

1 X p j aj b j PJ j∈J

!2  P2J .



Proof. Follows by Lemma 4.29 on taking into account that the functional  !2  21 X X X pi b2i − p i ai b i  Q (I) :=  pi a2i i∈I

i∈I

i∈I

is set-superadditive on Pf (N) (see Section 4.7).



The following corollary holds as well. Corollary 4.35. If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has the inequality  (4.73)

2n 2n X X 1 1 2  p i ai · pi b2i − P2n i=1 P2n i=1

" ≥

n X

1 Pn

i=1 p2i

p2i a22i

n X

1 Pn

i=1

" ×

!

2n 1 X p i ai b i P2n i=1

1 Pn i=1 p2i−1

i=1 p2i

n X



! p2i b22i

−

! p2i−1 a22i−1

Pn

i=1 p2i

−

1 Pn i=1 p2i−1

1 Pn i=1 p2i−1

n X

1

i=1

i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

!2  P2n 2

n X

n X

!2  12 p2i a2i b2i

Pn

i=1

p2i



i=1

! p2i−1 b22i−1

i=1

!2  21 p2i−1 a2i−1 b2i−1



Pn

i=1

p2i−1

.

i=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

71

R EFERENCES ´ The improvement of Cauchy-Buniakowski-Schwartz’s [1] S.S. DRAGOMIR AND Š.Z. ARSLANAGIC, inequality, Mat. Bilten, 16 (1992), 77–80. [2] S.S. DRAGOMIR AND B. MOND, On the superadditivity and monotonicity of Schwartz’s inequality in inner product spaces, Contributions, Ser. Math. Sci. Macedonian Acad. Sci., 15(1) (1994), 5–22. [3] R.P. AGARWAL AND S.S. DRAGOMIR, The property of supermultiplicity for some classical inequalities and applications, Computer Math. Applic., 35(6) (1998), 105–118. [4] S.S. DRAGOMIR AND B. MOND, On the superadditivity and monotonicity of Gram’s inequality and related results, Acta Math. Hungarica, 71(1-2) (1996), 75–90. ˘ ´ J.E. PECARI [5] D.S. MITRINOVIC, C´ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

72

S.S. D RAGOMIR

5. R EVERSE I NEQUALITIES 5.1. The Cassels’ Inequality. The following result was proved by J.W.S. Cassels in 1951 (see Appendix 1 of [2] or Appendix of [3]): ¯ = (b1 , . . . , bn ) be sequences of positive real numbers Theorem 5.1. Let ¯ a = (a1 , . . . , an ) , b and w ¯ = (w1 , . . . , wn ) a sequence of nonnegative real numbers. Suppose that     ai ai (5.1) m = min . and M = max bi bi i=1,n i=1,n Then one has the inequality P wi a2i ni=1 wi b2i (m + M )2 ≤ . P 2 4mM ( ni=1 wi ai bi )

Pn

i=1

(5.2)

The equality holds in (5.2) when w1 = M = ban1 .

1 , a1 b1

wn =

1 , an bn

w2 = · · · = wn−1 = 0, m =

an b1

and

Proof. 1. The original proof by Cassels (1951) is of interest. We shall follow the paper [5] in sketching this proof. We begin with the assertion that (5.3)

(1 + kw) (1 + k −1 w) (1 + k) (1 + k −1 ) , k > 0, w ≥ 0 ≤ 4 (1 + w)2

which, being an equivalent form of (5.2) for n = 2, shows that it holds for n = 2. To prove that the maximum of (5.2) is obtained when we have more than two wi ’s being nonzero, Cassels then notes that if for example, w1 , w2 , w3 6= 0 lead to an extremum M of XY , 22 then we would have the linear equations a2n X + b2n Y − 2M an bn Z = 0, k = 1, 2, 3. Nontrivial solutions exist if and only if the three vectors [a2n , b2n , an bn ] are linearly dependent. But this will be so only if, for some i 6= j (i, j = 1, 2, 3) ai = γaj , bi = γbj . And if that were true, we could, for example, drop the ai , bi terms and so deal with the same problem with one less variable. If only one wi 6= 0, then M = 1, the lower bound. So we need only examine all pairs wi 6= 0, wj 6= 0. The result (5.2) then quickly follows. 2. We will now use the barycentric method of Frucht [1] and Watson [4]. We will follow the paper [5]. We substitute wi = ub2i in the left hand side of (5.2), which may then be expressed as the ratio i

N D2 where N=

n  2 X ai i=1

bi

ui and D =

n   X ai i=1

bi

ui ,

Pn

ai =1. Butthe point with co-ordinates (D, N ) a2 must lie within the convex closure of the n points abii , b2i . The value of DN2 at points on the i o n n o ai parabola is one unit. If m = min bi and M = max abii , then the minimum must lie on assuming without loss of generality, that

i=1,n

i=1

i=1,n

the chord joining the point (m, m2 ) and (M, M 2 ) . Some easy calculus then leads to (5.2).



The following “unweighted” Cassels’ inequality holds.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

73

¯ satisfy the assumptions in Theorem 5.1, one has the inequality Corollary 5.2. If ¯ a and b Pn 2 Pn 2 (m + M )2 i=1 ai i=1 bi (5.4) ≤ . P 2 4mM ( ni=1 ai bi ) The following two additive versions of Cassels inequality hold. Corollary 5.3. With the assumptions of Theorem 5.1, one has ! 21 n n n X X X 2 2 (5.5) − wi ai bi 0≤ wi ai wi bi i=1

√ ≤

i=1

i=1

√ 2 n M− m X √ wi ai bi . 2 mM i=1

and 0≤

(5.6)

n X

wi a2i

i=1

≤

n X

wi b2i −

i=1 n X

(M − m)2 4mM

n X

!2 wi ai bi

i=1 !2

wi ai bi

.

i=1

Proof. Taking the square root in (5.2) we get 1 P P ( ni=1 wi a2i ni=1 wi b2i ) 2 M +m Pn 1≤ ≤ √ . 2 mM i=1 wi ai bi Subtracting 1 on both sides, a simple calculation will lead to (5.5). The second inequality follows by (5.2) on subtracting 1 and appropriate computation.



The following additive version of unweighted Cassels inequality also holds. ¯ one has the inequalities Corollary 5.4. With the assumption of Theorem 5.1 for ¯ a and b √ √ 2 ! 21 n n n n M− m X X X X 2 2 √ − ai b i ≤ ai bi (5.7) 0≤ ai bi 2 mM i=1 i=1 i=1 i=1 and (5.8)

0≤

n X

a2i

i=1

n X

b2i

−

n X

i=1

!2 ai b i

i=1

(M − m)2 ≤ 4mM

n X

!2 ai b i

.

i=1

5.2. The Pólya-Szegö Inequality. The following inequality was proved in 1925 by Pólya and Szegö [6, pp. 57, 213 – 214], [7, pp. 71– 72, 253 – 255]. ¯ = (b1 , . . . , bn ) be two sequences of positive real Theorem 5.5. Let ¯ a = (a1 , . . . , an ) and b numbers. If (5.9)

0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ for each i ∈ {1, . . . , n} ,

then one has the inequality Pn (5.10)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

a2i

Pn

2 i=1 bi P 2 ( ni=1 ai bi ) i=1

≤

(ab + AB)2 . 4abAB

http://jipam.vu.edu.au/

74

S.S. D RAGOMIR

The equality holds in (5.10) if and only if     A A B B A B p=n· + and q = n · + a a b b a b are integers and if p of the numbers a1 , . . . , an are equal to a and q of these numbers are equal to A, and if the corresponding numbers bi are equal to B and b respectively. Proof. Following [5], we shall present here the original proof of Pólya and Szegö. We may, without loss of generality, suppose that a1 ≥ · · · ≥ an , then to maximise the lefthand side of (5.10) we must have that the critical bi ’s be reversely ordered (for if bk > bm with k < m, then we can interchange bk and bm such that b2k + b2m = b2m + b2k and ak bk + am bm ≥ ak bm + am bk ), i.e., that b1 ≤ · · · ≤ bn . Pólya and Szegö then continue by defining nonnegative numbers ui and vi for i = 1, . . . , n−1 and n > 2 such that a2i = ui a21 + vi a2n and b2i = ui b21 + vi b2n .

(5.11)

Since ai bi > ui a1 b1 + vi an bn the left hand side of (5.10), Pn 2 Pn 2 (U a21 + V a2n ) (U b21 + V b2n ) i=1 ai i=1 bi ≤ , P 2 (U a1 b1 + V an bn )2 ( ni=1 ai bi ) P P where U = ni=1 ui and V = ni=1 vi . This reduces the problem to that with n = 2, which is solvable by elementary methods, leading to Pn 2 Pn 2 (a1 b1 + an bn )2 i=1 ai i=1 bi (5.12) ≤ , P 2 4a1 an b1 bn ( ni=1 ai bi ) where, since the ai ’s and bi ’s here are reversely ordered, (5.13)

a1 = max {ai } ,

an = min {ai } ,

b1 = min {bi } ,

bn = max {bi } .

i=1,n

i=1,n

i=1,n

i=1,n

If we now assume, as in (5.9), that 0 < a ≤ ai ≤ A, 0 < b ≤ bi ≤ B, i = (1, . . . , n) , then

(because

(a1 b1 + an bn )2 (ab + AB)2 ≤ 4a1 an b1 bn 4abAB (k+1)2 4k

≤

(α+1)2 4α

for k ≤ α), and the inequality (5.10) is proved.



Remark 5.6. The inequality (5.10) may also be obtained from the “unweighted” Cassels’ inequality Pn 2 Pn 2 (m + M )2 i=1 ai i=1 bi (5.14) ≤ , P 2 4mM ( ni=1 ai bi ) where 0 < m ≤

ai bi

≤ M for each i ∈ {1, . . . , n} .

The following additive versions of the Pólya-Szegö inequality also hold. Corollary 5.7. With the assumptions in Theorem 5.5, one has the inequality √ √ 2 ! 12 n n n n AB − ab X X X X 2 2 √ (5.15) 0≤ ai bi ai b i − ai bi ≤ 2 abAB i=1 i=1 i=1 i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

75

and (5.16)

0≤

n X i=1

a2i

n X

b2i

n X

−

i=1

!2 ai b i

i=1

n X

(AB − ab)2 ≤ 4abAB

!2 ai b i

.

i=1

5.3. The Greub-Rheinboldt Inequality. The following weighted version of the Pólya-Szegö inequality was obtained by Greub and Rheinboldt in 1959, [6]. ¯ = (b1 , . . . , bn ) be two sequences of positive real Theorem 5.8. Let ¯ a = (a1 , . . . , an ) and b numbers and w ¯ = (w1 , . . . , wn ) a sequence of nonnegative real numbers. Suppose that (5.17)

0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ (i = 1, . . . , n) .

Then one has the inequality Pn

P wi a2i ni=1 wi b2i (ab + AB)2 ≤ . P 2 4abAB ( ni=1 wi ai bi )

i=1

(5.18)

Equality holds in (5.18) when wi = a11b1 , wn = with a1 = A, an = a, b1 = b and bn = b.

1 , an bn

w2 = · · · = wn−1 = 0, m =

an , b1

M =

a1 bn

Remark 5.9. This inequality follows by Cassels’ result which states that Pn Pn 2 2 (m + M )2 i=1 wi ai i=1 wi bi (5.19) ≤ , Pn 2 4mM ( i=1 wi ai bi ) provided 0 < m ≤

ai bi

≤ M < ∞ for each i ∈ {1, . . . , n} .

The following additive versions of Greub-Rheinboldt also hold. Corollary 5.10. With the assumptions in Theorem 5.8, one has the inequalities ! 12 n n n X X X 2 2 (5.20) 0≤ wi ai wi bi − wi ai bi i=1

√ ≤

i=1

i=1

√ 2 n AB − ab X √ wi ai bi 2 abAB i=1

and (5.21)

0≤

n X i=1

wi a2i

n X

wi b2i −

i=1 2

(AB − ab) ≤ 4abAB

n X

!2 wi ai bi

i=1 n X

!2 wi ai bi

.

i=1

5.4. A Cassels’ Type Inequality for Complex Numbers. The following reverse inequality for the (CBS) −inequality holds [9]. Theorem 5.11. Let a, A ∈ K (K = C, R) such that Re (¯ aA) > 0. If x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) are sequences of complex numbers and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers with the property that n X (5.22) wi Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0, i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

76

S.S. D RAGOMIR

then one has the inequality " n # 12 Pn n X X wi Re [A¯ xi yi + a ¯xi y¯i ] 1 2 2 wi |xi | wi |yi | ≤ · i=1 (5.23) 1 2 [Re (¯ aA)] 2 i=1 i=1 n 1 |A| + |a| X ≤ · wi xi y¯i . 1 2 [Re (¯ aA)] 2 i=1

The constant

1 2

is sharp in the sense that it cannot be replaced by a smaller one.

Proof. We have, obviously, that Γ := =

n X i=1 n X

wi Re [(Ayi − xi ) (¯ xi − a ¯y¯i )]

wi Re [A¯ xi yi + a ¯xi y¯i ] −

i=1

n X

2

wi |xi | − Re (¯ aA)

i=1

n X

wi |yi |2

i=1

and then, by (5.22), one has n X

2

wi |xi | + Re (¯ aA)

i=1

n X

2

wi |yi | ≤

i=1

n X

wi Re [A¯ xi yi + a ¯xi y¯i ]

i=1

giving n X

1

(5.24)

[Re (¯ aA)]

1 2

2

wi |xi | + [Re (¯ aA)]

i=1

1 2

n X

2

wi |yi | ≤

i=1

Pn

i=1

wi Re [A¯ xi yi + a ¯xi y¯i ] 1

.

[Re (¯ aA)] 2

On the other hand, by the elementary inequality 1 αp2 + q 2 ≥ 2pq α holding for any p, q ≥ 0 and α > 0, we deduce ! 12 n n n n X X X 1 X 1 2 2 w |x | + [Re (¯ a A)] wi |yi |2 . (5.25) 2 wi |xi |2 wi |yi |2 ≤ i i 1 [Re (¯ aA)] 2 i=1 i=1 i=1 i=1 Utilising (5.24) and (5.25), we deduce the first part of (5.23). The second part is obvious by the fact that for z ∈ C, |Re (z)| ≤ |z| . Now, assume that the first inequality in (5.23) holds with a constant c > 0, i.e., Pn n n X X wi Re [A¯ xi yi + a ¯xi y¯i ] 2 2 , (5.26) wi |xi | wi |yi | ≤ c · i=1 1 2 [Re (¯ a A)] i=1 i=1 where a, A, x ¯, y ¯ satisfy (5.22). If we choose a = A = 1, y = x 6= 0, then obviously (5.23) holds and from (5.26) we may get n n X X wi |xi |2 ≤ 2c wi |xi |2 , i=1

i=1

1 . 2

giving c ≥ The theorem is completely proved.



The following corollary is a natural consequence of the above theorem.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

77

Corollary 5.12. Let m, M > 0 and x ¯, y ¯, w ¯ be as in Theorem 5.11 and with the property that (5.27)

n X

wi Re [(M yi − xi ) (¯ xi − m¯ yi )] ≥ 0,

i=1

then one has the inequality " n # 12 n n X X 1 M +mX 2 2 (5.28) wi Re (xi y¯i ) ≤ · √ wi |xi | wi |yi | 2 mM i=1 i=1 i=1 n 1 M + m X ≤ · √ wi xi y¯i . 2 mM i=1 The following corollary also holds. Corollary 5.13. With the assumptions in Corollary 5.12, then one has the following inequality: " n # 12 n n X X X 0≤ wi |xi |2 wi |yi |2 − (5.29) wi xi y¯i i=1 i=1 i=1 " n # 12 n n X X X ≤ wi |xi |2 wi |yi |2 − wi Re (xi y¯i ) i=1

i=1

√

√ 2 M− m √ ≤ 2 mM √ √ 2 M− m √ ≤ 2 mM

i=1 n X

wi Re (xi y¯i )

i=1

n X wi xi y¯i i=1

and (5.30)

2 n n X X 0≤ wi |xi |2 wi |yi |2 − wi xi y¯i i=1 i=1 i=1 " n #2 n n X X X ≤ wi |xi |2 wi |yi |2 − wi Re (xi y¯i ) n X

i=1

i=1

"

i=1 #2

n (M − m)2 X ≤ wi Re (xi y¯i ) 4mM i=1 2 n (M − m)2 X ≤ wi xi y¯i . 4mM i=1

5.5. A Reverse Inequality for Real Numbers. The following result holds [10, Proposition 5.1]. Theorem 5.14. Let a, A ∈ R and x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) be two sequences with the property that: (5.31)

ayi ≤ xi ≤ Ayi for each i ∈ {1, . . . , n} .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

78

S.S. D RAGOMIR

Then for any w ¯ = (w1 , . . . , wn ) a sequence of positive real numbers, one has the inequality !2 n n n X X X (5.32) wi x2i wi yi2 − wi xi yi 0≤ i=1

≤ The constant

1 4

i=1

i=1 n X

1 (A − a)2 4

!2 wi yi2

.

i=1

is sharp in (5.32).

Proof. Let us define I1 :=

A

n X

n X

wi yi2 −

i=1

and

n X

wi xi yi

i=1

wi yi2

i=1

wi xi yi − a

i=1

!

n X

I2 :=

!

n X

I1 = (a + A)

wi yi2

i=1

! wi yi2

i=1

(Ayi − xi ) (xi − ayi ) wi .

i=1

Then n X

n X

n X

n X

wi xi yi −

i=1

!2

n X

− aA

wi xi yi

i=1

!2 wi yi2

i=1

and I2 = (a + A)

n X

wi yi2

n X

i=1

wi xi yi −

i=1

n X

wi x2i

i=1

n X

wi yi2 − aA

n X

i=1

!2 wi yi2

i=1

giving I1 − I2 =

(5.33)

n X

wi x2i

i=1

n X

wi yi2 −

i=1

n X

!2 wi yi2

.

i=1

If (5.31) holds, then (Ayi − xi ) (xi − ayi ) ≥ 0 for each i ∈ {1, . . . , n} and thus I2 ≥ 0 giving !2 n n n X X X (5.34) wi x2i wi yi2 − wi yi2 i=1

i=1

i=1

" ≤

A

n X i=1

wi yi2

−

n X

! wi xi yi

i=1

n X

wi xi yi − a

i=1

n X

!# wi yi2

.

i=1

If we use the elementary inequality for real numbers u, v ∈ R 1 (5.35) uv ≤ (u + v)2 , 4 then we have for n n n n X X X X 2 u := A wi yi − wi xi yi , v := wi xi yi − a wi yi2 i=1

i=1

i=1

i=1

that A

n X i=1

wi yi2 −

n X

! wi xi yi

i=1

n X i=1

wi xi yi − a

n X i=1

! wi yi2

1 ≤ (A − a)2 4

n X

!2 wi yi2

i=1

and the inequality (5.32) is proved.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

Now, assume that (5.32) holds with a constant c > 0, i.e., !2 n n n X X X wi xi yi ≤ c (A − a)2 wi yi2 − (5.36) wi x2i i=1

i=1

n X

i=1

79

!2 wi yi2

,

i=1

where a, A, x ¯, y ¯ satisfy (5.31). We choose n = 2, w1 = w2 = 1 and let a, A, y1 , y2 , x, α ∈ R such that ay1 < x1 = Ay1 , ay2 = x2 < Ay2 . With these choices, we get from (5.36) that


2
2 a2 y12 + a2 y22 y12 + y22 − A2 y12 + a2 y22 ≤ c (A − a)2 y12 + y22 , which is equivalent to (A − a)2 y12 y22 ≤ c (A − a)2 y12 + y22


2

.

Since we may choose a 6= A, we deduce y12 y22 ≤ c y12 + y22


2

,

giving, for y1 = y2 = 1, c ≥ 41 .



The following corollary is obvious. Corollary 5.15. With the above assumptions for a, A, x ¯ and y ¯, we have the inequality !2 !2 n n n n X X X X 1 2 x2i yi2 − xi yi ≤ (A − a) yi2 . (5.37) 0≤ 4 i=1 i=1 i=1 i=1 Remark 5.16. Condition (5.31) may be replaced by the weaker condition (5.38)

n X

wi (Ayi − xi ) (xi − ayi ) ≥ 0

i=1

and the conclusion in Theorem 5.14 will still be valid, i.e., the inequality (5.32) holds. For (5.37) to be true it suffices that n X

(5.39)

(Ayi − xi ) (xi − ayi ) ≥ 0

i=1

holds true. 5.6. A Reverse Inequality for Complex Numbers. The following result holds [10, Proposition 5.1]. Theorem 5.17. Let a, A ∈ C and x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) ∈ Cn , w ¯ = (w1 , . . . , wn ) ∈ n R+ . If (5.40)

n X

wi Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0,

i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

80

S.S. D RAGOMIR

then one has the inequality 2 n n X X 2 2 0≤ wi |xi | wi |yi | − wi xi y¯i i=1 i=1 i=1 ! 2 n X 1 2 2 ≤ |A − a| wi |yi | . 4 i=1 n X

(5.41)

The constant

1 4

is sharp in (5.41).

