Chuyenhungvuong.net 100 Inequality Problems

Upload by wWw.chuyenhungvuong.net

100 INEQUALITY PROBLEMS ***** Cao Minh Quang 05/11/2006

Most of these problems were collected from the Mathematics and Youth Magazine in Vietnam. Translation from Vietnamese into English may have many errors. I am looking forward to hearing from your ideas. • Address: Cao Minh Quang, Mathematics teacher, Nguyen Binh Khiem specialized High School, Vinh Long town, Vinh Long, Vietnam. • Email: [email protected]

Upload by wWw.chuyenhungvuong.net

Upload by wWw.chuyenhungvuong.net 1. ( a, b > 0, a + b = 1) ,

a b 4 + 2 ≤ . a +1 b +1 5 2

First solution.

Applying the AM – GM Inequality we get

1 3 1 3 4a + 3 a 4a a2 + 1 = a2 + + ≥ 2 a2 . + = ⇒ 2 ≤ . 4 4 4 4 4 a + 1 4a + 3 Similarly, b 4b . ≤ b 2 + 1 4b + 3

Adding these two inequalities, a b 4b ⎞ ⎛ 4a + 2 ≤⎜ + ⎟= a + 1 b + 1 ⎝ 4a + 3 4b + 3 ⎠ 2

⎡ 1 ⎞⎤ ⎛ 1 ⎢ 2 − 3 ⎜ 4a + 3 + 4b + 3 ⎟ ⎥ . ⎝ ⎠⎦ ⎣

On the orther hand, ⎡⎛ 1 1 ⎞⎤ 1 ⎞ 4 2 ⎛ 1 ⎡⎣( 4a + 3) + ( 4b + 3) ⎤⎦ ⎢⎜ + + = . ⎟⎥ ≥ 4 ⇒ ⎜ ⎟≥ ⎝ 4a + 3 4b + 3 ⎠ 4 ( a + b ) + 6 5 ⎣⎝ 4a + 3 4b + 3 ⎠⎦

Thus, a b 2 4 + 2 ≤ 2 − 3. = . a +1 b +1 5 5 2

Second solution.

Applying the AM – GM Inequality we get 4

1 1 1 1 5 ⎛1⎞ a + 1 = a + + + + ≥ 5 5 a 2 ⎜ ⎟ = 5 4a 2 . 4 4 4 4 4 ⎝4⎠ 2

2

Similarly,

b2 + 1 ≥

55 2 4b . 4

Adding these two inequalities,

a b a b 4⎛ 1 1 + 2 ≤ + = ⎜⎜ 5 a 3 . . + a + 1 b + 1 5 5 4a 2 5 5 4b 2 5 ⎝ 2 2 4 4 2

5

1 1 b3. . 2 2

⎞ ⎟⎟ ≤ ⎠

1 1⎞ ⎛ 1 1 ⎞⎤ ⎡⎛ a + a + a + + ⎟ ⎜ b + b + b + + ⎟⎥ ⎢ ⎜ 4 2 2 + 2 2 = 4 ⎡ 3(a + b) + 2 ⎤ = 4 . ≤ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎢ ⎥ 5 ⎢⎜ 5 5 5⎣ 5 ⎥ ⎦ 5 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎠ ⎝ ⎠ ⎦⎥ ⎣⎢⎝ Third solution.

Applying the AM – GM Inequality we get

a + b ≥ 2 ab ⇒ ab ≤

1 . 4

Therefore,

a b ab 2 + a 2 b + a + b ab + 1 + 2 = 2 2 = = 2 2 2 2 a + 1 b + 1 a b + a + b + 1 ( a + b ) − 2ab + a 2 b 2 + 1 1 +1 ab + 1 ab + 1 4 4 = 2 2 = ≤ = . 2 a b − 2ab + 2 ⎛ 1⎞ 3 31 − 3 . 1 + 31 5 ⎜ ab − ⎟ − ab + 2 4 16 4⎠ 2 16 ⎝ 2. ( a, b, c > 0, a + b + c = 1) ,

a b c 3 + + ≤ . a +1 b +1 c +1 4 First solution.

Applying the AM – GM Inequality we get a a 1⎛ a a ⎞ = ≤ ⎜ + ⎟, a + 1 (a + b) + (a + c) 4 ⎝ a + b a + c ⎠ b b = ≤ b + 1 ( b + a ) + ( b + c)

1⎛ b b ⎞ + ⎜ ⎟, 4⎝ b + a b+ c⎠

c c 1⎛ c c ⎞ = ≤ ⎜ + ⎟. c + 1 (c + b) + (c + a ) 4 ⎝ c + b c + a ⎠ Adding these three inequalities,

a b c 1⎛ a + b b + c a + c⎞ 3 + + ≤ ⎜ + + ⎟= . a +1 b +1 c +1 4 ⎝ a + b b + c a + c ⎠ 4 Second solution.