Proof. Consider " A1 := Re

A

n X

wi |yi |2 −

i=1

n X

! wi xi y¯i

n X

i=1

wi x¯i yi − a ¯

i=1

n X

!# wi |yi |2

i=1

and A2 :=

n X

" wi |yi |2 − Re

n X

i=1

# wi (Ayi − xi ) (¯ xi − a ¯y¯i ) .

i=1

Then A1 =

n X

" 2

wi |yi | − Re A

i=1

n X

wi x¯i yi + a ¯

i=1

n X

# wi xi y¯i

i=1

2 !2 n n X X − wi xi y¯i − Re (¯ aA) wi |yi |2 i=1

i=1

and A2 =

n X i=1

" 2

wi |yi | − Re A

n X

wi x¯i yi + a ¯

i=1

n X

# wi xi y¯i

i=1

−

n X

n n X X 2 2 wi |xi | wi |yi | − Re (¯ aA) wi |yi |2

i=1

i=1

!2

i=1

giving (5.42)

A1 − A2 =

n X i=1

2 n n X X 2 2 wi |xi | wi |yi | − wi xi y¯i . i=1

If (5.40) holds, then A2 ≥ 0 and thus 2 n n n X X X 2 2 (5.43) wi |xi | wi |yi | − wi xi y¯i i=1 i=1 i=1 " ! n n X X ≤ Re A wi |yi |2 − wi xi y¯i i=1

i=1

i=1

n X

wi x¯i yi − a ¯

i=1

n X

!# wi |yi |2

.

i=1

If we use the elementary inequality for complex numbers z, t ∈ C (5.44)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

1 Re [z t¯] ≤ |z − t|2 , 4

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

81

then we have for z := A

n X

2

wi |yi | −

i=1

t :=

n X

n X

wi xi y¯i ,

i=1

wi xi y¯i − a

i=1

n X

wi |yi |2

i=1

that " (5.45)

Re

A

n X

wi |yi |2 −

i=1

n X

! wi xi y¯i

i=1

n X

wi x¯i yi − a ¯

i=1

n X

!# wi |yi |2

i=1

1 ≤ |A − a|2 4 and the inequality (5.41) is proved. Now, assume that (5.41) holds with a constant c > 0, i.e., 2 n n n X X X (5.46) wi |xi |2 wi |yi |2 − wi xi y¯i ≤ c |A − a|2 i=1

i=1

i=1

n X

n X

!2 wi |yi |2

i=1

!2 wi |yi |2

,

i=1

where x ¯, y ¯, a, A satisfy Pn(5.40).2 P P n Consider y ¯ ∈ C , i=1 |yi | wi = 1, a 6= A, m ¯ ∈ Cn , ni=1 wi |mi |2 = 1 with ni=1 wi yi mi = 0. Define A+a A+a xi := yi + mi , i ∈ {1, . . . , n} . 2 2 Then n n X A − a 2 X wi (Ayi − xi ) (¯ xi − a ¯ yi ) = wi (yi − mi ) (¯ yi − m ¯ i) = 0 2 i=1 i=1 and thus the condition (5.40) is fulfilled. From (5.46) we deduce 2 2  n n  X X A + a A − a A + a A − a 2 y + m w − y + m y ¯ w i i i i i i i ≤ c |A − a| 2 2 2 2 i=1 i=1 and since n X i=1

  2 A + a 2 A − a 2 A+a A−a wi yi + mi = − 2 2 2 2

and n  2  X A + a A + a 2 A−a yi + mi y¯i wi = 2 2 2 i=1 then by (5.46) we get

giving c ≥

1 4

|A − a|2 ≤ c |A − a|2 4 and the theorem is completely proved.



The following corollary holds.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

82

S.S. D RAGOMIR

Corollary 5.18. Let a, A ∈ C and x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) ∈ Cn be with the property that n X (5.47) Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0, i=1

then one has the inequality 0≤

(5.48)

n X i=1

The constant

1 4

2 n n X X 1 2 2 |xi | |yi | − xi y¯i ≤ |A − a|2 4 i=1 i=1

n X

!2 2

|yi |

.

i=1

is best in (5.48).

Remark 5.19. A sufficient condition for both (5.40) and (5.47) to hold is Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0

(5.49) for any i ∈ {1, . . . , n} .

5.7. Shisha-Mond Type Inequalities. As some particular case for bounds on differences of means, O. Shisha and B. Mond obtained in 1967 (see [23]) the following reverse of (CBS) − inequality: ¯ = (b1 , . . . , bn ) are such that there exists Theorem 5.20. Assume that ¯ a = (a1 , . . . , an ) and b a, A, b, B > 0 with the property that: a ≤ aj ≤ A and b ≤ bj ≤ B for any j ∈ {1, . . . , n}

(5.50)

then we have the inequality (5.51)

n X j=1

a2j

n X j=1

b2j −

n X j=1

!2 aj b j

r ≤

A − b

r !2 n n X X a aj b j b2j . B j=1 j=1

The equality holds in (5.51) if and only if there exists a subsequence (k1 , . . . , kp ) of (1, 2, . . . , n) such that   12   32 n A B =1+ , p a b akµ = A, bkµ = b (µ = 1, . . . , p) and ak = a, bk = B for every k distinct from all kµ . Using another result stated for weighted means in [23], we may prove the following reverse of the (CBS) −inequality. ¯ are positive sequences and there exists γ, Γ > 0 with the Theorem 5.21. Assume that ¯ a, b property that ai (5.52) 0<γ≤ ≤ Γ < ∞ for any i ∈ {1, . . . , n} . bi Then we have the inequality ! 21 n n n n X X X (Γ − γ)2 X 2 2 2 bi ai b i ≤ b. (5.53) 0≤ ai − 4 (γ + Γ) j=1 j i=1 i=1 i=1 The equality holds in (5.53) if and only if there exists a subsequence (k1 , . . . , kp ) of (1, 2, . . . , n) such that p n X Γ + 3γ X 2 akm ak 2 b km = bj , = Γ (m = 1, . . . , p) and =γ 4 (γ + Γ) j=1 b km bk m=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

83

for every k distinct from all km . Proof. In [23, p. 301], Shisha and Mond have proved the following weighted inequality ! 12 n n X X (C − c)2 2 − qj x j ≤ , (5.54) 0≤ qj x j 4 (c + C) j=1 j=1 Pn provided qj ≥ 0 (j = 1, . . . , n) with j=1 qj = 1 and 0 < c ≤ xj < C < ∞ for any j ∈ {1, . . . , n} . Equality holds in (5.54) if and only if there exists a subsequence (k1 , . . . , kp ) of (1, 2, . . . , n) such that p X C + 3c , (5.55) qk m = 4 (c + C) m=1 xkm = C (m = 1, 2, . . . , p) and xk = c for every k distinct from all km . If in (5.54) we choose xj =

b2j aj , qj = Pn 2 , j ∈ {1, . . . , n} ; bj k=1 bk

then we get ! 12 Pn Pn 2 a (Γ − γ)2 j j=1 aj bj Pj=1 P − ≤ , n n 2 2 4 (γ + Γ) k=1 bk k=1 bk giving the desired inequality (5.53). The case of equality follows by the similar case in (5.54) and we omit the details.



5.8. Zagier Type Inequalities. The following result was obtained by D. Zagier in 1995, [24]. Lemma 5.22. Let f, g : [0, ∞) → R be monotone decreasing nonnegative functions on [0, ∞). Then R∞ R∞ Z ∞ f (x) F (x) dx 0 g (x) G (x) dx 0 R ∞ , R∞ (5.56) f (x) g (x) dx ≥ max 0 F (x) dx, 0 G (x) dx 0 for any integrable functions F, G : [0, ∞) → [0, 1] . Proof. We will follow the proof in [24]. For all x ≥ 0 we have Z ∞ Z ∞ Z ∞ f (t) F (t) dt = f (x) F (t) dt + [f (t) − f (x)] F (t) dt 0 0 0 Z ∞ Z x ≤ f (x) F (t) dt + [f (t) − f (x)] dt 0

and hence, since

0

R∞ G (t) dt is bounded from above by both x and G (t) dt, 0 0 Z ∞ Z x f (t) F (t) dt · G (t) dt 0 0 Z ∞ Z ∞ Z x ≤ xf (x) F (t) dt + G (t) dt · [f (t) − f (x)] dt 0 0 0 Z ∞  Z x Z ∞ ≤ max F (t) dt, G (t) dt · f (t) dt. Rx

0

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

0

0

http://jipam.vu.edu.au/

84

S.S. D RAGOMIR

Now, R ∞ we multiply by R ∞−dg (x) and integrate by parts from 0 to ∞. The left hand side gives f (t) F (t) dt · −∞ g (t) G (t) dt, the right hand side gives −∞  Z ∞ Z ∞ Z ∞ F (t) dt, G (t) dt · f (t) g (t) dt, max 0

0

0

and the inequality remains true because the measure −dg (x) is nonnegative.



The following particular case is a reverse of the (CBS) −integral inequality obtained by D. Zagier in 1977, [25]. Corollary 5.23. If f, g : [0, ∞) → [0, ∞) are decreasing function on [0, ∞), then   Z ∞ Z ∞ Z ∞ (5.57) max f (0) g (t) dt, g (0) f (t) dt · f (t) g (t) dt 0 0 0 Z ∞ Z ∞ 2 ≥ f (t) dt g 2 (t) dt. 0

0

Remark 5.24. The following weighted version of (5.56) may be proved in a similar way, as noted by D. Zagier in [25] R∞ R∞ Z ∞ w (t) f (t) F (t) dt w (t) f (t) G (t) dt 0 R , w (t) f (t) g (t) dt ≥ 0 R ∞ (5.58) ∞ max 0 w (t) F (t) dt, 0 w (t) G (t) dt 0 provided w (t) > 0 on [0, ∞), f, g : [0, ∞) → [0, ∞) are monotonic decreasing and F, G : [0, ∞) → [0, 1] are integrable on [0, ∞). We may state and prove the following discrete inequality. ¯ = (b1 , . . . , bn ) , Theorem 5.25. Consider the sequences of real numbers ¯ a = (a1 , . . . , an ), b p ¯ = (p1 , . . . , pn ), q ¯ = (q1 , . . . , qn ) and w ¯ = (w1 , . . . , wn ) . If ¯ are decreasing and nonnegative; (i) ¯ a and b (ii) pi , qi ∈ [0, 1] and wi ≥ 0 for any i ∈ {1, . . . , n} , then we have the inequality Pn P n X wi pi ai ni=1 wi qi bi i=1 Pn Pn (5.59) wi ai bi ≥ . max { w p , i i i=1 i=1 wi qi } i=1 Proof. Consider the functions f, g, F, G, W : [0, ∞) → R given by   a1 , t ∈ [0, 1) b1 , t ∈ [0, 1)          a2 , t ∈ [1, 2)  b2 , t ∈ [1, 2) . .. .. f (t) = , g (t) = , .          an , t ∈ [n − 1, n)  bn , t ∈ [n − 1, n) 0 t ∈ [n, ∞) 0 t ∈ [n, ∞)   p1 , t ∈ [0, 1) q1 , t ∈ [0, 1)          p2 , t ∈ [1, 2)  q2 , t ∈ [1, 2) . .. .. F (t) = , G (t) = , .          pn , t ∈ [n − 1, n)  qn , t ∈ [n − 1, n) 0 t ∈ [n, ∞) 0 t ∈ [n, ∞)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

85

and  w1 ,      w2 , .. W (t) = .     wn ,  0

t ∈ [0, 1) t ∈ [1, 2) . t ∈ [n − 1, n) t ∈ [n, ∞)

We observe that, the above functions satisfy the hypothesis of Remark 5.24 and since, for example, Z ∞ Z ∞ n Z i X w (t) f (t) g (t) dt = w (t) f (t) g (t) dt + w (t) f (t) g (t) dt 0

=

i=1 n X

i−1

n

wk ak bk ,

k=1



then by (5.58) we deduce the desired inequality (5.59).

Remark 5.26. A similar inequality for sequences under some monotonicity assumptions for p ¯ and q ¯ was obtained in 1995 by J. Peˇcari´c in [26]. The following reverse of the (CBS) −discrete inequality holds. ¯ are decreasing nonnegative sequences with a1 , b1 6= 0 and w Theorem 5.27. Assume that ¯ a, b ¯ a nonnegative sequence. Then ( n ) n n n n X X X X X 2 2 (5.60) wi ai wi bi ≤ max b1 wi ai , a1 wi bi wi ai bi . i=1

i=1

i=1

The proof follows by Theorem 5.25 on choosing pi = {1, . . . , n} . We omit the details.

i=1 ai a1

i=1

∈ [0, 1] , qi =

bi b1

∈ [0, 1] , i ∈

Remark 5.28. When wi = 1, we recapture Alzer’s result from 1992, [27]. 5.9. A Reverse Inequality in Terms of the sup −Norm. The following result has been proved in [11]. Lemma 5.29. Let α ¯ = (α1 , . . . , αn ) and x ¯ = (x1 , . . . , xn ) be sequences complex numbers Pof n and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers such that i=1 pi = 1. Then one has the inequality n n n X X X (5.61) pi αi xi − pi αi pi xi i=1 i=1 i=1  !2  n n X X ≤ max |∆αi | max |∆xi |  i2 p i − ipi  , i=1,n−1

i=1,n−1

i=1

i=1

where ∆αi is the forward difference, i.e., ∆αi := αi+1 − αi . Inequality (5.61) is sharp in the sense that the constant C = 1 in the right membership cannot be replaced be a smaller one.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

86

S.S. D RAGOMIR

Proof. We shall follow the proof in [11]. We start with the following identity n n n n X X X 1X pi αi xi − pi αi pi xi = pi pj (αi − αj ) (xi − xj ) 2 i=1 i=1 i=1 i,j=1 X = pi pj (αi − αj ) (xi − xj ) . 1≤i<j≤n

As i < j, we can write that αj − αi =

j−1 X

∆αk

k=i

and xj − xi =

j−1 X

∆xk .

k=i

Using the generalised triangle inequality, we have successively j−1 j−1 n n n X X X X X X pi αi xi − pi αi pi xi = pi pj ∆αk ∆xk i=1 i=1 i=1 1≤i<j≤n k=i k=i j−1 j−1 X X X ≤ pi pj ∆αk ∆xk 1≤i<j≤n

≤

X

k=i

pi pj

1≤i<j≤n

j−1 X

k=i

|∆αk |

k=i

j−1 X

|∆xk |

k=i

:= A. Note that |∆αk | ≤ max |∆αs | 1≤s≤n−1

and |∆xk | ≤ max |∆xs | 1≤s≤n−1

for all k = i, . . . , j − 1 and then by summation j−1 X

|∆αk | ≤ (j − i) max |∆αs | 1≤s≤n−1

k=i

and

j−1 X

|∆xk | ≤ (j − i) max |∆xs | . 1≤s≤n−1

k=i

Taking into account the above estimations, we can write " # X A≤ pi pj (j − i)2 max |∆αs | max |∆xs | . 1≤s≤n−1

1≤i<j≤n

1≤s≤n−1

As a simple calculation shows that X

2

pi pj (j − i) =

1≤i<j≤n

n X i=1

i2 p i −

n X

!2 ipi

,

i=1

inequality (5.61) is proved.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

87

To prove the sharpness of the constant, let us assume that (5.61) holds with a constant C > 0, i.e., n n n X X X (5.62) pi αi xi − pi αi pi xi i=1 i=1 i=1  !2  n n X X ≤ C max |∆αi | max |∆xi |  i2 p i − ipi  . i=1,n−1

i=1,n−1

i=1

i=1

Now, choose the sequences αk = α + kβ (β 6= 0) and xk = x + ky (y 6= 0) , k ∈ {1, . . . , n} to get n n n n X 1 X X X pi αi xi − pi αi pi xi = pi pj (i − j) βy 2 i=1 i=1 i=1 i,j=1  !2  n n X X = |β| |y|  i2 p i − ipi  i=1

i=1

and  n X  max |∆αi | max |∆xi | i2 p i −

i=1,n−1

i=1,n−1

n X

i=1

!2  ipi

 n X  = |β| |y|  i2 p i −

i=1

i=1

n X

!2  ipi



i=1

and then, by (5.62), we get C ≥ 1.



The following reverse of the (CBS) −inequality holds [12]. ¯ = (b1 , . . . , bn ) be two sequences of real numbers Theorem 5.30. Let ¯ a = (a1 , . . . , an ) and b with ai 6= 0, (i = 1, . . . , n) . Then one has the inequality !2 n n n X X X 0≤ a2i b2i − ai b i i=1

i=1

i=1

   2 X n n X bk 2  ai i2 a2i − ≤ max ∆ ak k=1,n−1 i=1 i=1

n X

!2  ia2i

.

i=1

The constant C = 1 is sharp in the sense that it cannot be replaced by a smaller constant. Proof. Follows by Lemma 5.29 on choosing a2 pi = Pn i

2 k=1 ak

, αi =

bi bi , xi = , i ∈ {1, . . . , n} ai ai

and performing some elementary calculations. We omit the details.



5.10. A Reverse Inequality in Terms of the 1−Norm. The following result has been obtained in [13]. Lemma 5.31. Let α ¯ = (α1 , . . . , αn ) and x ¯ = (x1 , . . . , xn ) be sequences complex numbers Pof n and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers such that i=1 pi = 1. Then one

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

88

S.S. D RAGOMIR

has the inequality n n n n n−1 n−1 X 1X X X X X (5.63) pi αi xi − pi αi pi xi ≤ pi (1 − pi ) |∆αi | |∆xi | , 2 i=1 i=1 i=1 i=1 i=1 i=1 where ∆αi := αi+1 − αi is the forward difference. The constant 12 is sharp in the sense that it cannot be replaced by a smaller constant. Proof. We shall follow the proof in [13]. As in the proof of Lemma 5.29 in Section 5.9, we have n j−1 j−1 n n X X X X X X (5.64) pi αi xi − pi αi pi xi ≤ pi pj |∆αk | |∆xl | := A. i=1

i=1

i=1

1≤i<j≤n

k=i

l=i

It is obvious that for all 1 ≤ i < j ≤ n − 1, we have that j−1 X

|∆αk | ≤

n−1 X

k=i

k=1

j−1 X

n−1 X

|∆αk |

and |∆xl | ≤

l=i

|∆xl | .

l=1

Utilising these and the definition of A, we conclude that A≤

(5.65)

n−1 X

|∆αk |

k=1

n−1 X

X

|∆xl |

l=1

pi pj .

1≤i<j≤n

Now, let us observe that " n # X 1 X pi pj − pi pj pi pj = 2 i,j=1 i=j 1≤i<j≤n " n # n n X 1 X X = pi pj − p2i 2 i=1 j=1 i=1 X

(5.66)

n

1X = pi (1 − pi ) . 2 i=1 Making use of (5.64) – (5.66), we deduce the desired inequality (5.63). To prove the sharpness of the constant 21 , let us assume that (5.63) holds with a constant C > 0. That is n n n n n−1 n−1 X X X X X X (5.67) p α x − p α p x ≤ C p (1 − p ) |∆α | |∆xi | i i i i i i i i i i i=1

i=1

i=1

i=1

i=1

i=1

for all αi , xi , pi (i = 1, . . . , n) as above and n ≥ 1.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

89

Choose in (5.63) n = 2 and compute 2 X

pi αi xi −

2 X

i=1

2 X

pi αi

i=1

i=1

2 1X pi xi = pi pj (αi − αj ) (xi − xj ) 2 i,j=1 X = pi pj (αi − αj ) (xi − xj ) 1≤i<j≤2

= p1 p2 (α1 − α2 ) (x1 − x2 ) . Also 2 X

pi (1 − pi )

i=1

2 X

|∆αi |

i=1

2 X

|∆xi | = (p1 p2 + p1 p2 ) |α1 − α2 | |x1 − x2 | .

i=1

Substituting in (5.67), we obtain p1 p2 |α1 − α2 | |x1 − x2 | ≤ 2Cp1 p2 |α1 − α2 | |x1 − x2 | . If we assume that p1 , p2 > 0, α1 6= α2 , x1 6= x2 , then we obtain C ≥ 12 , which proves the sharpness of the constant 21 .  We are now able to state the following reverse of the (CBS) −inequality [12]. ¯ = (b1 , . . . , bn ) be two sequences of real numbers Theorem 5.32. Let ¯ a = (a1 , . . . , an ) and b with ai 6= 0 (i = 1, . . . , n) . Then one has the inequality !2 n n n X X X (5.68) 0≤ a2i b2i − ai b i i=1

i=1

i=1

" n−1   #2 X X ∆ bk a2i a2j . ≤ ak 1≤i<j≤n

k=1

The constant C = 1 is sharp in (5.68), in the sense that it cannot be replaced by a smaller constant. Proof. We choose a2i P pi = n

2 k=1 ak

,

αi = xi =

bi , i ∈ {1, . . . , n} ai

in (5.63) to get Pn 2 P 2 bi ( ni=1 ai bi ) i=1 0 ≤ Pn − P 2 2 ( nk=1 a2k ) k=1 ak  Pn 2  a2 Pn i 2 a 1 − i=1 i 1 k=1 ak Pn ≤ · 2 2 k=1 ak =

1 · 2

Pn

Pn 2 2 2 i=1 ai ( k=1 ak − ai ) P 2 ( nk=1 a2k )

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

 !2 n−1  X b ∆ j aj j=1

 n−1  X b j ∆ aj

!2

j=1

http://jipam.vu.edu.au/

90

S.S. D RAGOMIR

which is clearly equivalent to 0≤

n X

a2i

n X

b2i −



n X

1 ≤  2

!2 a2k

−

n X

1 2

ai b i 

 !2 n  X b ∆ j . a4i  aj

n X i=1

k=1



!2

i=1

i=1

i=1

Since

n X

!2 a2k

−

j=1

n X

 X

a4i  =

i=1

k=1

a2i a2j

1≤i<j≤n



the inequality (5.68) is thus proved.

5.11. A Reverse Inequality in Terms of the p−Norm. The following result has been obtained in [14]. Lemma 5.33. Let α ¯ = (α1 , . . . , αn ) and x ¯ = (x1 , . . . , xn ) be sequences Pof complex numbers and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers such that ni=1 pi = 1. Then one has the inequality n n n X X X (5.69) pi αi xi − pi αi pi xi i=1 i=1 i=1 ! 1q ! p1 n−1 n−1 X X X |∆xk |q , ≤ (i − j) pi pj |∆αk |p 1≤j
k=1

k=1

where p > 1, p1 + 1q = 1. The constant C = 1 in the right hand side of (5.69) is sharp in the sense that it cannot be replaced by a smaller constant. Proof. We shall follow the proof in [14]. As in the proof of Lemma 5.29 in Section 5.9, we have n n n i−1 i−1 X X X X X X (5.70) pi αi xi − pi αi pi xi ≤ pi pj |∆αk | |∆xl | := A. i=1

i=1

i=1

1≤j
k=j

l=j

Using Hölder’s discrete inequality, we can state that i−1 X

|∆αk | ≤ (i − j)

1 q

k=j

i−1 X

! p1 |∆αk |p

k=j

and i−1 X

|∆xl | ≤ (i − j)

1 p

where p > 1, (5.71)

+

1 q

! 1q |∆xl |q

,

l=j

l=j 1 p

i−1 X

= 1, and then we get

A≤

X

pi pj (i − j)

1≤j
J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

i−1 X k=j

! p1 |∆αk |p

i−1 X

! 1q |∆xk |q

.

k=j

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

Since

i−1 X

n−1 X

p

|∆αk | ≤

k=j

and

i−1 X

91

|∆αk |p

k=1

q

|∆xk | ≤

k=j

n−1 X

|∆xk |q ,

k=1

for all 1 ≤ j < i ≤ n, then by (5.70) and (5.71) we deduce the desired inequality (5.69). To prove the sharpness of the constant, let us assume that (5.69) holds with a constant C > 0. That is, n n n X X X (5.72) pi αi xi − pi αi pi xi i=1 i=1 i=1 ! 1q ! p1 n−1 n−1 X X X |∆xk |q . (i − j) pi pj ≤C |∆αk |p 1≤j
k=1

k=1

Note that, for n = 2, we have 2 2 2 X X X pi αi xi − pi αi pi xi = p1 p2 |α1 − α2 | |x1 − x2 | i=1

i=1

i=1

and X

(i − j) pi pj

1≤j
1 X

! p1 |∆αk |p

1 X

! 1q |∆xk |q

= p1 p2 |α1 − α2 | |x1 − x2 | .

k=1

k=1

Therefore, from (5.72), we obtain p1 p2 |α1 − α2 | |x1 − x2 | ≤ Cp1 p2 |α1 − α2 | |x1 − x2 | for all α1 6= α2 , x1 6= x2 , p1 p2 > 0, giving C ≥ 1.



We are able now to state the following reverse of the (CBS) −inequality. ¯ = (b1 , . . . , bn ) be two sequences of real numbers with Theorem 5.34. Let ¯ a = (a1 , . . . , an ) , b ai 6= 0, (i = 1, . . . , n) . Then one has the inequality !2 n n n X X X 0≤ a2i b2i − ai b i i=1

i=1

i=1

 p ! p1 n−1  X ∆ bk ak

≤

k=1

1 p

 q n−1  X b k ∆ ak k=1

! 1q X

(i − j) a2i a2j ,

1≤j
1 q

where p > 1, + = 1. The constant C = 1 is sharp in the above sense. Proof. Follows by Lemma 5.33 for a2 pi = Pn i

2 k=1 ak

, αi = x i =

bi , i ∈ {1, . . . , n} . ai 

The following corollary is a natural consequence of Theorem 5.34 for p = q = 2.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

92

S.S. D RAGOMIR

¯ we have Corollary 5.35. With the assumptions of Theorem 5.34 for ¯ a and b, 0≤

n X i=1

a2i

n X

n X

b2i −

i=1

!2 ai b i

i=1

 2 X n−1  X b k ∆ ≤ (i − j) a2i a2j . ak 1≤j
k=1

5.12. A Reverse Inequality Via an Andrica-Badea Result. The following result is due to Andrica and Badea [15, p. 16]. Lemma 5.36. Let x ¯ = (x1 , . . . , xn ) ∈ I n = [m, M ]n be a sequence of real numbers and let S be the subset of {1, . . . , n} that minimises the expression X 1 (5.73) p i − Pn , 2 i∈S

where Pn :=

Pn

i=1

pi > 0, p ¯ = (p1 , . . . , pn ) is a sequence of nonnegative real numbers. Then



(5.74)

n 1 X 2 maxn  pi xi − x ¯∈I Pn i=1

n 1 X pi xi Pn i=1

!2  =

! X

p i Pn −

i∈S

X i∈S

pi

(M − m)2 . Pn2

Proof. We shall follow the proof in [15, p. 161]. Define n n 1 X 2 1 X Dn (¯ x, p ¯ ) := pi xi − pi xi Pn i=1 Pn i=1 1 X = pi pj (xi − xj )2 . Pn 1≤i<j≤n

!2

Keeping in mind the convexity of the quadratic function, we have Dn (α¯ x + (1 − α) y ¯, p ¯) 1 X = 2 pi pj [αxi + (1 − α) yi − αxj − (1 − α) yj ]2 Pn 1≤i<j≤n 1 X = 2 pi pj [α (xi − xj ) + (1 − α) (yi − yj )]2 Pn 1≤i<j≤n   1 X ≤ 2 pi pj α (xi − xj )2 + (1 − α) (yi − yj )2 Pn 1≤i<j≤n = αDn (¯ x, p ¯ ) + (1 − α) Dn (¯ y, p ¯) , hence Dn (·, p ¯ ) is a convex function on I n . Using a well known theorem (see for instance [16, p. 124]), we get that the maximum of Dn (·, p ¯ ) is attained on the boundary of I n .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

93


Let S, S¯ be the partition of {1, . . . , n} such that the maximum of Dn (·, p ¯ ) is obtained for 0 0 0 0 ¯ x ¯0 = (x1 , . . . , xn ) , where xi = m if i ∈ S and xi = M if i ∈ S. In this case we have (5.75)

Dn (¯ x0 , p ¯) =

1 Pn2

X

pi pj (xi − xj )2

1≤i<j≤n

! X (M − m)2 X = p i Pn − pi . Pn2 i∈S i∈S The expression ! X

p i Pn −

X

i∈S

pi

i∈S

is a maximum when the set S minimises the expression X 1 p i − Pn . 2 i∈S

From (5.75) it follows that Dn (¯ x, p ¯ ) is also a maximum and the proof of the above lemma is complete.  The following reverse result of the (CBS) −inequality holds. ¯ = (b1 , . . . , bn ) be two sequences of real numbers Theorem 5.37. Let ¯ a = (a1 , . . . , an ) and b with ai 6= 0 (i = 1, . . . , n) and (5.76)

−∞ < m ≤

bi ≤ M < ∞ for each i ∈ {1, . . . , n} . ai

Let S be the subset of {1, . . . , n} that minimizes the expression n X X 1 2 2 (5.77) ai − ai , 2 i=1

i∈S

and denote S¯ := {1, . . . , n} \ S. Then we have the inequality (5.78)

0≤

n X i=1

a2i

n X

n X

b2i −

i=1

≤ (M − m)2

!2 ai b i

i=1

X i∈S

1 ≤ (M − m)2 4

a2i

X

a2i

i∈S¯ n X

!2 a2i

.

i=1

Proof. The proof of the second inequality in (5.78) follows by Lemma 5.36 on choosing pi = a2i , xi = abii , i ∈ {1, . . . , n} .

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

94

S.S. D RAGOMIR

The third inequality is obvious as X

a2i

i∈S

X

a2i =

i∈S¯

X

a2i

=

1 4

! X

a2j −

j=1

i∈S

1 ≤ 4

n X

X

a2i +

n X

i∈S n X

!2 a2j −

j=1

i∈S

a2i X

a2i

i∈S

!2 a2j

.

j=1

 5.13. A Refinement of Cassels’ Inequality. In 1914, P. Schweitzer [18] proved the following result. Theorem 5.38. If ¯ a = (a1 , . . . , an ) is a sequence of real numbers such that 0 < m ≤ ai ≤ M < ∞ (i ∈ {1, . . . , n}) , then ! ! n n 1X 1X 1 (M + m)2 ai ≤ . (5.79) n i=1 n i=1 ai 4mM In 1972, A. Lupa¸s [17] proved the following refinement of Schweitzer’s result which gives the best bound for n odd as well. Theorem 5.39. With the assumptions in Theorem 5.38, one has n  
 n+1   
n n X X M + n+1 m M + n2 m 1 2 2 2 (5.80) ai ≤ , a M m i i=1 i=1 where [·] is the integer part. In 1988, Andrica and Badea [15] established a weighted version of Schweitzer and Lupa¸s inequalities via the use of the following weighted version of the Grüss inequality [15, Theorem 2]. Theorem 5.40. If m1 ≤ ai ≤ M1 , m2 ≤ bi ≤ M2 (i ∈ {1, . . . , n}) and S is the subset of {1, . . . , n} which minimises the expression X 1 (5.81) p i − Pn , 2 i∈S P where Pn := ni=1 pi > 0, then n n n X X X (5.82) p i ai b i − p i ai · p i bi Pn i=1 i=1 i=1 ! X X ≤ (M1 − m1 ) (M2 − m2 ) p i Pn − pi . i∈S

i∈S

1 ≤ Pn2 (M1 − m1 ) (M2 − m2 ) . 4

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

95

Proof. Using the result in Lemma 5.36, Section 5.12, we have !2 ! n n X 1 X 2 1 X (M1 − m1 )2 X (5.83) p i ai − p i ai ≤ p i Pn − pi Pn i=1 Pn i=1 Pn2 i∈S i∈S and (5.84)

n 1 X 2 p i bi − Pn i=1

n 1 X p i bi Pn i=1

!2

X (M2 − m2 )2 X P − pi ≤ p i n Pn2 i∈S i∈S

!

and since (5.85)

n n 1 X 1 X p i ai b i − p i ai · Pn i=1 Pn i=1  n 1 X 2 p i ai − ≤ Pn i=1

n 1 X p i bi Pn i=1

!2

n 1 X p i ai Pn i=1

!2  

n X  1 pi b2i − Pn i=1

n 1 X p i bi Pn i=1

!2  ,

the first part of (5.82) holds true. The second part follows by the elementary inequality 1 (a + b)2 , a, b ∈ R 4 P P for the choices a := i∈S pi , b := Pn − i∈S pi . ab ≤



We are now able to state and prove the result of Andrica and Badea [15, Theorem 4], which is related to Schweitzer’s inequality. Theorem 5.41. If 0 < m ≤ ai ≤ M < ∞, i ∈ {1, . . . , n} and S is a subset of {1, . . . , n} that minimises the expression X Pn pi − , 2 i∈S

then we have the inequality ! n ! ! n 2 X X X pi X (M − m) (5.86) p i ai ≤ Pn2 + p i Pn − pi a Mm i=1 i=1 i i∈S i∈S ≤

(M + m)2 2 Pn . 4M m

Proof. We shall follow the proof in [15]. We obtain from Theorem 5.39 with bi = a1i , m1 = m, M1 = m, m2 = M1 , M2 = m1 , the following estimate !   n n X X pi 1 1 X 2 X p i ai − p i Pn − pi , Pn − ≤ (M − m) a m M i∈S i=1 i=1 i i∈S that leads, in a simple manner, to (5.86).



We may now prove the following reverse result for the weighted (CBS) − inequality that improves the additive version of Cassels’ inequality.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

96

S.S. D RAGOMIR

¯ = (b1 , . . . , bn ) be two sequences of positive real numTheorem 5.42. Let ¯ a = (a1 , . . . , an ) , b bers with the property that (5.87)

0<m≤

bi ≤ M < ∞ for each i ∈ {1, . . . , n} , ai

and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers such that Pn := S is a subset of {1, . . . , n} that minimises the expression n X X 1 (5.88) p a b − p a b i i i i i i 2

Pn

i=1

pi > 0. If

i=1

i∈S

then one has the inequality (5.89)

n X i=1

pi a2i

n X

pi b2i −

i=1

n X

!2 p i ai b i

i=1

n X X (M − m)2 X p i ai b i ≤ p i ai b i − p i ai b i Mm i=1 i∈S i∈S !2 n (M − m)2 X pi ai bi . ≤ 4M m i=1

!

Proof. Applying Theorem 5.41 for ai = xi , pi = qi xi we may deduce the inequality !2 ! n n n n 2 X X X X X X (M − m) (5.90) qi x2i qi − qi x i ≤ qi x i qi x i − qi x i , M m i=1 i=1 i=1 i=1 i∈S i∈S P provided qi ≥ 0, ni=1 qi > 0, 0 < m ≤ xi ≤ M < ∞, for i ∈ {1, . . . , n} and S is a subset of {1, . . . , n} that minimises the expression n X 1X (5.91) qi x i − qi x i . 2 i=1

i∈S

Now, if in (5.90) we choose qi = pi a2i , xi = desired result (5.89).

bi ai

∈ [m, M ] for i ∈ {1, . . . , n} , we deduce the 

The following corollary provides a refinement of Cassels’ inequality. Corollary 5.43. With the assumptions of Theorem 5.42, we have the inequality Pn Pn 2 2 i=1 pi ai i=1 pi bi (5.92) 1≤ P 2 ( ni=1 pi ai bi ) P P   (M − m)2 i∈S pi ai bi i∈S pi ai bi ≤1+ · Pn 1 − Pn Mm i=1 pi ai bi i=1 pi ai bi (M + m)2 ≤ . 4M m The case of the “unweighted” Cassels’ inequality is embodied in the following corollary as well.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

97

¯ satisfy (5.88). If S is a subset of {1, . . . , n} that minimises Corollary 5.44. Assume that ¯ a and b the expression n X X 1 (5.93) ai b i − ai b i 2 i∈S

i=1

then one has the inequality Pn 2 Pn 2 ai i=1 bi (5.94) 1 ≤ i=1 Pn 2 ( i=1 ai bi ) P P   (M − m)2 i∈S ai bi i∈S ai bi ≤1+ · Pn 1 − Pn Mm i=1 ai bi i=1 ai bi (M + m)2 ≤ . 4M m In particular, we may obtain the following refinement of the Pólya-Szegö’s inequality. Corollary 5.45. Assume that (5.95)

0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ for i ∈ {1, . . . , n} .

If S is a subset of {1, . . . , n} that minimises the expression (5.93), then one has the inequality Pn 2 Pn 2 ai i=1 bi (5.96) 1 ≤ i=1 Pn 2 ( i=1 ai bi ) P P   (AB − ab)2 i∈S ai bi i∈S ai bi ≤1+ · Pn 1 − Pn abAB i=1 ai bi i=1 ai bi (AB + ab)2 ≤ . 4abAB 5.14. Two Reverse Results Via Diaz-Metcalf Results. In [19], J.B. Diaz and F.T. Metcalf proved the following inequality for sequences of complex numbers. ¯ = (b1 , . . . , bn ) be sequences of complex numbers Lemma 5.46. Let ¯ a = (a1 , . . . , an ) and b such that ak 6= 0, k ∈ {1, . . . , n} and         bk bk bk bk (5.97) m ≤ Re + Im ≤ M, m ≤ Re − Im ≤ M, ak ak ak ak where m, M ∈ R and k ∈ {1, . . . , n} . Then one has the inequality " n # n n X X X (5.98) |bk |2 + mM |ak |2 ≤ (m + M ) Re ak¯bk k=1

k=1

k=1 n X ¯ ≤ |M + m| ak bk . k=1

Using the above result we may state and prove the following reverse inequality.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

98

S.S. D RAGOMIR

¯ are as in (5.97) and m, M > 0, then one has the inequality Theorem 5.47. If ¯ a and b n X

n X (M + m)2 2 |ak | |bk |2 ≤ 4mM k=1 k=1

(5.99)

Re

n X

!2 ak¯bk

k=1

n 2 (M + m) X ¯ ≤ ak b k . 4mM k=1 2

Proof. Using the elementary inequality αp2 +

1 2 q ≥ 2pq, α > 0, p, q ≥ 0 α

we have √ (5.100)

n X

n n n X X 1 X 2 2 2 mM |ak | + √ |bk | ≥ 2 |ak | |bk |2 mM k=1 k=1 k=1 k=1

! 12 .

On the other hand, by (5.98), we have

(5.101)

" n # n n X X √ 1 X (M + m) √ |bk |2 + mM |ak |2 ≤ √ Re ak¯bk mM k=1 mM k=1 n k=1 M + m X ¯ ≤ √ ak b k . mM k=1 

Combining (5.100) and (5.101), we deduce the desired result (5.99). The following corollary is a natural consequence of the above lemma. ¯ and m, M satisfy the hypothesis of Theorem 5.47, then Corollary 5.48. If ¯ a and b

(5.102)

0≤

≤

≤

≤

! 21

n X ¯ − ai b i |ai | |bi | i=1 i=1 i=1 ! ! 12 n n n X X X 2 2 ¯ − Re ak b k |ai | |bi | i=1 i=1 k=1 √  √ 2 ! n M − m X ¯ √ ak bk Re 2 mM k=1 √  √ 2 n M − m X ¯ √ ak b k 2 mM n X

2

n X

2

k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

99

and 2 n n X X 2 2 ¯ 0≤ |ai | |bi | − ai b i i=1 i=1 i=1 ! 2 n n n X X X ≤ |ai |2 |bi |2 − Re ai¯bi i=1 i=1 i=1 ! 2 n X (M − m)2 ¯ ai b i ≤ Re 4mM i=1 n 2 (M − m)2 X ¯ ai b i . ≤ 4mM n X

(5.103)

(5.104)

(5.105)

(5.106)

i=1

Another result obtained by Diaz and Metcalf in [19] is the following one. ¯ m and M be complex numbers such that Lemma 5.49. Let ¯ a, b,     bk bk (5.107) Re (m) + Im (m) ≤ Re + Im ≤ Re (M ) + Im (M ) ; ak ak     bk bk Re (m) − Im (m) ≤ Re − Im ≤ Re (M ) − Im (M ) ; ak ak for each k ∈ {1, . . . , n} . Then (5.108)

n X

2

¯ |bk | + Re mM

n
X

k=1

k=1

" 2

|ak | ≤ Re (M + m)

n X

# ak¯bk

k=1 n X ≤ |M + m| ak¯bk . k=1

The following reverse result for the (CBS) −inequality may be stated as well.
¯ > 0, then we have the Theorem 5.50. With the assumptions in Lemma 5.49, and if Re mM inequality: " n # 21   P n X X Re (M + m) nk=1 ak¯bk 2 2 (5.109) |ak | |bk | ≤ 
 1 ¯ 2 2 Re mM k=1 k=1 P |M + m| nk=1 ak¯bk . ≤ 
 1 ¯ 2 2 Re mM The proof is similar to the one in Theorem 5.47 and we omit the details. Remark 5.51. Similar additive versions may be stated. They are left as an exercise for the interested reader. ˇ 5.15. Some Reverse Results Via the Cebyšev Functional. For x¯= (x1 , . . . , xn ) , y¯= (y1 , . . . , yn ) two sequences of real numbers and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers Pn ˇ with i=1 pi = 1, define the Cebyšev functional (5.110)

Tn (¯ p; x ¯, y ¯) :=

n X i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

pi xi yi −

n X i=1

pi xi

n X

pi yi .

i=1

http://jipam.vu.edu.au/

100

S.S. D RAGOMIR

For x¯ and p ¯ as above consider the norms: k¯ xk∞ := max |xi | i=1,n

n X

k¯ xkp¯,α :=

! α1 pi |xi |α

, α ∈ [1, ∞).

i=1

The following result holds [20]. Theorem 5.52. Let x¯, y¯, p ¯ be as above and c¯= (c, . . . , c) a constant sequence with c ∈ R. Then one has the inequalities (5.111)

0 ≤ |Tn (¯ p; x ¯, y ¯)|  k¯ y−y ¯µ,p kp¯,1 · inf k¯ x−¯ c k∞ ;   c∈R      k¯ y−y ¯µ,p kp¯,β · inf k¯ x−¯ ckp¯,α , α > 1, α1 + ≤ c∈R        k¯ y−y ¯µ,p k∞ · inf k¯ x−¯ ckp¯,1 ;

1

= 1;

β

c∈R

  k¯ y−y ¯µ,p kp¯,1 min k¯ xk∞ , k¯ x−x ¯µ,p k∞ ;       n o  k¯ y−y ¯µ,p kp¯,β min k¯ xkp¯,α , k¯ x−x ¯µ,p kp¯,α , ≤  1  α > 1, α + β1 =o1;   n    k¯ y−y ¯µ,p k∞ · min k¯ xkp¯,1 , k¯ x−x ¯µ,p kp¯,1 ; where xµ,p :=

n X

pi xi ,

i=1

yµ,p :=

n X

pi yi

i=1

and x¯µ,p , y¯µ,p are the sequences with all components equal to xµ,p , yµ,p . Proof. Firstly, let us observe that for any c ∈ R, one has Sonin’s identity (5.112)

Tn (¯ p; x ¯, y ¯) = Tn (¯ p; x ¯−¯ c, y ¯−y ¯µ,p ) =

n X

pi (xi − c) yi −

i=1

n X

! p j yj

.

j=1

Taking the modulus and using Hölder’s inequality, we have (5.113)

|Tn (¯ p; x ¯, y ¯)| ≤

n X

pi |xi − c| |yi − yµ,p |

i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

101

P  max |xi − c| ni=1 pi |yi − yµ,p |   i=1,n       P 1 1 Pn α α n β β ≤ ( i=1 pi |xi − c| ) , i=1 pi |yi − yµ,p |   1 1   α > 1, α + β = 1;  Pn    i=1 pi |xi − c| max |yi − yµ,p | i=1,n

 k¯ x−¯ ck∞ k¯ y−y ¯µ,p kp¯,1 ;      k¯ x−¯ ckp¯,α k¯ y−y ¯µ,p kp¯,β , α > 1, α1 + =      k¯ x−¯ ckp¯,1 k¯ y−y ¯µ,p k∞ .