Applying the AM – GM Inequality we get 1 1 1 1 a 1 a +1 = a + + + ≥ 44 a 3 ⇒ ≤ 4 a3 33 , 3 3 3 a +1 4 3

Similarly, b 1 ≤ b +1 4

4

b 3 33 ,

c 1 ≤ c+1 4

4

c3 33 .

Adding these three inequalities, a b c 3⎛ 1 1 1⎞ + + ≤ ⎜⎜ 4 a.a.a. + 4 b.b.b. + 4 c.c.c. ⎟⎟ ≤ a +1 b +1 c +1 4 ⎝ 3 3 3⎠

1⎞ ⎛ 1⎞ ⎛ 1 ⎞⎤ ⎡⎛ a + a + a + ⎟ ⎜ b + b + b + ⎟ ⎜ c + c + c + ⎟⎥ ⎢ ⎜ 3 3 + 3 + 3 =3. ≤ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ 4 ⎢⎜ 4 4 4 ⎟ ⎜ ⎟ ⎜ ⎟⎥ 4 ⎠ ⎝ ⎠ ⎝ ⎠ ⎦⎥ ⎣⎢⎝ Third solution.

Applying the AM – GM Inequality we get 1 1 ⎞ ⎛ 1 + + ⎡⎣( a + 1) + ( b + 1) + ( c + 1) ⎤⎦ ⎜ ⎟≥9 ⎝ a +1 b +1 c +1⎠

1 1 ⎞ 9 9 ⎛ 1 ⇒⎜ + + = . ⎟≥ ⎝ a + 1 b + 1 c + 1 ⎠ ( a + 1) + ( b + 1) + ( c + 1) 4

Therefore, a b c 1 1 ⎞ 9 3 ⎛ 1 + + = 3−⎜ + + ⎟ ≤ 3− = . a +1 b +1 c +1 4 4 ⎝ a +1 b +1 c +1⎠

Forth solution

Applying the AM – GM Inequality we get

1 2 a 2 a 3a 1 1 a +1 = a + + ≥ 2 + ⇒ ≤ + − . 3 3 3 3 a +1 2 2 3

1 . a 2 2 + 3 3

Similarly, b 3b 1 1 ≤ + − . b +1 2 2 3

1 c 3c 1 1 1 , . ≤ + − . 2 2 3 b 2 c +1 c 2 + 2 2 + 3 3 3 3

Adding these three inequalities, ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ a b c 3 a b c 3 1 1 1 1 ⎜ ⎟≤ + + ≤ ⎜⎜ + + + − + + ⎟ a +1 b +1 c +1 2 ⎝ 3 3 3 ⎟⎠ 2 3 ⎜ a 2 b 2 b 2⎟ + 2 + 2 + ⎟ ⎜2 3 3 3 3 3 3⎠ ⎝

1⎞ ⎛ 1⎞ ⎛ 1 ⎞⎤ ⎡⎛ a + ⎟ ⎜ b + ⎟ ⎜ c + ⎟⎥ ⎢ ⎜ 3 3 + 3 + 3 +3− ≤ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ 2 ⎢⎜ 2 ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎥ 2 ⎛ a + 2 ⎜ ⎢⎣ ⎝ ⎥ ⎠ ⎝ ⎠ ⎝ ⎠⎦ 3 ⎝

3. ( a, b, c ≥ 1) ,

1 1 1⎞ + + ⎟≥9 ⎝a b c⎠

( 2 + abc ) ⎛⎜ Solution.

We have,

9 b + 3

c⎞ 2 ⎟ + 3. 3⎠ 3

≤3−

9 3 = . 4 4

( a − 1)( b − 1) ≥ 0 ⇔ ab + 1 ≥ a + b , ( ab − 1)( c − 1) ≥ 0 ⇔ abc + 1 ≥ ab + c . Adding these two inequalities, we obtain abc + 2 ≥ a + b + c .

Thus,

1 1 1⎞ ⎛1 1 1⎞ + + ⎟ ≥ ( a + b + c) ⎜ + + ⎟ ≥ 9 . ⎝a b c⎠ ⎝a b c⎠

( 2 + abc ) ⎛⎜

(

)

4. x, y, z > 0, xy xy + yz yz + zx zx = 1 , x6 y6 z6 1 + + ≥ . 3 3 3 3 3 3 x +y y +z z +x 2

Solution.

We set a = x 3 , b = y3 , c = z 3 and observe that

ab + bc + ca = 1 .