1 β

= 1;

Taking the inf over c ∈ R in (5.113), we deduce the second inequality in (5.111). Since  xkp¯,α ,  k¯ inf k¯ x−¯ ckp¯,α ≤ for any α ∈ [1, ∞] c∈R  k¯ x−x ¯µ,p kp¯,α 

the last part of (5.110) is also proved. For p ¯ and x¯ as above, define Tn (¯ p; x ¯) :=

n X

pi x2i −

i=1

n X

!2 pi xi

.

i=1

The following corollary holds [20]. Corollary 5.53. With the above assumptions we have (5.114)

0 ≤ |Tn (¯ p; x ¯)|  k¯ x−x ¯µ,p kp¯,1 · inf k¯ x−¯ c k∞ ;   c∈R      k¯ x−x ¯µ,p kp¯,β · inf k¯ x−¯ ckp¯,α , α > 1, α1 + ≤ c∈R        k¯ x−x ¯µ,p k∞ · inf k¯ x−¯ ckp¯,1 ;

1 β

= 1;

c∈R

  k¯ x−x ¯µ,p kp¯,1 min k¯ xk∞ , k¯ x−x ¯µ,p k∞ ;       n o  k¯ x − x ¯ k min k¯ x k , k¯ x − x ¯ k µ,p p µ,p p ≤ p ¯ ,α ¯ ,β ¯ ,α ,   α > 1, α1 + β1 =o1;   n    k¯ x−x ¯µ,p k∞ · min k¯ xkp¯,1 , k¯ x−x ¯µ,p kp¯,1 . Remark 5.54. If pi := n1 , i = 1, . . . , n, then from Theorem 5.52 and Corollary 5.53 we recapture the results in [22]. The following reverse of the (CBS) −inequality holds [20].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

102

S.S. D RAGOMIR

¯ be two sequences of real numbers with ai 6= 0, i ∈ {1, . . . , n} . Then Theorem 5.55. Let ¯ a, b one has the inequality !2 n n n X X X (5.115) ai b i 0≤ a2i b2i − i=1

i=1 i=1  n " #  n X bi X a a ≤ inf max − c |ai | ak k i bk bi c∈R i=1,n ai i=1 k=1  bi   max  " # ai  n n X  i=1,n X ak ai × ≤ |ai | ak Pn bk bi  bi  a b i=1 k=1  k k k=1 .   max − Pn a2 i=1,n ai k=1

k

Proof. By Corollary 5.53, we may state that (5.116)

0 ≤ Tn (¯ p; x ¯) ≤ k¯ x−x ¯µ,p kp¯,1 · inf k¯ x−¯ c k∞ c∈R  xk∞ ,  k¯ ≤ k¯ x−x ¯µ,p kp¯,1 ×  k¯ x−x ¯µ,p k∞ .

For the choices

a2 pi = Pn i

2 k=1 ak

, xi =

bi , i = 1, . . . , n; ai

we get Pn Tn (¯ p; x ¯) =

i=1

a2i

Pn

Pn 2 i=1 bi − ( i=1 Pn 2 2 ( k=1 ak )

2

ai b i )

,

n X k¯ x−x ¯µ,p kp¯,1 = pi xi − pj xj i=1 j=1 n n X X 1 1 2 bi = Pn ai − Pn aj b j 2 2 ai k=1 ak i=1 k=1 ak j=1 n n n X X X 1 2 2 = Pn a b a − a a b i i j j k i 2 2 ( k=1 ak ) i=1 j=1 k=1 n n X X 1 = Pn |ai | ak (ak bi − ai bk ) 2 2 ( k=1 ak ) i=1 k=1 n n X X ak ai 1 , = Pn |ai | ak 2 bk bi ( k=1 a2k ) i=1 k=1 bi bi k¯ x−¯ ck∞ = max − c , k¯ xk∞ = max i=1,n ai i=1,n ai Pn b i j=1 aj bj k¯ x−x ¯µ,p k∞ = max − Pn . 2 i=1,n ai k=1 ak n X

and

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

103



Utilising the inequality (5.116) we deduce the desired result (5.115). The following result also holds [20]. Theorem 5.56. With the assumption in Theorem 5.55 and if α > 1, the inequality: !2 n n n X X X 2 2 (5.117) 0≤ ai bi − ai b i i=1

i=1

1 α

+

1 β

= 1, then we have

i=1

 β1 n ! α1 β n n X X X a a |ai |2−β ≤ |ai |2−α |bi − cai |α ak k i  inf b b c∈R k i 



i=1

 ≤

n X i=1

i=1

k=1

1 β



 n β X a a |ai |2−β ak k i  bk bi k=1    α1 n  P  2−α α  |ai | |bi |    i=1 ×   1    Pn ak ai α α P 1  n 2−α  Pn .  i=1 |ai | 2 k=1 ak bk bi k=1 ak

Proof. By Corollary 5.53, we may state that 0 ≤ Tn (¯ p; x ¯)

(5.118)

≤ k¯ x−x ¯µ,p kp¯,β · inf k¯ x−¯ ckp¯,α c∈R  xkp¯,α ,  k¯ ≤ k¯ x−x ¯µ,p kp¯,β ×  k¯ x−x ¯µ,p kp¯,α , for α > 1, α1 + β1 = 1. For the choices

a2 pi = Pn i

2 k=1 ak

, xi =

bi , i = 1, . . . , n; ai

we get  k¯ x−x ¯µ,p kp¯,β = 

n X i=1

 =

n X i=1

β  β1 n X pi xi − pj xj  j=1

Pn β  β1 Pn 2 bi k=1 ak − ai j=1 aj bj a2 Pn i 2 Pn  2 a a a i k=1 k k=1 k  β  β1 n n X X a a 2−β  |ai | ak k i  , bk bi

k¯ x−¯ ckp¯,α =

1 = P 1+ 1 ( nk=1 a2k ) β i=1 ! α1 n X 1 α = P pi |xi − c| 1 n ( k=1 a2k ) α i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

k=1

n X

! α1 2−α

|ai |

α

|bi − cai |

,

i=1

http://jipam.vu.edu.au/

104

S.S. D RAGOMIR

1

k¯ xkp¯,α = P 1 ( nk=1 a2k ) α

n X

|ai |

α

|bi |

i=1

and n X

1

! α1 2−α

n α ! α1 X a a . |ai |2−α ak k i bk bi

k¯ x−x ¯µ,p kp¯,α = P 1+ 1 ( nk=1 a2k ) α i=1 k=1 Utilising the inequality (5.118), we deduce the desired result (5.117).



Finally, the following result also holds [20]. Theorem 5.57. With the assumptions in Theorem 5.55 we have the following reverse of the (CBS) −inequality: !2 n n n X X X (5.119) 0≤ a2i b2i − ai b i i=1

i=1 i=1n " n # n b X X X i a2k − aj bj inf ≤ max |ai | |bi − cai | c∈R i=1,n ai j=1 i=1 k=1 n n b X X i a2k − aj b j ≤ max i=1,n ai j=1 k=1  n P   |ai bi |    i=1 ×  n n P ak ai  P 1  .  |ai | ak  Pn 2 b b a k i i=1 k=1 k=1 k

Proof. By Corollary 5.53, we may state that (5.120)

0 ≤ Tn (¯ p; x ¯) ≤ k¯ x−x ¯µ,p k∞ · inf k¯ x−¯ ckp¯,1 c∈R  xkp¯,1 ,  k¯ ≤ k¯ x−x ¯µ,p k∞  k¯ x−x ¯µ,p kp¯,1 .

For the choices

a2 pi = Pn i

2 k=1 ak

, xi =

bi , i = 1, . . . , n; ai

we get k¯ x−x ¯µ,p k∞

n X pj xj = max xi − i=1,n j=1 Pn b a b j j i j=1 = max − Pn 2 i=1,n ai k=1 ak n n X X 1 bi 2 P = n max ak − aj b j , 2 k=1 ak i=1,n ai k=1 j=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

k¯ x−¯ ckp¯,1 = =

n X i=1 n X i=1

= Pn

pi |xi − c| a2i

bi − c Pn 2 k=1 ak ai n X

1

k=1

k¯ xkp¯,1 =

n X

pi |xi | =

i=1

n X i=1

105

a2k

|ai | |bi − cai | ,

i=1

n X bi 1 Pn = Pn |ai bi | 2 2 k=1 ak ai k=1 ak i=1 a2i

and 1 k¯ x−x ¯µ,p kp¯,1 = Pn 2 ( k=1 a2k )

n X i=1

n X a a |ai | ak k i . bk bi k=1



Utilising the inequality (5.120) we deduce (5.119).

5.16. Another Reverse Result via a Grüss Type Result. The following Grüss type inequality has been obtained in [21]. ¯ = (b1 , . . . , bn ) be two sequences of real numbers and Lemma 5.58. Let ¯ a = (a1 , . . . , an ), b assume that there are γ, Γ ∈ R such that −∞ < γ ≤ ai ≤ Γ < ∞ for each i ∈ {1, . . . , n} .

(5.121)

P Then for any p ¯ = (p1 , . . . , pn ) a nonnegative sequence with the property that ni=1 pi = 1, one has the inequality n n n n n X 1 X X X X (5.122) p i bi − p k bk . p i ai b i − p i ai pi bi ≤ (Γ − γ) 2 i=1 i=1 i=1 i=1 k=1 The constant

1 2

is sharp in the sense that it cannot be replaced by a smaller constant.

Proof. We will give here a simpler direct proof based on Sonin’s identity. A simple calculation shows that: !   n n n n n X X X X X γ+Γ (5.123) p i ai b i − p i ai p i bi = p i ai − bi − p k bk . 2 i=1 i=1 i=1 i=1 k=1 By (5.121) we have Γ−γ γ + Γ ai − ≤ for all i ∈ {1, . . . , n} 2 2 and thus, by (5.123), on taking the modulus, we get n n n n n X X X X X γ + Γ p i ai b i − p i ai p i bi ≤ p i ai − p k bk bi − 2 i=1 i=1 i=1 i=1 k=1 n n X X 1 ≤ (Γ − γ) p i bi − p k bk . 2 i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

k=1

http://jipam.vu.edu.au/

106

S.S. D RAGOMIR

To prove the sharpness of the constant 12 , let us assume that (5.122) holds with a constant c > 0, i.e., n n n n n X X X X X (5.124) p i ai b i − p i ai pi bi ≤ c (Γ − γ) p i bi − p k bk . i=1

i=1

i=1

i=1

k=1

provided ai satisfies (5.121). If we choose n = 2 in (5.124) and take into account that 2 X

p i ai b i −

2 X

p i ai

i=1

i=1

2 X

pi bi = p1 p2 (a1 − a2 ) (b1 − b2 )

i=1

provided p1 + p2 = 1, p1 , p2 ∈ [0, 1] , and since 2 2 X X p i bi − pk bk = p1 |(p1 + p2 ) b1 − p1 b1 − p2 b2 | + p2 |(p1 + p2 ) b2 − p1 b1 − p2 b2 | i=1

k=1

= 2p1 p2 |b1 − b2 | we deduce by (5.124) p1 p2 |a1 − a2 | |b1 − b2 | ≤ 2c (Γ − γ) |b1 − b2 | p1 p2 .

(5.125)

If we assume that p1 , p2 6= 0, b1 6= b2 and a1 = Γ, a2 = γ, then by (5.125) we deduce c ≥ 21 , which proves the sharpness of the constant 21 .  The following corollary is a natural consequence of the above lemma. Corollary 5.59. Assume that ¯ a = (a1 , . . . , an ) satisfies the assumption (5.121) and p ¯ is a probability sequence. Then !2 n n n n X X X X 1 (5.126) 0≤ pi a2i − p i ai ≤ (Γ − γ) p i ai − p k ak . 2 i=1 i=1 i=1 k=1 The constant

1 2

is best possible in the sense mentioned above.

The following reverse of the (CBS) −inequality may be stated. Theorem 5.60. Assume that x ¯ = (x1 , . . . , xn ) and y ¯ = (y1 , . . . , yn ) are sequences of real numbers with yi 6= 0 (i = 1, . . . , n) . If there exists the real numbers m, M such that m≤

(5.127)

xi ≤ M for each i ∈ {1, . . . , n} , yi

then we have the inequality (5.128)

0≤

n X

x2i

i=1

n X i=1

yi2 −

n X

!2 xi yi

i=1

n n X X xi yi 1 . ≤ (M − m) |yi | yk · xk y k 2 i=1 k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

Proof. If we choose pi = deduce Pn 2 x Pni=1 i2 − k=1 yk

y2 Pn i

k=1

yk2

, ai = n X

1 Pn

k=1

yk2

xi yi

107

for i = 1, . . . , n and γ = m, Γ = M in (5.126), we !2

xi yi

i=1

n n x X X 1 1 1 i 2 ≤ (M − m) Pn yi − Pn xk yk 2 2 yi 2 k=1 yk i=1 k=1 yk k=1 n n n X X X 1 1 2 x y = (M − m) Pn |y | x y − y i k k i i k 2 2 2 ( k=1 yk ) i=1 k=1 k=1 n n X X xi yi 1 1 . = (M − m) Pn |yi | yk · 2 2 x y 2 k k ( k=1 yk ) i=1 k=1 

giving the desired inequality (5.128).

R EFERENCES [1] R. FRUCHT, Sobre algunas disigualdades: Observación relative a la solución del Problema No. 21, indicada par el Ing. Ernesto M. Saleme (in Spanish), Math. Notae, 3 (1943), 41–46. [2] G.S. WATSON, Serial Correlation in Regression Analysis, Ph.D. Thesis, Dept. of Experimental Statistics, North Carolina State College, Raleigh; Univ. of North Carolina, Mimograph Ser., No. 49, 1951. [3] G.S. WATSON, Serial correlation in regression analysis, I, Biometrika, 42 (1955), 327–341. [4] G.S. WATSON, A method for discovering Kantorovich-type inequalities and a probabilistic interpretation, Linear Algebra Appl., 97 (1987), 211–217. [5] G.S. WATSON, G. ALPARGU and G.P.H. STYAN, Some comments on six inequalities associated with the inefficiency of ordinary least squares with one regressor, Linear Algebra and its Appl., 264 (1997), 13–54. [6] G. PÓLYA and G. SZEGÖ, Aufgaben und Lehrsütze ans der Analysis, Band I: Reihen, Integralrechnung, Funktiontheorie (in German), 4th Ed., Springer Verlag, Berlin, 1970 (Original version: Julius Springer, Berlin, 1925). [7] G. PÓLYA and G. SZEGÖ, Problems and Theorems in Analysis, Volume 1: Series, Integral Calculus, Theory of Functions (in English), translated from german by D. Aeppli, corrected printing of the revised translation of the fourth German edition, Springer Verlag, New York, 1972. [8] W. GREUB AND W. RHEINBOLDT, On a generalisation of an inequality of L.V. Kantorovich, Proc. Amer. Math. Soc., 10 (1959), 407–415. [9] S.S. DRAGOMIR, A generalisation of Cassel’s and Greub-Reinboldt’s inequalities in inner product spaces, RGMIA Res. Rep. Coll., 6 (2003), Supplement. [ONLINE: http://rgmia.vu.edu. au/v6(E).html] [10] S.S. DRAGOMIR, A counterpart of Schwartz’s inequality in inner product spaces, RGMIA Res. Rep. Coll., 6 (2003), Supplement, Article 18. [ONLINE: http://rgmia.vu.edu.au/v6(E) .html] [11] S.S. DRAGOMIR AND G.L. BOOTH, Grüss-Lupa¸s type inequality and its applications for the estimation of p-moments of guessing mappings, Math. Comm., 5 (2000), 117–126.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

108

S.S. D RAGOMIR

[12] S.S. DRAGOMIR, Some counterpart inequalities in terms of forward difference for (CBS) −inequality, submitted. [13] S.S. DRAGOMIR, A Grüss type inequality for sequences of vectors in normed linear spaces and applications, submitted. [14] S.S. DRAGOMIR, Another Grüss type inequality for sequences of vectors in normed linear spaces and applications, J. Comp. Anal. & Applic., 4(2) (2002), 155–172. [15] D. ANDRICA AND C. BADEA, Grüss’ inequality for positive linear functionals, Periodica Mat. Hungarica, 19(2) (1988), 155–167. [16] A.W. ROBERTS AND D.E. VARBERG, Convex Functions, Academic Press, New York, 1973. [17] A. LUPAS, ¸ A remark on the Schweitzer and Kantorovich inequalities, Publ. Elek. Fak. Univ. Beograde, Ser. Mat. i Fiz., 383 (1972), 13–14. [18] P. SCHWEITZER, An inequality about the arithmetic mean (Hungarian), Math. Phys. Lapok (Budapest), 23 (1914), 257–261. [19] J.B. DIAZ AND F.T. METCALF, Stronger forms of a class of inequalities of G. Pólya – G. Szegö and L.V. Kantorovich, Bull. Amer. Math. Soc., 69 (1963), 415–418. ˇ [20] P. CERONE AND S.S. DRAGOMIR, New inequalities for the Cebyšev functional involving two n−tuples of real numbers and applications, RGMIA Res. Rep. Coll., 5(3) (2002), Article 4. [ONLINE: http://rgmia.vu.edu.au/v5n3.html] [21] P. CERONE AND S.S. DRAGOMIR, A refinement of the Grüss inequality and applications, RGMIA Res. Rep. Coll., 5(2) (2002), Article 2. [ONLINE: http://rgmia.vu.edu.au/v5n2.html] ˇ [22] S.S. DRAGOMIR, New inequalities for the weighted Cebyšev functionals and applications for the (CBS) −inequality, in preparation. [23] O. SHISHA AND B. MOND, Bounds on differences of means, Inequalities, Academic Press Inc., New York, 1967, 293–308. [24] D. ZAGIER, A converse to Cauchy’s inequality, Amer. Math. Month., 102(10) (1995), 919–920. [25] D. ZAGIER, An inequality converse to that of Cauchy (Dutch), Indag. Math., 39(4) (1997), 349– 351. ˇ ´ The inequalities of Zagier and Cebyšev, ˇ [26] J. PECARI C, Arch. Math., (Basel), 64(5) (1995), 415–417. [27] H. ALZER, On a converse Cauchy inequality of D. Zagier, Arch. Math., (Basel), 58(2) (1995), 157–159.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

109

6. R ELATED I NEQUALITIES 6.1. Ostrowski’s Inequality for Real Sequences. In 1951, A.M. Ostrowski [2, p. 289] gave the following result related to the (CBS) −inequality for real sequences (see also [1, p. 92]). ¯ = (b1 , . . . , bn ) be two non-proportional sequences Theorem 6.1. Let ¯ a = (a1 , . . . , an ) and b of real numbers. Let x ¯ = (x1 , . . . , xn ) be a sequence of real numbers such that n X

(6.1)

ai xi = 0 and

n X

i=1

bi xi = 1.

i=1

Then (6.2)

n X

x2i

Pn

≥ Pn

i=1

i=1

2 i=1 ai Pn 2 b − ( i=1 i i=1

Pn 2

ai

2

ai b i )

with equality if and only if bk

Pn

a2i − ak

Pn

i=1 ai bi P P 2 n n 2 2 i=1 ai i=1 bi − ( i=1 ai bi ) i=1

xk = Pn

(6.3) for any k ∈ {1, . . . , n} .