The inequality is equivalent to

a2 b2 c2 1 + + ≥ . a+ b b+c c+a 2 Applying the Cauchy – Buniakowski Inequality we get

⎡ a2 b2 c2 ⎤ 2 ⎢ a + b + b + c + c + a ⎥ ⎡⎣( a + b) + ( b + c) + ( c + a) ⎤⎦ ≥ ( a + b + c) ⎣ ⎦ a2 b2 c2 1 1 ⇒ + + ≥ ( a + b + c) ≥ a+ b b+c c+a 2 2

(

)

ab + bc + ca =

1 . 2

5. ( x, y > 0, xy = 1) ,

x 2 + 3x + y 2 + 3y +

9 ≥ 11 . x + y2 + 1 2

Solution.

Applying the AM – GM Inequality we get

(x 6. ( m, n ∈

2

+ y 2 + 1) +

9 9 + 3x + 3y − 1 ≥ 4 4 ( x 2 + y 2 + 1) . 2 .3x.3y − 1 = 11 . 2 x + y +1 x + y2 + 1 2

\ {0} , m ≤ a, b, c ≤ n ) , a b c 3 ( n − m) . + + ≤ + b + c a + c a + b 2 2m ( n + m) 2

Solution.

Without loss of generality, we can assume that a ≥ b ≥ c , we set x = b + c, y = c + a, z = a + b . We observe that x ≤ y ≤ z , therefore,

y ⎞⎛ z⎞ ⎛ x ⎞⎛ y⎞ ⎛ ⎜1 − ⎟ ⎜1 − ⎟ + ⎜1 − ⎟ ⎜1 − ⎟ ≥ 0 x ⎠⎝ y⎠ ⎝ y ⎠⎝ z⎠ ⎝ ⇔

⎛x z⎞ y+z x+z x+y + + ≤ 2⎜ + ⎟ + 2 x y z ⎝y x⎠

(z − x) + 6 1 . b c ⎞ ⎛ a ⇔ 2⎜ + + () ⎟+3≤ 2 xy ⎝ b+c c+a a+ b ⎠ 2

On the other hand,

0≤

z−x a−c n−m z−x a−c a−c n−c n−m and 0 ≤ . = ≤ = ≤ ≤ ≤ x b+c 2m z a+ b a+c n+c n+m

Thus,

(z − x) xz

2

(n − m) 2 . ≤ ( ) 2m ( n + m ) 2

From (1) and (2), we have a b c 3 ( n − m) . + + ≤ + b + c a + c a + b 2 2m ( n + m) 2

7. ( 0 < x1 , x 2 , ..., x n ≤ 1, n ≥ 2 ) , x1 + x 2 + ... + x n 1 . ≤ n 1 + n (1 − x1 )(1 − x 2 ) ... (1 − x n )

Solution.

We set 1 − x i = ai , ( i = 1, 2, ..., n ) and observe that 0 ≤ ai < 1 , ( i = 1, 2, ..., n ) . The inequality is equivalent to

(1 − a1 ) + (1 − a2 ) + ... + (1 − an ) ≤ n

1 , 1 + na1a2 ...an

or ( a1 + a2 + ... + an )(1 + a1a2 ...an ) ≥ n 2 a1a2 ...an . Applying the AM – GM Inequality we get

a1 + a2 + ... + an ≥ n n a1a2 ...an and 1 + na1a2 ...an ≥ 1 + ( n − 1) a1a2 ...an ≥ n n ( a1a2 ...an )

n −1

.

Therefore,

( a1 + a2 + ... + an )(1 + a1a2 ...an ) ≥ n2 n ( a1a2 ...an )

n

= n2 ( a1 + a2 + ... + an )(1 + a1a2 ...an ) ≥ n2a1a2 ...an .

8. ( 0 < x < y ≤ z ≤ 1,3x + 2y + z ≤ 4 ) ,

3x 2 + 2y 2 + z 2 ≤ Solution.

10 . 3

We have 3x + 2y + z ≤ 4 , thus, 3x 2 + 2xy + xz ≤ 4x . On the other hand, 0 < x < y ≤ z ≤ 1 , therefore 2y ( y − x ) ≤ 2 ( y − x ) and z ( z − x ) ≤ z − x . Adding these two inequalities, we obtain

3x 2 + 2y 2 + z 2 ≤ x + 2y + z . Applying the Cauchy – Buniakowski Inequality we get

(3x

2

2 10 2 ⎛1 ⎞ + 2y2 + z2 ) ≤ ( x + 2y + z ) ≤ ⎜ + 2 + 1⎟ ( 3x2 + 2y2 + z2 ) ⇒ 3x2 + 2y2 + z2 ≤ . 3 ⎝3 ⎠

9. ( 0 ≤ x ≤ y ≤ z ≤ 1) ,

x 2 ( y − z ) + y 2 ( z − y ) + z 2 (1 − z ) ≤

108 . 529

Solution.