Proof. We shall follow the proof in [1, p. 93 – p. 94]. Let n n n X X X 2 2 (6.4) A= ai , B = bi , C = ai b i i=1

i=1

i=1

and (6.5)

yi =

Abi − Cai for any i ∈ {1, . . . , n} . AB − C 2

It is easy to see that the sequence y ¯ = (y1 , . . . , yn ) as defined by (6.5) satisfies (6.1). Any sequence x ¯ = (x1 , . . . , xn ) that satisfies (6.1) fulfills the equality n X

xi yi =

n X

i=1

xi ·

i=1

(Abi − Cai ) A = ; 2 AB − C AB − C 2

so, in particular n X

A . AB − C 2

yi2 =

i=1

Any sequence x ¯ = (x1 , . . . , xn ) that satisfies (6.1) therefore satisfies (6.6)

n X

x2i −

i=1

and thus

n X

yi2 =

i=1 n X i=1

x2i

≥

n X

(xi − yi )2 ≥ 0,

i=1 n X i=1

yi2 =

A AB − C 2

and the inequality (6.2) is proved. From (6.6) it follows that equality holds in (6.1) iff xi = yi for each i ∈ {1, . . . , n}, and the theorem is completely proved. 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

110

S.S. D RAGOMIR

6.2. Ostrowski’s Inequality for Complex Sequences. The following result that points out a natural generalisation of Ostrowski’s inequality for complex numbers holds [3]. ¯ = (b1 , . . . , bn ) and x Theorem 6.2. Let ¯ a = (a1 , . . . , an ) , b ¯ = (x1 , . . . , xn ) be sequences of
¯ ¯ ¯ ¯ complex numbers. If ¯ a and b, where b = b1 , . . . , bn , are not proportional and n X

(6.7)

i=1 n X

(6.8)



then one has the inequality n X (6.9) |xi |2 ≥ Pn i=1

i=1

xi a ¯i = 0;

¯ xi bi = 1, Pn

i=1

i=1

|ai |2

Pn

i=1

|ai |2

P 2 |bi |2 − ni=1 ai¯bi

with equality iff "

# Pn b a ¯ k k xi = µ bi − Pnk=1 2 · ai , i ∈ {1, . . . , n} |a | k k=1

(6.10) and µ ∈ C with

Pn

k=1

|µ| = Pn

(6.11)

k=1

|ak |2

Pn

k=1

|ak |2

P 2 . |bk |2 − nk=1 ak¯bk

Proof. Recall the (CBS) −inequality for complex sequences n 2 n n X X X |uk |2 |vk |2 ≥ uk v¯k (6.12) k=1

k=1

k=1

with equality iff there is a complex number α ∈ C such that uk = αvk , k = 1, . . . , n.

(6.13) If we apply (6.12) for

Pn zi c¯i uk = zk − Pni=1 2 · ck , |ci | Pi=1 n di c¯i ¯ ¯ vk = dk − Pni=1 2 · ck , where ¯ c 6= 0 and ¯ c, d, z ∈ Cn , |c | i=1 i we have (6.14)

2 n 2 Pn Pn n X X z c ¯ d c ¯ i i i i zk − Pni=1 2 · ck dk − Pni=1 2 · ck i=1 |ci | i=1 |ci | k=1 k=1 ! ! 2 Pn Pn n X z c ¯ d c ¯ i i i i i=1 i=1 ≥ zk − Pn · ck dk − Pn · ck 2 2 i=1 |ci | i=1 |ci | k=1

with equality iff there is a β ∈ C such that Pn zi c¯i (6.15) zk = Pni=1 2 · ck + β i=1 |ci |

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

! Pn d c ¯ i i dk − Pni=1 2 · ck . i=1 |ci |

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

111

Since a simple calculation shows that 2 P Pn Pn n 2 n 2 Pn 2 X z c ¯ ¯k | i=1 i i k=1 |zk | k=1 |ck | − | k=1 zk c · c = , zk − Pn k Pn 2 2
2 |c | i |c | i=1 k=1 k=1 k 2 P P P Pn n 2 n n 2 n 2 X ¯i ¯k | i=1 di c k=1 |dk | k=1 |ck | − | k=1 dk c dk − Pn
2 Pn 2 · ck = |ck |2 i=1 |ci | k=1

k=1

and n X k=1

! ! Pn Pn ¯i ¯i i=1 zi c i=1 di c zk − Pn dk − Pn 2 · ck 2 · ck i=1 |ci | i=1 |ci | Pn ¯ Pn |ck |2 − Pn zk c¯k · Pn ck d¯k k=1 zk dk · k=1 k=1 k=1 = Pn 2
2 i=1 |ci |

then by (6.12) we deduce  n 2   n n 2  n n n X X X X  X X 2 2 2 2  (6.16) |zk | |ck | − zk c¯k |dk | |ck | − dk c¯k  k=1 k=1 k=1 k=1 k=1 k=1 2 n n n n X X X X ≥ |ck |2 − zk c¯k ck d¯k zk d¯k · k=1

k=1

k=1

k=1

with equality iff there is a β ∈ C such that (6.15) holds. ¯ satisfy (6.7) and (6.8), then by (6.16) and (6.15) for the choices ¯ If ¯ a, x ¯, b z=x ¯, ¯ c=¯ a and ¯ ¯ d = b, we deduce (6.9) with equality iff there is a µ ∈ C such that ! Pn ¯ a b i i xk = µ bk − Pni=1 2 · ak , i=1 |ai | and, by (6.8), ! Pn n X ¯ a b i=1 i i ¯ µ b − · a · b Pn k k k = 1. 2 i=1 |ai |

(6.17)

k=1

Since (6.17) is clearly equivalent to (6.15), the theorem is completely proved.



6.3. Another Ostrowski’s Inequality. In his book from 1951, [2, p. 130], A.M. Ostrowski proved the following inequality as well (see also [1, p. 94]). ¯ x Theorem 6.3. Let ¯ a, b, ¯ be sequences of real numbers so that ¯ a 6= 0 and n X (6.18) x2k = 1 k=1 n X

ak xk = 0.

(6.19)

k=1

Then Pn (6.20)

2 k=1 ak

P 2 b2n − ( nk=1 ak bk ) k=1 Pn ≥ 2 k=1 ak

Pn

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

n X

!2 bk x k

.

k=1

http://jipam.vu.edu.au/

112

S.S. D RAGOMIR

¯ are non-proportional, then equality holds in (6.20) iff If ¯ a and b P P bk ni=1 a2i − ak ni=1 ai bi (6.21) xk = q · hP i1 , 1 Pn Pn 2 Pn 2 2 n 2 2 2 ( k=1 ak ) i=1 ai bi ) i=1 ai i=1 bi − ( k ∈ {1, . . . , n} , q ∈ {−1, 1} . We may extend this result for sequences of complex numbers as follows [4]. ¯ x ¯ are not proporTheorem 6.4. Let ¯ a, b, ¯ be sequences of complex numbers so that ¯ a 6= 0, ¯ a, b tional, and n X |xk |2 = 1 (6.22) k=1 n X

(6.23)

xk a ¯k = 0.

k=1

Then Pn

2 k=1 |ak |

(6.24)

2 Pn 2 n 2 ¯ X |b | − a b k=1 k k=1 k k ¯ ≥ x b Pn k k . 2 k=1 |ak | k=1

Pn

The equality holds in (6.24) iff xk = β

(6.25)

! Pn b a ¯ i i bk − Pni=1 2 · ak , i=1 |ai |

k ∈ {1, . . . , n} ;

where β ∈ C is such that Pn |β| =  Pn

(6.26)

2 k=1 |ak |

2

k=1

|ak |


21

Pn 2  12 ¯ |b | − a b k=1 k k=1 k k

Pn

.

2

Proof. In Subsection 6.2, we proved the following inequality:  n 2   n n 2  n n n X X X X X X  2 2 2 2  (6.27) |zk | |ck | − zk c¯k |dk | |ck | − dk c¯k  k=1 k=1 k=1 k=1 k=1 k=1 n 2 n n n X X X X ≥ zk d¯k · |ck |2 − zk c¯k ck d¯k k=1

k=1

k=1

k=1

¯ sequences of complex numbers, with equality iff there is a β ∈ C such that for any ¯ z, ¯ c, d ! Pn Pn z c ¯ d c ¯ i i i i (6.28) zk = Pni=1 2 · ck + β dk − Pni=1 2 · ck i=1 |ci | i=1 |ci | for each k ∈ {1, . . . , n} . ¯=b ¯ and take into consideration that (6.22) and If in (6.27) we choose ¯ z=x ¯, ¯ c=¯ a and d (6.23) hold, then we get  n 2  n 2 !2 n n n n n X X X X X X X |xk |2 |ak |2  |bk |2 |ak |2 − bk a ¯k  ≥ xk¯bk |ak |2 k=1

k=1

k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

k=1

k=1

k=1

k=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

which is clearly equivalent to (6.24). By (6.28) the equality holds in (6.24) iff ! Pn b a ¯ i i xk = β bk − Pni=1 2 · ak , i=1 |ai |

113

k ∈ {1, . . . , n} .

Since x ¯ should satisfy (6.22), we get 1=

n X

|xk |2

k=1

2 Pn n X b a ¯ i i i=1 = |β| · ak bk − Pn 2 i=1 |ai | k=1 " n # Pn 2 X | b a ¯ | k k = |β|2 |bk |2 − Pk=1 n 2 k=1 |ak | k=1 2

from where we deduce that β satisfies (6.26).



6.4. Fan and Todd Inequalities. In 1955, K. Fan and J. Todd [5] proved the following inequality (see also [1, p. 94]). ¯ = (b1 , . . . , bn ) be two sequences of real numbers Theorem 6.5. Let ¯ a = (a1 , . . . , an ) and b such that ai bj 6= aj bi for i 6= j. Then  2 Pn 2  X n n aj n X  i=1 ai (6.29) ≤ Pn 2 Pn 2 Pn   . 2 2 a b − a b ( i=1 ai ) ( i=1 bi ) − ( i=1 ai bi ) j i i j j=1 i=1 j6=i

Proof. We shall follow the proof in [1, p. 94 – p. 95]. Define  −1 X aj n xi := 2 aj b i − ai b j j6=i

(1 ≤ i ≤ n) .

The terms in the sum on the right-hand side n X

 x i ai =

i=1

n 2

−1 X n i=1

 n X  j=1 j6=i

 ai aj   aj b i − ai b j

can be grouped in pairs of the form  −1   ai aj aj ai n + 2 aj b i − ai b j ai b j − aj b i

(i 6= j)

and the sum of each such pair vanishes. Hence, we deduce n n X X ai xi = 0 and bi xi = 1. i=1

i=1

Applying Ostrowski’s inequality (see Section 6.1) we deduce the desired result (6.29).



A weighted version of the result is also due to K. Fan and J. Todd [5] (see also [1, p. 95]). We may state the result as follows.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

114

S.S. D RAGOMIR

Theorem 6.6. Let pij (i, j ∈ {1, . . . , n} , i 6= j) be real numbers such that for any i, j ∈ {1, . . . , n} with i 6= j.

pij = pji ,

(6.30) P

Denote P := 1≤i<j≤n pij and assume that P 6= 0. Then for any two sequences of real numbers ¯ = (b1 , . . . , bn ) satisfying ai bj 6= aj bi (i 6= j) , we have ¯ a = (a1 , . . . , an ) and b  2 Pn 2 n n 1 X X pij aj  i=1 ai (6.31) ≤ Pn 2 Pn 2 Pn   . 2 2 P a b − a b a b − ( a b ) j i i j i i j=1 i=1 i i=1 i i=1 i=1 j6=i

6.5. Some Results for Asynchronous Sequences. If S (R) is the linear space of real sequences, S+ (R) is the subset of nonnegative sequences and Pf (N) denotes the set of finite parts of N, then for the functional T : Pf (N) × S+ (R) × S 2 (R) → R, ! 21 X X X
2 2 ¯ (6.32) T I, p ¯, ¯ a, b := p i ai p i bi − p i ai b i i∈I

i∈I

i∈I

we may state the following result [6, Theorem 3]. Theorem 6.7. If |¯ a| = (|ai |)i∈N and ¯ b = (|bi |)i∈N are asynchronous, i.e., (|ai | − |aj |) (|bi | − |bj |) ≤ 0 for all i, j ∈ N, then (6.33)


¯ ≥ T I, p ¯, ¯ a, b

P

i∈I

pi |ai | P

P

i∈I

i∈I

pi |bi |

pi

−

X

pi |ai bi | ≥ 0.

i∈I

Proof. We shall follow the proof in [6]. Consider the inequalities ! 12 X

pi

X

i∈I

i∈I

X

X

pi a2i

≥

X

pi |ai |

i∈I

and

! 12 i∈I

pi

pi b2i

≥

X

pi |bi |

i∈I

i∈I

which by multiplication give ! 12 X i∈I

pi a2i

X

pi b2i

P ≥

i∈I

i∈I

pi |ai | P

P

i∈I

i∈I

pi

pi |bi |

.

ˇ Now, by the definition of T and by Cebyšev’s inequality for asynchronous sequences, we have P P X
p |a | p |b | i∈I i i ¯ ≥ i∈I i Pi T I, p ¯, ¯ a, b − p i ai b i p i∈I i i∈I P P pi |ai | i∈I pi |bi | X ≥ i∈I P − pi |ai | |bi | i∈I pi i∈I ≥0 and the theorem is proved.



The following result also holds [6, Theorem 4].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

115

Theorem 6.8. If |¯ a| and ¯ b are synchronous, i.e., (|ai | − |aj |) (|bi | − |bj |) ≥ 0 for all i, j ∈ N, then one has the inequality

¯ ≤ T I, p ¯ 1 , (6.34) 0 ≤ T I, p ¯, ¯ a, b ¯, ¯ ab, where 1 = (ei )i∈N , ei = 1, i ∈ N. ˇ Proof. We have, by Cebyšev’s inequality for the synchronous sequences ¯ a2 = (a2i )i∈N and ¯ 2 = (b2i ) , that b i∈N ! 21 X X X
¯ = T I, p ¯, ¯ a, b pi a2i pi b2i − p i ai b i i∈I i∈I i∈I ! 12 X X X 2 2 − p i ai b i ≤ p i ai b i pi i∈I i∈I i∈I
¯ 1 = T I, p ¯, ¯ ab, 

and the theorem is proved.

6.6. An Inequality via A − G − H Mean Inequality. The following result holds [6, Theorem 5]. ¯ be sequences of positive real numbers. Define Theorem 6.9. Let ¯ a and b a2 b2i i (6.35) ∆i = P P 2 2 i∈I ai i∈I bi where i ∈ I and I is a finite part of N. Then one has the inequality "   # P 21P 2
2 P Y ai ∆i i∈I ai i∈I bi a b i∈I i i P P (6.36) ≥ 2 2 bi i∈I ai i∈I bi i∈I P P 2 b2 i∈I ai ≥ P a3 Pi∈I b3i . i

i

i∈I bi

i∈I ai

The equality holds in all the inequalities from (6.36) iff there exists a positive number k > 0 such that ai = kbi for all i ∈ I. Proof. We shall follow the proof in [6]. We will use the AGH−inequality (6.37)

1 X pi xi ≥ PI i∈I

! P1 Y i∈I

I

xpi i

≥P

PI

pi i∈I xi

,

P where pi > 0, xi ≥ 0 for all i ∈ I, where PI := i∈I pi > 0. We remark that the equality holds in (6.37) iff xi = xj for each i, j ∈ I. Choosing pi = a2i and xi = abii (i ∈ I) in (6.37), then we get (6.38)

P P   P a2i 2 2 Y a b bi i∈I ai i∈I i i i∈I ai P ≥ ≥ P a3i 2 ai i∈I ai i∈I i∈I bi

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

116

S.S. D RAGOMIR

and by pi = b2i and xi =

ai , bi

we also have

P P   P b2i 2 2 Y a b ai i∈I bi i∈I bi i∈I i i P ≥ ≥ P b3i . 2 b i i∈I bi i∈I

(6.39)

i∈I ai

If we multiply (6.38) with (6.39) we easily deduce the desired inequality (6.36). The case of equality follows by the same case in the arithmetic mean – geometric mean – harmonic mean inequality. We omit the details.  The following corollary holds [6, Corollary 5.1]. ¯ as above, one has the inequality Corollary 6.10. With ¯ a and b P  12 P P a3i P b3i 2 2 i∈I bi i∈I ai i∈I ai i∈I bi   (6.40) ≥

P P 2 2 . a b a b i i i i i∈I i∈I The equality holds in (6.40) iff there is a k > 0 such that ai = kbi , i ∈ {1, . . . , n} . 6.7. A Related Result via Jensen’s Inequality for Power Functions. The following result also holds [6, Theorem 6]. ¯ be sequences of positive real numbers and p ≥ 1. If I is a finite Theorem 6.11. Let ¯ a and b part of N, then one has the inequality "P #1
2 P 2−p p P p 2−p p a b a b a b i i i i∈I i P i∈I2 P P Pi∈I 2i i . (6.41) ≤ 2 2 a b a i∈I i i∈I i i∈I i i∈I bi The equality holds in (6.41) if and only if there exists a k > 0 such that ai = kbi for all i ∈ I. If p ∈ (0, 1) , the inequality in (6.41) reverses. Proof. We shall follow the proof in [6]. By Jensen’s inequality for the convex mapping f : R+ → R, f (x) = xp , p ≥ 1 one has P

i∈I pi xi PI

(6.42)

p

P ≤

i∈I

pi xpi

PI

,

P where PI := i∈I pi , pi > 0, xi ≥ 0, i ∈ I. The equality holds in (6.42) iff xi = xj for all i, j ∈ I. Now, choosing in (6.42) pi = a2i , xi = abii , we get ! p1 P P 2−p p a b a b i i i i∈I i Pi∈I 2 ≤ P (6.43) 2 a a i∈I i i∈I i and for pi = b2i , xi = (6.44)

ai , bi

the inequality (6.42) also gives P ai b i Pi∈I 2 ≤ i∈I bi

! p1 P p 2−p a b i∈I i i P . 2 i∈I bi

By multiplying the inequalities (6.43) and (6.44), we deduce the desired result from (6.42). The case of equality follows by the fact that in (6.42) the equality holds iff (xi )i∈I is constant. If p ∈ (0, 1) , then a reverse inequality holds in (6.42) giving the corresponding result in (6.41). 

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

117

Remark 6.12. If p = 2, then (6.41) becomes the (CBS) −inequality. 6.8. Inequalities Derived from the Double Sums Case. Let A = (aij )i,j=1,n and B = (bij )i,j=1,n be two matrices of real numbers. The following inequality is known as the (CBS) −inequality for double sums !2 n n n X X X 2 (6.45) aij bij ≤ aij b2ij i,j=1

i,j=1

i,j=1

with equality iff there is a real number r such that aij = rbij for any i, j ∈ {1, . . . , n} . The following inequality holds [7, Theorem 5.2]. ¯ = (b1 , . . . , bn ) be sequences of real numbers. Then Theorem 6.13. Let ¯ a = (a1 , . . . , an ) and b

(6.46)

!2 n X ak + k=1

n X

!2 bk

k=1

n n n X X X
− 2n ak bk ≤ n a2k + b2k − 2 ak bk . k=1 k=1 k=1 k=1 n X

Proof. We shall follow the proof from [7]. Applying (6.45) for aij = ai − bj , bij = bi − aj and taking into account that !2 !2 n n n n X X X X (ai − bj ) (bi − aj ) = 2n ak b k − ak − bk i,j=1 n X

(ai − bj )2 = n

i,j=1

k=1 n X

k=1


a2k + b2k − 2

k=1

k=1 n X

ak

k=1

n X

bk

k=1

and n X

2

(bi − aj ) = n

i,j=1

n X

a2k

+

b2k

k=1

we may deduce the desired inequality (6.46).




−2

n X k=1

ak

n X

bk ,

k=1



The following result also holds [7, Theorem 5.3]. ¯ = (b1 , . . . , bn ) , ¯ ¯ = (d1 , . . . , dn ) Theorem 6.14. Let ¯ a = (a1 , . . . , an ), b c = (c1 , . . . , cn ) and d be sequences of real numbers. Then one has the inequality: 2   Pn Pn i=1 ai di i=1 ai ci  (6.47) det  P Pn n i=1 bi ci i=1 bi di   Pn 2   Pn 2 Pn Pn i=1 ci di i=1 ai bi i=1 ci i=1 ai .  × det  ≤ det  P Pn Pn 2 Pn 2 n i=1 ci di i=1 di i=1 ai bi i=1 bi Proof. We shall follow the proof in [7]. Applying (6.45) for aij = ai bj − aj bi , bij = ci dj − cj di and using Cauchy-Binet’s identity [1, p. 85] ! n ! n n n n X X X X X 1 (6.48) (ai bj − aj bi ) (ci dj − cj di ) = ai c i bi di − ai di bi c i 2 i,j=1 i=1 i=1 i=1 i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

118

S.S. D RAGOMIR

and Lagrange’s identity [1, p. 84] (6.49)

n n n X X 1X 2 2 (ai bj − aj bi ) = ai b2i − 2 i,j=1 i=1 i=1

n X

!2 ai b i

,

i=1



we deduce the desired result (6.47).

6.9. A Functional Generalisation for Double Sums. The following result holds [7, Theorem 5.5]. Theorem 6.15. Let A be a subset of real numbers R, f : A → R and ¯ a = (a1 , . . . , an ), ¯ b = (b1 , . . . , bn ) sequences of real numbers with the property that (i) ak bi , a2i , b2k ∈ A for any i, k ∈ {1, . . . , n} ; (ii) f (a2k ) , f (b2k ) ≥ 0 for any k ∈ {1, . . . , n} ; (iii) f 2 (ak bi ) ≤ f (a2k ) f (b2i ) for any i, k ∈ {1, . . . , n} . Then one has the inequality " n #2 n n X X
X
2 2 (6.50) f (ak bi ) ≤ n f ak f b2k . k,i=1

k=1

k=1

Proof. We will follow the proof in [7]. Using the assumption (iii) and the (CBS) −inequality for double sums, we have n n X X (6.51) f (ak bi ) ≤ |f (ak bi )| k,i=1

≤

k,i=1 n X



 1 f a2k f b2i 2

k,i=1

 !2 n  X 1 
 ≤ f a2k 2 

n X 
 1 f b2i 2

k,i=1

"

n X

=

n
X 2

f ak

k,i=1

" =n

n X k=1

!2  12  

k,i=1

# 12
2

f bi

k,i=1

# 12 "
2

f ak

n X

# 12
2

f bk

k=1



which is clearly equivalent to (6.50). The following corollary is a natural consequence of the above theorem [7, Corollary 5.6]. Corollary 6.16. Let A, f and ¯ a be as above. If (i) ak ai ∈ A for any i, k ∈ {1, . . . , n} ; (ii) f (a2k ) ≥ 0 for any k ∈ {1, . . . , n} ; (iii) f 2 (ak ai ) ≤ f (a2k ) f (a2i ) for any i, k ∈ {1, . . . , n} , then one has the inequality n n X X
(6.52) f (a a ) ≤ n f a2k . k i k,i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

k=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

119

The following particular inequalities also hold [7, p. 23]. (1) If ϕ : N → N is Euler’s indicator and s (n) denotes the sum of all relatively prime ¯ = (b1 , . . . , bn ) numbers including and less than n, then for any ¯ a = (a1 , . . . , an ), b sequences of natural numbers, one has the inequalities "

n X

(6.53)

#2

k,i=1 n X

(6.54)

≤n

ϕ (ak bi )

ϕ (ak ai ) ≤ n

k,i=1

" (6.55)

n X

n X

a2k

ϕ

n
X

k=1 n X


ϕ b2k ;

k=1


ϕ a2k ;

k=1

#2 s (ak bi )

k,i=1 n X

(6.56)

2

≤n

2

n X

s

a2k

n
X

k=1

s (ak ai ) ≤ n

k,i=1

n X


s b2k ;

k=1


s a2k .

k=1

¯ = (b1 , . . . , bn ) are sequences of real numbers, then (2) If a > 1 and ¯ a = (a1 , . . . , an ), b " (6.57)

n X

#2 expa (ak bi )

k,i=1 n X

(6.58)

≤n

expa (ak ai ) ≤ n

2

n X

expa

a2k

n
X

k=1 n X


expa b2k ;

k=1


expa a2k ;

k=1

k,i=1

¯ are sequences of real numbers such that ak , bk ∈ (−1, 1) (k ∈ {1, . . . , n}) , (3) If ¯ a and b then one has the inequalities: " (6.59)

n X

1 (1 − ak bi )m k,i=1 n X

#2 ≤ n2

n X

k=1 n X

n X 1 1 m m, (1 − a2k ) k=1 (1 − b2k )

1 1 m, m ≤ n (1 − ak ai ) (1 − a2k ) k,i=1 k=1

(6.60)

where m > 0.

6.10. A (CBS) −Type Result for Lipschitzian Functions. The following result was obtained in [8, Theorem]. Theorem 6.17. Let f : I ⊆ R → R be a Lipschitzian function with the constant M, i.e., it satisfies the condition (6.61)

|f (x) − f (y)| ≤ M |x − y| for any x, y ∈ I.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

120

S.S. D RAGOMIR

¯ = (b1 , . . . , bn ) are sequences of real numbers with ai bj ∈ I for any If ¯ a = (a1 , . . . , an ), b i, j ∈ {1, . . . , n} , then n n X X (6.62) 0≤ f (ai bj ) |f (ai bj )| − |f (aj bi )| f (ai bj ) ≤

i,j=1 n X

n X

i,j=1

i,j=1

i,j=1

f 2 (aj bi ) −

f (ai bj ) f (aj bi )

 n n X X 2 2 ≤M ai b2i − i=1

i=1

n X

!2  ai b i

.

i=1

Proof. We shall follow the proof in [8]. Since f is Lipschitzian with the constant M, we have (6.63)

0 ≤ ||f (ai bj )| − |f (aj bi )|| ≤ |f (ai bj ) − f (aj bi )| ≤ M |ai bj − aj bi |

for any i, j ∈ {1, . . . , n} , giving (6.64)

0 ≤ |(|f (ai bj )| − |f (aj bi )|) (f (ai bj ) − f (aj bi ))| ≤ (f (ai bj ) − f (aj bi ))2 ≤ M 2 (ai bj − aj bi )2

for any i, j ∈ {1, . . . , n} . The inequality (6.64) is obviously equivalent to |f (ai bj )| f (ai bj ) + |f (aj bi )| f (aj bi ) − |f (ai bj )| f (ai bj ) − |f (aj bi )| f (aj bi ) (6.65) ≤ f 2 (ai bj ) − 2f (ai bj ) f (aj bi ) + f 2 (aj bi )
≤ M 2 a2i b2j − 2ai bi aj bj + a2j b2i for any i, j ∈ {1, . . . , n} . Summing over i and j from 1 to n in (6.65) and taking into account that: n n X X |f (ai bj )| f (ai bj ) = |f (aj bi )| f (ai bj ) , i,j=1 n X

|f (ai bj )| f (aj bi ) =

i,j=1 n X

f 2 (ai bj ) =

i,j=1

i,j=1 n X i,j=1 n X

|f (aj bi )| f (ai bj ) , f 2 (aj bi ) ,

i,j=1



we deduce the desired inequality.

The following particular inequalities hold [8, p. 27 – p. 28]. (1) Let x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) be sequences of real numbers such that 0 ≤ |xi | ≤ M1 , 0 ≤ |yi | ≤ M2 , i ∈ {1, . . . , n} . Then for any r ≥ 1 one has !2 n n n X X X 0≤ x2r yi2r − |xi yi |r (6.66) i i=1

i=1

i=1

 n n X X 2(r−1)  2 2 ≤ r (M1 M2 ) xi yi2 − i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

i=1

n X

!2  |xi yi |

.

i=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

121

(2) If 0 < m1 ≤ |xi | , 0 < m2 ≤ |yi | , i ∈ {1, . . . , n} and r ∈ (0, 1) , then !2 n n n X X X (6.67) x2r yi2r − |xi yi |r 0≤ i i=1

≤

i=1

r

2

(m1 m2 )2(r−1)

i=1

 n n X X 2  xi yi2 − i=1

i=1

n X

!2  |xi yi |

.

i=1

(3) If 0 ≤ |xi | ≤ M1 , 0 ≤ |yi | ≤ M2 , i ∈ {1, . . . , n} , then for any natural number k one has n n X X 2k+1 0≤ x2k+1 |x | yi2k+1 |yi |2k+1 (6.68) i i i=1 i=1 n n X X 2k+1 2k+1 2k+1 − x2k+1 |y | y |x | i i i i i=1 i=1 !2 n n n X X X 2(2k+1) 2(2k+1) ≤ xi yi − x2k+1 yi2k+1 i i=1

i=1

i=1

 n n X X 2 4k  2 ≤ (2k + 1) (M1 M2 ) xi yi2 − i=1

i=1

n X

!2  |xi yi |

.

i=1

(4) If 0 < m1 ≤ xi , 0 < m2 ≤ yi , for any i ∈ {1, . . . , n} , then one has the inequality 2 "X  # 2 n   n X xi xi (6.69) − ln 0≤n ln yi yi i=1 i=1  !2  n n n X X X 1  ≤ x2i yi2 − xi yi  . 2 (m1 m2 ) i=1 i=1 i=1 6.11. An Inequality via Jensen’s Discrete Inequality. The following result holds [9]. ¯ = (b1 , . . . , bn ) be two sequences of real numbers with Theorem 6.18. Let ¯ a = (a1 , . . . , an ), b ai 6= 0, i ∈ {1, . . . , n} . If f : I ⊆ R → R is a convex (concave) function on I and abii ∈ I for each i ∈ {1, . . . , n} , and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers, then   Pn bi  Pn  2 w a f i=1 i i ai wi ai bi i=1 Pn (6.70) f Pn ≤ (≥) . 2 2 i=1 wi ai i=1 wi ai Proof. We shall use Jensen’s discrete inequality for convex (concave) functions ! n n 1 X 1 X (6.71) f pi xi ≤ (≥) pi f (xi ) , Pn i=1 Pn i=1 P where pi ≥ 0 with Pn := ni=1 pi > 0 and xi ∈ I for each i ∈ {1, . . . , n} . If in (6.71) we choose xi = abii and pi = wi a2i , then by (6.71) we deduce the desired result (6.70).  The following corollary holds [9].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

122

S.S. D RAGOMIR

¯ be sequences of positive real numbers and assume that w Corollary 6.19. Let ¯ a and b ¯ is as above. If p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequality !p−1 n !p n n X X X (6.72) wi ai bi ≤ (≥) wi a2i wi ai2−p bpi . i=1

i=1

i=1

Proof. Follows by Theorem 6.18 applied for convex (concave) function f : [0, ∞) → R, f (x) = xp , p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) .  Remark 6.20. If p = 2, then by (6.72) we deduce the (CBS) −inequality. 6.12. An Inequality via Lah-Ribari´c Inequality. The following reverse of Jensen’s discrete inequality was obtained in 1973 by Lah and Ribari´c [10]. Lemma 6.21. Let f : I → R be a convex function, xi ∈ [m, M ] ⊆ I for each i ∈ {1, . . . , n} and p ¯ = (p1 , . . . , pn ) be a positive n−tuple. Then P Pn n 1 M − P1n ni=1 pi xi 1 X i=1 pi xi − m Pn f (m) + f (M ) . (6.73) pi f (xi ) ≤ M −m M −m Pn i=1 Proof. We observe for each i ∈ {1, . . . , n} , that (M − xi ) m + (xi − m) M (6.74) xi = . M −m If in the definition of convexity, i.e., α, β ≥ 0, α + β > 0   αa + βb αf (a) + βf (b) (6.75) f ≤ α+β α+β we choose α = M − xi , β = xi − m, a = m and b = M, we deduce, by (6.75), that   (M − xi ) m + (xi − m) M (6.76) f (xi ) = f M −m (M − xi ) f (m) + (xi − m) f (M ) ≤ M −m for each i ∈ {1, . . . , n} . If we multiply (6.76) by pi > 0 and sum over i from 1 to n, we deduce (6.73).



The following result holds. ¯ = (b1 , . . . , bn ) be two sequences of real numbers with Theorem 6.22. Let ¯ a = (a1 , . . . , an ) , b ai 6= 0, i ∈ {1, . . . , n} . If I ⊆ R → R is a convex (concave) function on I and abii ∈ [m, M ] ⊆ I for each i ∈ {1, . . . , n} and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers, then   Pn wi ai bi bi 2 Pi=1 w a f M − n 2 i=1 i i ai i=1 wi ai Pn ≤ (≥) f (m) + 2 M −m i=1 wi ai

Pn (6.77)

Proof. Follows by Lemma 6.21 for the choices pi = wi a2i , xi =

Pn wi ai bi Pi=1 n 2 i=1 wi ai

−m

M −m bi , ai

f (M ) .

i ∈ {1, . . . , n} .



The following corollary holds. ¯ = (b1 , . . . , bn ) be two sequences of positive real Corollary 6.23. Let ¯ a = (a1 , . . . , an ) , b numbers and such that bi (6.78) 0<m≤ ≤ M < ∞ for each i ∈ {1, . . . , n} . ai

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

123

If w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers and p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequality (6.79)

n X

wi a2−p bpi i

i=1

n n M m (M p−1 − mp−1 ) X M p − mp X 2 + wi ai ≤ (≥) wi ai bi . M −m M − m i=1 i=1

Proof. If we write the inequality (6.77) for the convex (concave) function f (x) = xp , p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , we get n wi ai bi Pn 2−p p Pi=1 M − n 2 wi ai bi i=1 wi ai i=1 Pn ≤ (≥) · mp + 2 M −m i=1 wi ai

P

Pn wi ai bi Pi=1 n 2 i=1 wi ai

−m

M −m

· M p, 

which, after elementary calculations, is equivalent to (6.79). Remark 6.24. For p = 2, we get n X

(6.80)

wi b2i + M m

i=1

n X

wi a2i ≤ (M + m)

i=1

n X

wi ai bi ,

i=1

which is the well known Diaz-Metcalf inequality [11]. 6.13. An Inequality via Dragomir-Ionescu Inequality. The following reverse of Jensen’s inequality was proved in 1994 by S.S. Dragomir and N.M. Ionescu [12]. ˚ xi ∈I˚ (i ∈ {1, . . . , n}) Lemma 6.25. Let f : I ⊆ R → R be a differentiable convex function on I, Pn and pi ≥ 0 (i ∈ {1, . . . , n}) such that Pn := i=1 pi > 0. Then one has the inequality ! n n 1 X 1 X (6.81) pi f (xi ) − f pi xi 0≤ Pn i=1 Pn i=1 ≤

n n n 1 X 1 X 1 X pi xi f 0 (xi ) − pi xi · pi f 0 (xi ) . Pn i=1 Pn i=1 Pn i=1

Proof. Since f is differentiable convex on ˚I, one has f (x) − f (y) ≥ (x − y) f 0 (y) ,

(6.82) for any x, y ∈˚I. If we choose x = (6.83)

f

1 Pn

Pn

pi xi and y = yk , k ∈ {1, . . . , n} , we get ! ! n n X X 1 1 pi xi − f (yk ) ≥ pi xi − yk f 0 (yk ) . Pn i=1 Pn i=1 i=1

Multiplying (6.83) by pk ≥ 0 and summing over k from 1 to n, we deduce the desired result (6.81).  The following result holds [9]. ¯ = (b1 , . . . , bn ) be two sequences of real numbers with Theorem 6.26. Let ¯ a = (a1 , . . . , an ), b ai 6= 0, i ∈ {1, . . . , n} . If f : I ⊆ R → R is a differentiable convex (concave) function on I˚ ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real and abii ∈I˚ for each i ∈ {1, . . . , n} , and w

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

124

S.S. D RAGOMIR

numbers, then (6.84)

!2     Pn n X wi ai bi b i 2 2 2 i=1 0≤ wi ai wi ai f − wi ai · f Pn 2 a i i=1 wi ai i=1 i=1 i=1   X   n n n n X X X bi bi 2 0 2 0 ≤ wi ai wi ai bi f − wi ai bi wi ai f . ai ai i=1 i=1 i=1 i=1 n X

n X

bi , ai

Proof. Follows from Lemma 6.25 on choosing pi = wi a2i , xi =

i ∈ {1, . . . , n} .



The following corollary holds [9]. ¯ = (b1 , . . . , bn ) be two sequences of positive real Corollary 6.27. Let ¯ a = (a1 , . . . , an ), b numbers with ai 6= 0, i ∈ {1, . . . , n} . If p ∈ [1, ∞), then one has the inequality 0≤

(6.85)

n X

wi a2i

i=1

" ≤p

n X

wi a2−p bpi − i

n X

i=1

n X

wi a2i

i=1

n X

wi a2−p bi − i

i=1

!2−p

n X

wi a2i

i=1 n X

!p wi ai bi

i=1

wi ai bi

n X

i=1

# wi a3−p bp−1 . i i

i=1

If p ∈ (0, 1) , then 0≤

(6.86)

n X

!2−p wi a2i

i=1 n X

! wi ai bi

−

i=1

" ≤p

n X

wi ai bi

i=1

n X

wi a3−p bpi − i

i=1

n X

wi a2i

i=1 n X

wi a2i

i=1

n X

wi a2−p bpi i

i=1 n X

# wi a2−p bi . i

i=1

6.14. An Inequality via a Refinement of Jensen’s Inequality. We will use the following lemma which contains a refinement of Jensen’s inequality obtained in [13]. Lemma 6.28. P Let f : I ⊆ R →R be a convex function on the interval I and xi ∈ I, pi ≥ 0 with Pn := ni=1 pi > 0. Then the following inequality holds: (6.87)

f

n 1 X pi xi Pn i=1

!

n X

xi1 + · · · + xik+1 ≤ k+1 pi1 · · · pik f Pn i ,...,i =1 k+1 1 k+1   n 1 X xi1 + · · · + xik pi · · · pik f ≤ k Pn i ,...,i =1 1 k 1

1

≤ ··· ≤





k

n 1 X pi f (xi ) , Pn i=1

where k ≥ 1, k ∈ N. Proof. We shall follow the proof in [13].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

125

The first inequality follows by Jensen’s inequality for multiple sums   x +···+x  Pn i1 ik+1 i1 ,...,ik+1 =1 pi1 · · · pik+1 k+1  Pn f p · · · p i i 1 k+1 i1 ,...,ik+1 =1  x +···+x  i1 ik+1 · · · p f p i i k+1 i1 ,...,ik+1 =1 1 k+1 Pn i1 ,...,ik+1 =1 pi1 · · · pik+1

Pn = since

x

i1 +···+xik+1

Pn

i1 ,...,ik+1 =1 pi1 · · · pik+1 k+1 Pn i1 ,...,ik+1 =1 pi1 · · · pik+1

and

n X

 =

Pnk

n X

pi xi

i=1

pi1 · · · pik+1 = Pnk+1 .

i1 ,...,ik+1 =1

Now, applying Jensen’s inequality for xi1 + xi2 · · · + xik−1 + xik xi + xi3 · · · + xik + xik+1 , y2 = 2 , k k xi + xi1 + xi2 + · · · + xik−1 · · · , yk+1 = k+1 k we have   y1 + y2 + · · · + yk + yk+1 f (y1 ) + f (y2 ) + · · · + f (yk ) + f (yk+1 ) f ≤ , k+1 k+1 which is equivalent to   xi1 + · · · + xik+1 (6.88) f k+1  x +x ···+x +x   x +x +x +···+x  i1 i2 ik−1 ik ik+1 i1 i2 ik−1 f + · · · + f k k ≤ . k+1 Multiplying (6.88) with the nonnegative real numbers pi1 , . . . , pik+1 and summing over i1 , . . . , ik+1 from 1 to n we deduce   n X xi1 + · · · + xik+1 (6.89) pi1 · · · pik+1 f k+1 i1 ,...,ik+1 =1    n 1  X xi1 + · · · + xik ≤ pi · · · pik+1 f k + 1 i ,...,i =1 1 k 1 k+1    n X xik+1 + xi1 + xi2 + · · · + xik−1  + ··· + pi1 · · · pik+1 f k i1 ,...,ik+1 =1   n X xi1 + · · · + xik = Pn pi1 · · · pik f k i ,...,i =1 y1 =

1

k

which proves the second part of (6.87).



The following result holds.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

126

S.S. D RAGOMIR

Theorem 6.29. Let f : I ⊆ R →R be a convex function on the interval I, ¯ a = (a1 , . . . , an ), bi ¯ b = (b1 , . . . , bn ) real numbers such that ai 6= 0, i ∈ {1, . . . , n} and ai ∈ I, i ∈ {1, . . . , n} . If w ¯ = (w1 , . . . , wn ) are positive real numbers, then  Pn  wi ai bi i=1 (6.90) f Pn 2 i=1 wi ai  bi bik+1  1 n + · · · + X a aik+1 1 2 2  i1  ≤ Pn w · · · w a · · · a f i i 1 i i k+1 1 k+1 k+1 k + 1 ( i=1 wi a2i ) i1 ,...,ik+1 =1 b  bik i1 n + · · · + X 1 ai aik  wi1 · · · wik a2i1 · · · a2ik f  1 ≤ Pn k k ( i=1 wi a2i ) i1 ,...,ik =1   n X 1 bi 2 ≤ · · · ≤ Pn wi ai f . 2 ai i=1 wi ai i=1 The proof is obvious by Lemma 6.28 applied for pi = wi a2i , xi = The following corollary holds.

bi , ai

i ∈ {1, . . . , n} .

¯ and w Corollary 6.30. Let ¯ a, b ¯ be sequences of positive real numbers. If p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequalities !p n X wi ai bi i=1

Pn   n p−k−1 X bik+1 p bi1 ≤ ( i=1 wi a2i ) 2 2 wi1 · · · wik+1 ai1 · · · aik+1 + ··· + (≥) (k + 1)p ai1 aik+1 i1 ,...,ik+1 =1 Pn  p n p−k X bi1 bik ≤ ( i=1 wi a2i ) 2 2 wi1 · · · wik ai1 · · · aik + ··· + (≥) kp a aik i 1 i1 ,...,ik =1 !p−1 n n X X ≤ ≤ ··· wi a2i wi a2−p bpi . i (≥) (≥) i=1

i=1

Remark 6.31. If p = 2, then we deduce the following refinement of the (CBS) − inequality  2 !2 P n n k+1 k+1 1−k X X ( ni=1 wi a2i ) X Y  bi` aij  wi ai bi ≤ w · · · w i1 ik+1  2 (k + 1) j=1 i=1 i1 ,...,ik+1 =1 `=1 j6=`

P 2−k ( ni=1 wi a2i ) ≤ k2

n X



k

k

2

X Y  wi1 · · · wik  bi` aij 

i1 ,...,ik =1

`=1

j=1 j6=`

n 1X ≤ ··· ≤ wi wj (bi aj + ai bj )2 4 i,j=1

≤

n X

wi a2i

i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

n X

wi b2i .

i=1

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

127

6.15. Another Refinement via Jensen’s Inequality. The following refinement of Jensen’s inequality holds (see [15]). Lemma 6.32. Let P f : [a, b] → R be a differentiable convex function on (a, b) and xi ∈ (a, b) , pi ≥ 0 with Pn := ni=1 pi > 0. Then one has the inequality ! ! n n n n 1 X X 1 X 1 X 1 (6.91) pi f (xi ) − f pi xi ≥ pi f (xi ) − f pj xj Pn Pn i=1 Pn i=1 P n j=1 i=1 ! n n n 1 X 1 X 0 1 X − f pj xj · pi xi − pj xj ≥ 0. Pn Pn Pn j=1

i=1

j=1

Proof. Since f is differentiable convex on (a, b) , then for each x, y ∈ (a, b), one has the inequality (6.92)

f (x) − f (y) ≥ (x − y) f 0 (y) .

Using the properties of the modulus, we have (6.93)

f (x) − f (y) − (x − y) f 0 (y) = |f (x) − f (y) − (x − y) f 0 (y)| ≥ ||f (x) − f (y)| − |x − y| |f 0 (y)||

for each x, y ∈ (a, b) . P If we choose y = P1n nj=1 pj xj and x = xi , i ∈ {1, . . . , n} , then we have ! ! ! n n n X 1 X 1 X 1 (6.94) f (xi ) − f pj xj − xi − pj xj f 0 pj xj Pn j=1 Pn j=1 Pn j=1 ! ! n n n 1 X 1 X 0 1 X ≥ f (xi ) − f pj xj − xi − pj xj f p j xj Pn j=1 Pn j=1 Pn j=1 for any i ∈ {1, . . . , n} . If we multiply (6.94) by pi ≥ 0, sum over i from 1 to n, and divide by Pn > 0, we deduce ! n n 1 X 1 X pi f (xi ) − f pj xj Pn i=1 Pn j=1 ! ! n n n X X X 1 1 1 − pi xi − pj xj f 0 pj xj Pn i=1 Pn j=1 Pn j=1 ! n n 1 X 1 X ≥ pi f (xi ) − f pj xj Pn i=1 Pn j=1 ! n n X X 1 0 1 − xi − pj x j f pj xj Pn j=1 Pn j=1 ! n n 1 X 1 X ≥ pi f (xi ) − f pj xj Pn Pn j=1 i=1 ! n n n X X X 1 1 0 1 − pi xi − pj xj · f pj xj . Pn Pn Pn i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

j=1

j=1

http://jipam.vu.edu.au/

128

S.S. D RAGOMIR

Since n n 1 X 1 X pi xi − pj xj Pn i=1 Pn j=1

! = 0,



the inequality (6.91) is proved. In particular, we have the following result for unweighted means. Corollary 6.33. With the above assumptions for f and xi , one has the inequality   f (x1 ) + · · · + f (xn ) x1 + · · · + xn (6.95) −f n n   n 1 X x + · · · + x 1 n xi − f ≥ n n i=1   n n X 0 x1 + · · · + xn 1 X 1 · − f x − x i j ≥ 0. n n n i=1 j=1 The following refinement of the (CBS) −inequality holds. Theorem 6.34. If ai , bi ∈ R, i ∈ {1, . . . , n} , then one has the inequality;

(6.96)

n X i=1

a2i

n X

b2i −

i=1

n X

!2 ai b i

i=1

n X a2i b2i 2 P 2 ≥ Pn 2 Pn n 2 j=1 aj bj j=1 bj i=1 bi i=1 n n n X X X a b −2 ak bk · |bi | bj i i ≥ 0. aj b j 1

k=1

i=1

j=1

Proof. If we apply Lemma 6.32 for f (x) = x2 , we get n 1 X 2 (6.97) pi xi − Pn i=1

!2 n X 1 pi xi Pn i=1 !2 n n 1 X 2 1 X ≥ pi xi − pj xj Pn j=1 Pn i=1 n n n 1 X 1 X X 1 − 2 pk xk · pi xi − pj xj ≥ 0. Pn Pn Pn j=1 i=1 k=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

If in (6.97), we choose pi = b2i , xi =

ai , bi

129

i ∈ {1, . . . , n} , we get

Pn 2 P 2 ai ( ni=1 ai bi ) i=1 (6.98) Pn 2 − Pn 2 2 ( i=1 bi ) i=1 bi 2 n X 1 a 2 ≥ Pn 2 bi · 2i − i=1 bi i=1 bi Pn k=1 ak bk −2 Pn 2 · b i=1 i

!2 Pn a b j j Pj=1 n 2 b j=1 j .P Pn 2 ai Pn n 2 i=1 bi bi − j=1 aj bj j=1 bj , Pn 2 b i=1 i 

which is clearly equivalent to (6.96).

6.16. An Inequality via Slater’s Result. Suppose that I is an interval of real numbers with ◦ ◦ interior I and f : I → R is a convex function on I. Then f is continuous on I and has finite ◦ ◦ left and right derivatives at each point of I . Moreover, if x, y ∈I and x < y, then D− f (x) ≤ D+ f (x) ≤ D− f (y) ≤ D+ f (y) which shows that both D− f and D+ f are nondecreasing ◦

functions on I . It is also known that a convex function must be differentiable except for at most countably many points. For a convex function f : I → R, the of f denoted by ∂f is the set of all  ◦ subdifferential  functions ϕ : I → [−∞, ∞] such that ϕ I ⊂ R and f (x) ≥ f (a) + (x − a) ϕ (a) for any x, a ∈ I.

(6.99)

It is also well known that if f is convex on I, then ∂f is nonempty, D+ f, D− f ∈ ∂f and if ϕ ∈ ∂f, then D− f (x) ≤ ϕ (x) ≤ D+ f (x)

(6.100) ◦

for every x ∈I . In particular, ϕ is a nondecreasing function. ◦ If f is differentiable convex on I , then ∂f = {f 0 } . The following inequality is well known in the literature as Slater’s inequality [16]. Lemma 6.35. Let f P : I → R be a nondecreasing (nonincreasing) convex function, xi ∈ I, Pn n p > 0 and p ϕ (x pi ≥ 0 with Pn := i ) 6= 0 where ϕ ∈ ∂f. Then one has the i=1 i i=1 i inequality:  Pn  n pi xi ϕ (xi ) 1 X i=1 Pn (6.101) pi f (xi ) ≤ f . Pn i=1 i=1 pi ϕ (xi ) Proof. Firstly, observe that since, for example, f is nondecreasing, then ϕ (x) ≥ 0 for any x ∈ I and thus Pn pi xi ϕ (xi ) Pi=1 (6.102) ∈ I, n i=1 pi ϕ (xi ) since it is a convex combination of xi with the positive weights x ϕ (xi ) Pn i , i = 1, . . . , n. i=1 pi ϕ (xi ) A similar argument applies if f is nonincreasing.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

130

S.S. D RAGOMIR

Now, if we use the inequality (6.99), we deduce f (x) − f (xi ) ≥ (x − xi ) ϕ (xi ) for any x, xi ∈ I, i = 1, . . . , n.

(6.103)

Multiplying (6.103) by pi ≥ 0 and summing over i from 1 to n, we deduce f (x) −

(6.104)

n n n 1 X 1 X 1 X pi f (xi ) ≥ x · pi ϕ (xi ) − pi xi ϕ (xi ) Pn i=1 Pn i=1 Pn i=1

for any x ∈ I. If in (6.104) we choose Pn pi xi ϕ (xi ) x = Pi=1 , n i=1 pi ϕ (xi ) which, we have proved that it belongs to I, we deduce the desired inequality (6.101).



If one would like to drop the assumption of monotonicity for the function f, then one can state and prove in a similar way the following result. Lemma 6.36. Let f : I → R be a convex function, xi ∈ I, pi ≥ 0 with Pn > 0 and Pn i=1 pi ϕ (xi ) 6= 0, where ϕ ∈ ∂f. If Pn pi xi ϕ (xi ) Pi=1 ∈ I, (6.105) n i=1 pi ϕ (xi ) then the inequality (6.101) holds. The following result in connection to the (CBS) −inequality holds. Theorem 6.37. Assume that f : R+ → R is a convex function on R+ := [0, ∞), ai , bi ≥ 0 with Pn 2  bi  ai 6= 0, i ∈ {1, . . . , n} and i=1 ai ϕ ai 6= 0 where ϕ ∈ ∂f. (i) If f is monotonic nondecreasing (nonincreasing) in [0, ∞) then    Pn bi   n n X X i=1 ai bi ϕ ai bi 2 2   . (6.106) ai ϕ ≤ ai · f  P n a 2 i a ϕ bi i=1 i=1 i=1

i

ai

(ii) If   bi a b ϕ i i i=1 ai Pn 2  bi  ≥ 0, i=1 ai ϕ ai

Pn (6.107) then (6.106) also holds.

Remark 6.38. Consider the function f : [0, ∞) → R, f (x) = xp , p ≥ 1. Then f is convex and monotonic nondecreasing and ϕ (x) = pxp−1 . Applying (6.106), we may deduce the following inequality: !p+1 !p n n n X X X 3−p p−1 2−p p (6.108) p ai b i ≤ a2i ai b i i=1

i=1

i=1

for p ≥ 1, ai , bi ≥ 0, i = 1, . . . , n.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

131

6.17. An Inequality via an Andrica-Ra¸sa Result. The following Jensen type inequality has been obtained in [17] by Andrica and Ra¸sa. Lemma 6.39. Let f : [a, b] → R be a twice differentiable function and assume that m = inf f 00 (t) > −∞ and M = sup f 00 (t) < ∞. t∈(a,b)

t∈(a,b)

Pn

If xi ∈ [a, b] and pi ≥ 0 (i = 1, . . . , n) with i=1 pi = 1, then one has the inequalities:  !2  ! n n n n X X X X 1  (6.109) pi x2i − pi xi  ≤ pi f (xi ) − f pi xi m 2 i=1 i=1 i=1 i=1  !2  n n X X 1 ≤ M pi x2i − pi xi  . 2 i=1 i=1 Proof. Consider the auxiliary function fm (t) := f (t) − 12 mt2 . This function is twice differen00 tiable and fm (t) ≥ 0, t ∈ (a, b) , showing that fm is convex. Applying Jensen’s inequality for fm , i.e., ! n n X X pi fm (xi ) ≥ fm pi xi , i=1

i=1

we deduce, by a simple calculation, the first inequality in (6.109). The second inequality follows in a similar way for the auxiliary function fM (t) = 21 M t2 − f (t) . We omit the details.  The above result may be naturally used to obtain the following inequality related to the (CBS) −inequality. ¯ = (b1 , . . . , bn ) be two sequences of real numbers with Theorem 6.40. Let ¯ a = (a1 , . . . , an ) , b the property that there exists γ, Γ ∈ R such that ai (6.110) −∞ ≤ γ ≤ ≤ Γ ≤ ∞, for each i ∈ {1, . . . , n} , bi and bi 6= 0, i = 1, . . . , n. If f : (γ, Γ) → R is twice differentiable and m = inf f 00 (t) and M = sup f 00 (t) , t∈(γ,Γ)

then we have the inequality  n n 1 X 2 X 2 (6.111) m ai bi − 2 i=1 i=1 ≤

n X

b2i

i=1

n X i=1

t∈(γ,Γ)

n X

!2  ai b i



i=1

  ai 2 bi f − bi

 n n 1 X 2 X 2 ≤ M ai bi − 2 i=1 i=1

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

n X i=1 n X

!2  P  n ai bi 2 i=1 bi f Pn 2 i=1 bi  !2

ai b i

.

i=1

http://jipam.vu.edu.au/

132

S.S. D RAGOMIR

b2

Proof. We may apply Lemma 6.39 for the choices pi = Pn i b2 and xi = k=1 k "P  Pn 2 # n 2 ai b i a 1 m Pni=1 i2 − Pi=1 n 2 2 k=1 bk k=1 bk Pn 2  ai   Pn  i=1 bi f bi ai b i i=1 Pn 2 ≤ − f Pn 2 k=1 bk k=1 bk "P P  2 # n n 2 a a b 1 i i ≤ M Pni=1 i2 − Pi=1 n 2 2 k=1 bk k=1 bk

ai bi

to get



giving the desired result (6.111). The following corollary is a natural consequence of the above theorem. ¯ are sequences of nonnegative real numbers and Corollary 6.41. Assume that ¯ a, b ai (6.112) 0<ϕ≤ ≤ Φ < ∞ for each i ∈ {1, . . . , n} . bi Then for any p ∈ [1, ∞) one has the inequalities  !2  n n n X X X 1 (6.113) p (p − 1) ϕp−2  a2i b2i − ai b i  2 i=1 i=1 i=1 !2−p !p ! p n n n n X X X X ≤ b2i api b2−p − b2i ai b i i i=1

i=1

i=1

 n n X X 1 p−2  2 ≤ p (p − 1) Φ ai b2i − 2 i=1 i=1

i=1 n X

!2  ai b i



i=1

if p ∈ [2, ∞) and (6.114)

 !2  n n n X X X 1 p (p − 1) Φp−2  a2i b2i − ai bi  2 i=1 i=1 i=1 ! !2−p !p p n n n n X X X X ≤ b2i api b2−p − b2i ai b i i i=1

i=1

i=1

 n n X X 1 ≤ p (p − 1) ϕp−2  a2i b2i − 2 i=1 i=1

i=1 n X

!2  ai b i



i=1

if p ∈ [1, 2] . 6.18. An Inequality via Jensen’s Result for Double Sums. The following result for convex functions via Jensen’s inequality also holds [18]. Lemma 6.42. Let f : R → R be a convex (concave) function and x ¯ = (x ¯ = P1n, . . . , xn ) , p (p1 , . . . , pn ) real sequences with the property that pi ≥ 0 (i = 1, . . . , n) and i=1 pi = 1. Then one has the inequality: hP i P "P # j−1 Pn 2 n 2 p p f (∆x · ∆x ) i j k l 1≤i<j≤n k,l=i pi xi − ( i=1 pi xi ) (6.115) f Pi=1 ≤ (≥) , P P Pn 2 2 n n n 2 2 i=1 i pi − ( i=1 ipi ) i=1 i pi − ( i=1 ipi )

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

133

where ∆xk := xk+1 − xk (k = 1, . . . , n − 1) is the forward difference. Proof. We have, by Jensen’s inequality for multiple sums that "P # # "P Pn 2 2 n 2 p p (x − x ) p x − ( p x ) i j i j i i i i 1≤i<j≤n f Pi=1 (6.116) =f P Pi=1 2 2 n n 2 1≤i<j≤n pi pj (j − i) i=1 i pi − ( i=1 ipi )  P 2 (xj −xi )2 p p (j − i) 2 i j 1≤i<j≤n (j−i)  =f P 2 p p (j − i) i j 1≤i<j≤n   P (xj −xi )2 2 1≤i<j≤n pi pj (j − i) f (j−i)2 ≤ =: I. P 2 1≤i<j≤n pi pj (j − i) On the other hand, for j > i, one has (6.117)

xj − xi =

j−1 X

(xk+1 − xk ) =

k=i

(xj − xi ) =

j−1 X k=i

∆xk

k=i

and thus 2

j−1 X

!2 ∆xk

=

j−1 X

∆xk · ∆xl .

k,l=i

Applying once more the Jensen inequality for multiple sums, we deduce # " Pj−1 # " Pj−1 ∆x · ∆x (xj − xi )2 k l k,l=i k,l=i f (∆xk · ∆xl ) (6.118) f =f ≤ (≥) 2 2 (j − i) (j − i) (j − i)2 and thus, by (6.118), we deduce (6.119)

Pj−1

f (∆xk ·∆xl )

2 k,l=i 1≤i<j≤n pi pj (j − i) (j−i)2 I ≤ (≥) P 2 1≤i<j≤n pi pj (j − i) hP i P j−1 1≤i<j≤n pi pj k,l=i f (∆xk · ∆xl ) = , Pn 2 Pn 2 i=1 i pi − ( i=1 ipi )

P

and then, by (6.116) and (6.119), we deduce the desired inequality (6.115).



The following inequality connected with the (CBS) −inequality may be stated. ¯ = Theorem 6.43. Let f : R → R be a convex (concave) function and ¯ a = (a1 , . . . , an ) , b (b1 , . . . , bn ) , w ¯ = (w1 , . . . , wn ) sequences of real numbers such that bi 6= 0, wi ≥ 0 (i = 1, . . . , n) and not all wi are zero. Then one has the inequality "P # Pn Pn 2 n 2 2 w a w b − ( w a b ) i i i i i i i i=1 (6.120) f Pni=1 Pni=1 2 2 P 2 n 2 w b i w b − ( i i i=1 i i i=1 i=1 iwi bi )  i hP    P j−1 al ak 2 2 w w b b f ∆ · ∆ i j i j 1≤i<j≤n k,l=i bk bl ≤ (≥) . Pn P P 2 n n 2 2 2 i=1 wi bi i=1 i wi bi − ( i=1 iwi bi ) Proof. Follows by Lemma 6.42 on choosing pi = wi b2i and xi = details.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

ai , bi

i = 1, . . . , n. We omit the 

http://jipam.vu.edu.au/

134

S.S. D RAGOMIR

ˇ 6.19. Some Inequalities for the Cebyšev Functional. For two sequences of real numbers ¯ ¯ a = (a1 , . . . , an ) , b = (b1 , . . . , bn ) and p ¯ = (p1 , . . . , pn ) with pi ≥ 0 (i ∈ {1, . . . , n}) and P n ˇ i=1 pi = 1, consider the Cebyšev functional n n n X X X
¯ := ¯, ¯ a, b p i ai b i − p i ai p i bi . T p

(6.121)

i=1

i=1

i=1

By Korkine’s identity [1, p. 242] one has the representation n X
¯ =1 T p ¯, ¯ a, b pi pj (ai − aj ) (bi − bj ) 2 i,j=1 X = pi pj (aj − ai ) (bj − bi ) .

(6.122)

1≤i<j≤n

Using the (CBS) −inequality for double sums one may state the following result 

¯ 2 ≤ T (¯ ¯ b ¯ , (6.123) T p ¯, ¯ a, b p, ¯ a, ¯ a) T p ¯ , b, where, obviously (6.124)

n 1X pi pj (ai − aj )2 2 i,j=1 X = pi pj (aj − ai )2 .

T (¯ p, ¯ a, ¯ a) =

1≤i<j≤n

The following result holds [14]. ¯ = (b1 , . . . , bn ) are real numbers such that for Lemma 6.44. Assume that ¯ a = (a1 , . . . , an ) , b each i, j ∈ {1, . . . , n} , i < j, one has (6.125)

m (bj − bi ) ≤ aj − ai ≤ M (bj − bi ) ,

where m, M are given real numbers. P If p ¯ = (p1 , . . . , pn ) is a nonnegative sequence with ni=1 pi = 1, then one has the inequality

¯ ≥ T (¯ ¯ b ¯ . (6.126) (m + M ) T p ¯, ¯ a, b p, ¯ a, ¯ a) + mM T p ¯ , b, Proof. If we use the condition (6.125), we get (6.127)

[M (bi − bj ) − (ai − aj )] [(ai − aj ) − m (bi − bj )] ≥ 0

for i, j ∈ {1, . . . , n} , i < j. If we multiply in (6.127), then, obviously, for any i, j ∈ {1, . . . , n} , i < j we have (6.128)

(aj − ai )2 + mM (bj − bi )2 ≤ (m + M ) (aj − ai ) (bj − bi ) .

Multiplying (6.128) by pi pj ≥ 0, i, j ∈ {1, . . . , n} , i < j, summing over i and j, i < j from 1 to n and using the identities (6.122) and (6.124), we deduce the required inequality (6.125).  The following result holds [14]. ¯ p Theorem 6.45. If ¯ a, b, ¯ are as in Lemma 6.44 and M ≥ m > 0, then one has the inequality providing a reverse for (6.123)

 ¯ b ¯ . ¯ 2 ≥ 4mM T (¯ (6.129) T p ¯, ¯ a, b p , ¯ a , ¯ a ) T p ¯ , b, (m + M )2

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

135

Proof. We use the following elementary inequality 1 (6.130) αx2 + y 2 ≥ 2xy, x, y ≥ 0, α > 0 α to get, for the choices √ 
 1 1 ¯ b ¯ 2 ≥ 0, y = [T (¯ ¯ , b, p, ¯ a, ¯ a)] 2 ≥ 0 α = mM > 0, x = T p the following inequality: √ 
 1
1 ¯ b ¯ 2 [T (¯ ¯ b ¯ + √ 1 T (¯ p, ¯ a, ¯ a) ≥ 2 T p ¯ , b, p, ¯ a, ¯ a)] 2 . (6.131) mM T p ¯ , b, mM Using (6.130) and (6.131), we deduce

 1 1 (m + M ) ¯ ≥ T p ¯ b ¯ 2 [T (¯ √ T p ¯, ¯ a, b ¯ , b, p, ¯ a, ¯ a)] 2 2 mM which is clearly equivalent to (6.129).



The following corollary also holds [14]. Corollary 6.46. With the assumptions of Theorem 6.45, we have: 
 1
1 ¯ b ¯ 2 [T (¯ ¯ 0≤ T p ¯ , b, p, ¯ a, ¯ a)] 2 − T p (6.132) ¯, ¯ a, b √ √ 2 M− m
¯ √ ≤ T p ¯, ¯ a, b 2 mM and

 ¯ b ¯ − T p ¯ 2 (6.133) 0 ≤ T (¯ p, ¯ a, ¯ a) T p ¯ , b, ¯, ¯ a, b
 (M − m)2  ¯ 2. T p ¯, ¯ a, b 4mM The following result is useful in practical applications [14]. ≤

Theorem 6.47. Let f, g : [α, β] → R be continuous on [α, β] and differentiable on (α, β) with g 0 (x) 6= 0 for x ∈ (α, β) . Assume f 0 (x) f 0 (x) (6.134) −∞ < γ = inf , sup = Γ < ∞. 0 x∈(α,β) g 0 (x) x∈(α,β) g (x) If x ¯ is a real sequence with xi ∈ [α, β] and xi 6= xj for i 6= j and if we denote by f (¯ x) := (f (x1 ) , . . . , f (xn )) , then we have the inequality: (γ + Γ) T (¯ p, f (¯ x) , g (¯ x)) ≥ T (¯ p, f (¯ x) , f (¯ x)) + γΓT (¯ p, g (¯ x) , g (¯ x)) Pn for any p¯ with pi ≥ 0 (i ∈ {1, . . . , n}) , i=1 pi = 1.

(6.135)

Proof. Applying the Cauchy Mean-Value Theorem, there exists ξij ∈ (α, β) (i < j) such that (6.136)

f (xj ) − f (xi ) f 0 (ξij ) = 0 ∈ [γ, Γ] g (xj ) − g (xi ) g (ξij )

for i, j ∈ {1, . . . , n} , i < j. Then    f (xj ) − f (xi ) f (xj ) − f (xi ) (6.137) Γ− − γ ≥ 0, 1 ≤ i < j ≤ n, g (xj ) − g (xi ) g (xj ) − g (xi ) which, by a similar argument to the one in Lemma 6.44 will give the desired result (6.135).  The following corollary is natural [14].

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

136

S.S. D RAGOMIR

Corollary 6.48. With the assumptions in Theorem 6.47 and if Γ ≥ γ > 0, then one has the inequalities: (6.138)

(6.139)

[T (¯ p, f (¯ x) , g (¯ x))]2 ≥

4γΓ T (¯ p, f (¯ x) , f (¯ x)) T (¯ p, g (¯ x) , g (¯ x)) , (γ + Γ)2 1

1

0 ≤ [T (¯ p, f (¯ x) , f (¯ x))] 2 [T (¯ p, g (¯ x) , g (¯ x))] 2 − T (¯ p, f (¯ x) , g (¯ x))  √ √ 2 Γ− γ √ T (¯ p, f (¯ x) , g (¯ x)) ≤ 2 γΓ

and (6.140)

0 ≤ T (¯ p, f (¯ x) , f (¯ x)) T (¯ p, g (¯ x) , g (¯ x)) − T 2 (¯ p, f (¯ x) , g (¯ x)) (Γ − γ)2 2 T (¯ p, f (¯ x) , g (¯ x)) . ≤ 4γΓ

ˇ 6.20. Other Inequalities for the Cebyšev Functional. For two sequences of real numbers ¯ ¯ a = (a , . . . , a ) , b = (b , . . . , b ) and p ¯ = (p1 , . . . , pn ) with pi ≥ 0 (i ∈ {1, . . . , n}) and 1 n 1 n Pn ˇ i=1 pi = 1, consider the Cebyšev functional n n n X X
X ¯ T p ¯, ¯ a, b = p i ai b i − p i ai p i bi .

(6.141)

i=1

i=1

i=1

By Sonin’s identity [1, p. 246] one has the representation n
X

¯ = ¯ , T p ¯, ¯ a, b pi (ai − An (¯ p, ¯ a)) bi − An p ¯, b

(6.142)

i=1

where An (¯ p, ¯ a) :=

n X

p j aj ,

j=1

An

n X
¯ p ¯ , b := p j bj . j=1

Using the (CBS) −inequality for weighted sums, we may state the following result 

¯ 2 ≤ T (¯ ¯ b ¯ , (6.143) T p ¯, ¯ a, b p, ¯ a, ¯ a) T p ¯ , b, where, obviously T (¯ p, ¯ a, ¯ a) =

(6.144)

n X

pi (ai − An (¯ p, ¯ a))2 .

i=1

The following result holds [14]. ¯ = (b1 , . . . , bn ) are real numbers, p Lemma 6.49. Assume that ¯ a =P (a1 , . . . , an ) , b ¯ = (p1 , . . . , pn )
n ¯ for each i ∈ {1, . . . , n} . If are nonnegative numbers with i=1 pi = 1 and bi 6= An p ¯, b (6.145)

−∞ < l ≤

ai − An (¯ p, ¯ a)
≤ L < ∞ for all i ∈ {1, . . . , n} , ¯ b i − An p ¯, b

where l, L are given real numbers, then one has the inequality

¯ ≥ T (¯ ¯ b ¯ . (6.146) (l + L) T p ¯, ¯ a, b p, ¯ a, ¯ a) + LlT p ¯ , b,

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

137

Proof. Using (6.145) we have (6.147)

ai − An (¯ p, ¯ a)
L− ¯ b i − An p ¯, b

!

ai − An (¯ p, ¯ a)
¯ −l b i − An p ¯, b

! ≥0

for each i ∈ {1, . . . , n} .

¯ 2 ≥ 0, we get If we multiply (6.147) by bi − An p ¯, b (6.148)

¯ (ai − An (¯ p, ¯ a))2 + Ll bi − An p ¯, b



2

¯ ≤ (L + l) (ai − An (¯ p, ¯ a)) bi − An p ¯, b





for each i ∈ {1, . . . , n} . Finally, if we multiply (6.148) by pi ≥ 0, sum over i from 1 to n and use the identity (6.142) and (6.144), we obtain (6.146).  Using Lemma 6.49 and a similar argument to that in the previous section, we may state the following theorem [14]. Theorem 6.50. With the assumption of Lemma 6.49 and if L ≥ l > 0, then one has the inequality (6.149)


 ¯ 2≥ T p ¯, ¯ a, b


4lL ¯ b ¯ . T (¯ p , ¯ a , ¯ a ) T p ¯ , b, (L + l)2

The following corollary is natural [14]. Corollary 6.51. With the assumptions in Theorem 6.50 one has
 1
1  ¯ b ¯ 2 −T p ¯ (6.150) 0 ≤ [T (¯ p, ¯ a, ¯ a)] 2 T p ¯ , b, ¯, ¯ a, b √ √ 2 L− l
¯ , √ ≤ T p ¯, ¯ a, b 2 lL and (6.151)



 ¯ b ¯ − T p ¯ 2 0 ≤ T (¯ p, ¯ a, ¯ a) T p ¯ , b, ¯, ¯ a, b ≤


 (L − l)2  ¯ 2. T p ¯, ¯ a, b 4lL

ˇ 6.21. Bounds for the Cebyšev Functional. The following result holds. ¯ = (b1 , . . . , bn ) (with bi 6= bj for i 6= j) be two Theorem 6.52. Let ¯ a = (a1 , . . . , an ) , b sequences of real numbers with the property that there exists the real constants m, M such that for any 1 ≤ i < j ≤ n one has (6.152)

m≤

aj − ai ≤ M. bj − bi

Then we have the inequality


¯ b ¯ ≤T p ¯ ≤ MT p ¯ b ¯ , mT p ¯ , b, ¯, ¯ a, b ¯ , b, P for any nonnegative sequence p ¯ = (p1 , . . . , pn ) with ni=1 pi = 1. (6.153)

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

138

S.S. D RAGOMIR

Proof. From (6.152), by multiplying with (bj − bi )2 > 0, one has m (bj − bi )2 ≤ (aj − ai ) (bj − bi ) ≤ M (bj − bi )2 for any 1 ≤ i < j ≤ n, giving by multiplication with pi pj ≥ 0, that X X pi pj (aj − ai ) (bj − bi ) m pi pj (bi − bj )2 ≤ 1≤i<j≤n

1≤i<j≤n

≤M

X

pi pj (bi − bj )2 .

1≤i<j≤n

Using Korkine’s identity (see for example Subsection 6.19), we deduce the desired result (6.153).  The following corollary is natural. ¯ in Theorem 6.52 is strictly increasing and there Corollary 6.53. Assume that the sequence b exists m, M such that ∆ak (6.154) m≤ ≤ M, k = 1, . . . , n − 1; ∆bk where ∆ak := ak+1 − ak is the forward difference, then (6.153) holds true. Proof. Follows from Theorem 6.52 on taking into account that for j > i and from (6.154) one has j−1 j−1 j−1 X X X m ∆bk ≤ ∆ak ≤ M ∆bk , k=i

k=i

k=i

giving m (bj − bi ) ≤ aj − ai ≤ M (bj − bi ) .



Another possibility is to use functions that generate similar inequalities. Theorem 6.54. Let f, g : [α, β] → R be continuous on [α, β] and differentiable on (α, β) with g 0 (x) 6= 0 for x ∈ (α, β) . Assume that f 0 (x) , x∈(α,β) g 0 (x)

−∞ < m = inf

f 0 (x) = M < ∞. 0 x∈(α,β) g (x) sup

If x ¯ = (x1 , . . . , xn ) is a real sequence with xi ∈ [α, β] and xi 6= xj for i 6= j and if we denote f (¯ x) := (f (x1 ) , . . . , f (xn )) , then we have the inequality (6.155)

mT (¯ p, g (¯ x) , g (¯ x)) ≤ T (¯ p, f (¯ x) , g (¯ x)) ≤ M T (¯ p, g (¯ x) , g (¯ x)) .

Proof. Applying the Cauchy Mean-Value Theorem, for any j > i there exists ξij ∈ (α, β) such that f (xj ) − f (xi ) f 0 (ξij ) = 0 ∈ [m, M ] . g (xj ) − g (xi ) g (ξij ) Then, by Theorem 6.52 applied for ai = f (xi ) , bi = g (xi ) , we deduce the desired inequality (6.155).  The following inequality related to the (CBS) −inequality holds. Theorem 6.55. Let ¯ a, x ¯, y ¯ be sequences of real numbers such that xi 6= 0 and i 6= j, (i, j ∈ {1, . . . , n}) . If there exist the real numbers γ, Γ such that aj − ai (6.156) γ ≤ yj ≤ Γ for 1 ≤ i < j ≤ n, − xyii xj

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

yi xi

6=

yj xj

for

http://jipam.vu.edu.au/

A S URVEY ON C AUCHY-B UNYAKOVSKY-S CHWARZ T YPE D ISCRETE I NEQUALITIES

then we have the inequality  n n X X 2  (6.157) xi yi2 − γ i=1

i=1

n X

!2  ≤

xi yi

n X

i=1

x2i

n X

i=1

ai x i y i −

i=1

Proof. Follows by Theorem 6.52 on choosing pi = omit the details.

x2 Pn i

k=1

x2k

ai x2i

i=1

 n n X X ≤ Γ x2i yi2 − i=1

n X

139

xi yi

i=1

!2  xi yi

.

i=1

i=1

, bi =

n X

n X

yi , xi

m = γ and M = Γ. We 

The following different approach may be considered as well. ¯ = (b1 , . . . , bn ) are sequences of real numbers, Theorem 6.56. Assume that ¯ a = (a1 , . . . , an ) , b P p ¯ = (p1
, . . . ,P pn ) is a sequence of nonnegative real numbers with ni=1 pi = 1 and bi 6= ¯ := n pi bi . If An p ¯, b i=1 (6.158)

−∞ < l ≤

ai − An (¯ p, ¯ a)
¯ ≤ L < ∞ for each i ∈ {1, . . . , n} , b i − An p ¯, b

where l, L are given real numbers, then one has the inequality


¯ b ¯ ≤T p ¯ ≤ LT p ¯ b ¯ . (6.159) lT p ¯ , b, ¯, ¯ a, b ¯ , b,

¯ 2 > 0, we deduce Proof. From (6.158), by multiplying with bi − An p ¯, b



¯ 2 ≤ (ai − An (¯ ¯ l b i − An p ¯, b ¯, b p, ¯ a)) bi − An p

¯ 2, ≤ L b i − An p ¯, b for any i ∈ {1, . . . , n} . By multiplying with pi ≥ 0, and summing over i from 1 to n, we deduce n n X

2 X

¯ ¯ l p i b i − An p ¯, b ≤ pi (ai − An (¯ p, ¯ a)) bi − An p ¯, b i=1

i=1

≤L

n X

¯ p i b i − An p ¯, b



2

.

i=1

Using Sonin’s identity (see for example Section 6.20), we deduce the desired result (6.159).  The following result in connection with the (CBS) −inequality may be stated. ¯ be sequences of real numbers such that xi 6= 0 and Theorem 6.57. Let ¯ a, x ¯, b   yi 1 y 6= Pn 2 An x2 , xi x i=1 xi for i ∈ {1, . . . , n} . If there exists the real numbers φ, Φ such that   ai − Pn1 x2 An x2 , a i=1 i  ≤ Φ,  (6.160) φ≤ y yi 1 2 P − n x2 An x , x xi i=1 i   y y1 yn 2 2 2 where x = (x1 , . . . , xn ) and x = x1 , . . . , xn , then one has the inequality (6.157).

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

140

S.S. D RAGOMIR

Proof. Follows by Theorem 6.56 on choosing pi = the details.

x2 Pn i

i=1

x2i

, bi =

yi , xi

l = φ and L = Φ. We omit 

R EFERENCES ´ J.E. PECARI ˇ [1] D.S. MITRINOVIC, C´ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993. [2] A.M. OSTROWSKI, Vorlesungen über Differential und Integralrechnung II, Birkhäuser, Basel, 1951. [3] S.S. DRAGOMIR, Ostrowski’s inequality in complex inner product spaces, submitted. [4] S.S. DRAGOMIR, Another Ostrowski type inequality, submitted. [5] K. FAN AND J. TODD, A determinantal inequality, J. London Math. Soc., 30 (1955), 58–64. [6] S.S. DRAGOMIR AND V.M. BOLDEA, Some new remarks on Cauchy-Buniakowski-Schwartz’s inequality, Coll. Sci. Pap. Faculty of Science Kragujevac, 21 (1999), 33–40. [7] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s Inequality for Real Numbers (Romanian), “Caiete Metodico-Stiintifice”, ¸ No. 57, 1989, pp. 24, Faculty of Mathematics, University of Timi¸soara, Romania. ´ A Cauchy-Buniakowski[8] S.S. DRAGOMIR, D.M. MILOŠEVIC´ AND Š.Z. ARSLANAGIC, Schwartz inequality for Lipschitzian functions, Coll. Sci. Pap. of the Faculty of Science, Kragujevac, 14 (1991), 25–28. [9] S.S. DRAGOMIR, Generalisation of the (CBS) −inequality via Jensen type inequalities, submitted. ´ Converse of Jensen’s inequality for convex functions, Univ. Beograd [10] P. LAH AND M. RIBARIC, Publ. Elek. Fak. Ser. Math, Fiz., No. 412-460, (1973), 201–205. [11] J.B. DIAZ AND F.T. METCALF, Stronger forms of a class of inequalities of G. Pólya-G. Szegö and L.V. Kantorovich, Bull. Amer. Math. Soc., 69 (1963), 415–419. [12] S.S. DRAGOMIR AND N.M. IONESCU, Some converse of Jensen’s inequaliy and applications, Anal. Num. Theor. Approx., 23 (1994), 71–78. ˇ [13] J.E. PECARI C´ AND S.S. DRAGOMIR, A refinement of Jensen’s inequality and applications, Studia Univ. Babe¸s-Bolyai, Mathematica, 34(1) (1989), 15–19. ˇ [14] S.S. DRAGOMIR, Some inequalities for the Cebyšev functional, The Australian Math. Soc. Gazette., 29(3) (2002), 164–168. [15] S.S. DRAGOMIR AND F.P. SCARMOZZINO, A refinement of Jensen’s discrete inequality for differentiable convex functions, Preprint, RGMIA Res. Rep. Coll., 5(4) (2002), Article 12. [ONLINE: http://rgmia.vu.edu.au/v5n4.html] [16] M.S. SLATER, A companion inequality to Jensen’s inequality, J. Approx. Theory, 32 (1981), 160– 166. [17] D. ANDRICA AND I. RASA, ¸ The Jensen inequality: refinements and applications, Anal. Num. Theor. Approx., (Romania), 14 (1985), 155–167. [18] S.S. DRAGOMIR, On an inequality for convex functions, submitted.

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/

I NDEX J−convex function, 16, 17 ˇ Cebyšev functional, 99, 134, 136 ˇ Cebyšev’s inequality, 10, 114, 115 1-norm, 87

Euler’s indicator, 18, 119

additive, 10, 63, 64, 73–75 Alzer, 45, 85 Andrica-Badea inequality, 92, 94, 95 arithmetic mean, 36, 63, 67, 116 Arithmetic-Geometric inequality, 11 asynchronous, 10, 114 sequence, 114

geometric mean, 36, 63, 67, 116 Grüss inequality, 94 Greub-Rheinboldt inequality, 75

Fan and Todd inequalities, 113 forward difference, 85, 88

harmonic mean, 36, 37, 116 identity, 6, 41, 42, 44, 59, 61, 86, 100, 137 Binet-Cauchy, 6 Cauchy-Binet, 117 Korkine’s, 134 Lagrange, 5, 7, 44, 59, 61, 118 Sonin, 136 index set mapping, 54, 57, 61, 62, 67 inner product spaces, 5, 43, 60

barycentric method, 72 Binet-Cauchy’s identity, 6 Callebaut’s inequality, 21, 43 Cassels’ inequality, 72, 75 additive version, 73 refinement, 94, 96 unweighted, 72–74, 96 Cauchy, 5 Cauchy-Bunyakovsky-Schwarz, 5 Cauchy-Schwarz, 45 Cauchy Mean-Value Theorem, 135 Cauchy-Bunyakovsky-Schwarz, 5 Cauchy-Schwarz, 45 CBS-inequality, 5, 12–19, 26, 35, 44, 117, 122 complex numbers, 6 complex sequences, 110 double sums, 117, 118 for Lipschitzian functions, 119 for weights, 33, 51 functionals associated to, 68 generalisation, 7, 12, 14–17, 22 generalisation via Young’s inequality, 13 inequality related to, 64 real numbers, 5, 7, 24, 41, 52 real sequence, 109 refinement, 11, 32, 37, 40, 41, 57, 58, 61, 63–65, 126, 128 discrete, 42 reverse, 75, 82, 84, 85, 87, 89, 91, 93, 95, 99, 101, 104 type, 20 weights, 10 concave function, 121–123 convex mapping, 116

Jensen’s inequality, 116, 125 discrete, 121 reverse, 122 refinement, 124, 127 reverse, 123 Korkine’s identity, 134 Lagrange’s identity, 5, 7, 44, 59, 61, 118 Lah-Ribari´c inequality, 122 Lipschitzian functions, 119 Lupa¸s inequality, 94 matrices, 117 maximum, 72, 92, 93 McLaughlin’s inequality, 41 Milne’s inequality, 43 minimum, 72 monotonicity, 49, 51, 53, 54, 56, 58, 85 multiple sums, 125 n-tuple, 122 non-proportional, 109, 110, 112 Ostrowski’s inequality, 109–111 p-norm, 90 Pólya-Szegö, 74 Pólya-Szegö’s inequality, 73–75, 97 parameter, 10 Popoviciu’s inequality, 18 positive real numbers, 9, 11, 12, 31, 36, 37, 72, 73, 75, 78, 96, 115, 116, 122, 124, 126 Power functions, 116 power series, 19 proportional, 5, 7, 12

Daykin-Eliezer-Carlitz, 42, 43 De Bruijn’s inequality, 40 Diaz-Metcalf inequality, 97, 99, 123 discriminant, 5, 24, 26 double sums, 117, 118, 134 Dragomir-Ionescu inequality, 123 Dunkl-Williams’ inequality, 43

quadratic polynomial, 5, 23

141

142

S.S. D RAGOMIR

refinement, 35 Cassels’ inequality, 94, 96 CBS inequality, 11, 32, 37, 40, 41, 44, 57, 58, 61, 63–65, 126, 128 due to Alzer and Zheng, 45 due to Daykin-Eliezer-Carlitz, 42 in terms of moduli, 28, 31 Jensen’s inequality, 124, 127 non-constant sequences, 37 Pólya-Szegö inequality, 97 Schwitzer’s result, 94 sequence less than the weights, 33 sequence whose norm is 1, 29 via Dunkl-Williams’ Inequality, 43 reverse inequality, 75, 82, 84, 85, 87, 89–91, 93, 97, 99, 101, 104 complex numbers, 79 Jensen, 122, 123 real numbers, 77 ˇ via Cebyšev Functional, 99 via Andrica-Badea, 92 via Diaz-Metcalf, 97 rotation, 41

superadditivity, 52, 54–57, 59, 61, 62 supermultiplicative, 63 supermultiplicity, 63, 67 synchronous, 115 sequence, 115 triangle inequality, 7, 86 Wagner’s inequality, 22, 24 weights, 10, 33, 51, 52, 55, 56, 63 Young’s Inequality, 11–13 Zagier inequalities, 83, 84 Zheng, 45

Schweitzer inequality, 94 sequence, 36, 42, 50, 55, 77, 85, 87, 100, 109 asynchronous, 114 complex, 110 complex numbers, 6, 12, 24, 44, 75, 85, 87, 90, 97, 110, 112 decreasing nonnegative, 85 natural numbers, 119 nonnegative, 9, 10, 63, 85, 114, 134 nonnegative numbers, 11 nonnegative real numbers, 12, 21, 22, 33, 51, 52, 72, 75, 85, 87, 90, 92, 96, 121–124 nonzero complex numbers, 44 nonzero real numbers, 17 positive, 82 positive and real numbers, 8 positive numbers, 7, 42 positive real numbers, 9, 11, 12, 31, 37, 72, 73, 75, 78, 96, 115, 116, 122, 124, 126 real, 109, 114, 135 real numbers, 7, 8, 10, 18–20, 22, 28, 29, 31–37, 40, 41, 51, 52, 59, 84, 87, 89, 91–94, 99, 102, 105, 109, 111, 113, 114, 117–123, 134, 136 synchronous, 115 set-additive, 67, 68 set-superadditive, 67–70 set-superadditivity, 68 set-supermultiplicative, 67 Shisha-Mond inequalities, 82 Sonin’s identity, 136 strong monotonicity, 56, 58 strong superadditivity, 55–57 sup-norm, 85

J. Inequal. Pure and Appl. Math., 4(3) Art. 63, 2003

http://jipam.vu.edu.au/