We set T = x 2 ( y − z ) + y 2 ( z − y ) + z 2 (1 − z ) . Applying the AM – GM Inequality and the Cauchy – Buniakowski Inequality we get 3

1 1 ⎛ y + y + 2z − 2y ⎞ 2 T ≤ 0 + ⎡⎣ y.y ( 2z − 2y ) ⎤⎦ + z 2 (1 − z ) ≤ ⎜ ⎟ + z (1 − z ) 2 2⎝ 3 ⎠ 2

⎛ 4 ⎞ ⎛ 23 ⎞ ⎛ 54 ⎞ ⎛ 23 ⎞ ⎛ 23 ⎞ ⎛ 23 ⎞ = z2 ⎜ z + 1 − z ⎟ = z2 ⎜ 1 − z ⎟ = ⎜ ⎟ ⎜ z ⎟ ⎜ z ⎟ ⎜ 1 − z ⎟ ⎝ 27 ⎠ ⎝ 27 ⎠ ⎝ 23 ⎠ ⎝ 54 ⎠ ⎝ 54 ⎠ ⎝ 27 ⎠ 23 23 ⎤ ⎡ 23 2 z + z +1− z ⎥ ⎛ 54 ⎞ ⎢ 54 54 27 = 108 . ≤⎜ ⎟ ⎢ ⎥ 3 ⎝ 23 ⎠ ⎢ ⎥ 529 ⎣ ⎦

(

)

10. a, b, c ∈ , a 2 + b 2 + c 2 ≤ 8 , ab + bc + 2bc ≥ −8 .

Solution. 2

2 b ⎛ ⎞ 3b ≥ 0. We have 8 + ( ab + bc + 2bc ) ≥ a + b + c + ab + bc + 2bc = ⎜ a + + c ⎟ + 2 4 ⎝ ⎠ 2

2

2

Therefore, ab + bc + ca ≥ −8 . 11. ( a, b > 0 ) ,

( a − b ) ( 3a + b )( a + 3b ) . a+ b ≥ ab + 2 8 ( a + b ) ( a2 + 6ab + b 2 ) 2

Solution.

If a = b , the inequality is true. If a ≠ b , the inequality is equivalent to

( a − b ) ( 3a + b )( a + 3b ) a+ b − ab ≥ 2 8 ( a + b ) ( a2 + 6ab + b 2 ) 2

⇔

(

a− b

)

2

⎡ ⎢1 − ⎢ ⎢⎣

(

a+ b

) ( 3a + b )( a + 3b ) ⎤⎥ ≥ 0 2

4 ( a + b ) ( a2 + 6ab + b 2 ) ⎥ ⎥⎦

(

)

⇔ 4 ( a + b ) ( a2 + 6ab + b2 ) − a + b + 2 ab ( 3a2 + 3b 2 + 10ab ) ≥ 0 . We set x = a + b, y = ab and observe that a2 + b2 = x 2 − 2y 2 . The inequality is equivalent to 4x ( x 2 + 4y 2 ) − ( x + 2y ) ( 3x 2 + 4y 2 ) ≥ 0 ⇔ ( x − 2y ) ≥ 0 ⇔ x ≥ 2y ⇔ a + b ≥ 2 ab . 3

12. ( x ∈

), sin x + sin 2x + sin 3x <

3 3 . 2

Solution.

We have,

sin x + sin 2x + sin 3x = sin 2x + 2 sin 2x cos x = sin 2x + 4 sin x cos2 x ≤ cos2 x cos2 x ≤ 1 + 4 sin x cos x ≤ 1 + 4 sin x cos x = 1 + 8 sin x. . 2 2 2

2

⎛ cos 2 x cos 2 x 2 sin x + + ⎜ 2 2 ≤ 1+ 8 ⎜ 3 ⎜⎜ ⎝

4

2

3

⎞ ⎟ 8 3 3 3 < . ⎟ = 1+ 9 2 ⎟⎟ ⎠

13. ( x1 , x 2 , ..., x n > 0, n ≥ 3) ,

x12 + x 2 x 3 x 2 + x3x 4 x 2 + x n x1 x 2 + x1x 2 + 2 + ... + n −1 + n ≥n. x1 ( x 2 + x 3 ) x 2 ( x 3 + x 4 ) x n −1 ( x n + x1 ) x n ( x1 + x 2 ) Solution.

We set,

A=

x12 + x 2 x 3 x2 + x3x 4 x 2 + x n x1 x 2 + x1x 2 + 2 + ... + n −1 + n . x1 ( x 2 + x 3 ) x 2 ( x 3 + x 4 ) x n −1 ( x n + x1 ) x n ( x1 + x 2 )

Applying the AM – GM Inequality we have

⎡ x12 + x 2 x 3 ⎤ ⎡ x 22 + x 3 x 4 ⎤ ⎡ x 2n −1 + x n x1 ⎤ ⎡ x 2n + x1x 2 ⎤ A+n = ⎢ + 1⎥ + ⎢ + 1⎥ + ... + ⎢ + 1⎥ + ⎢ + 1⎥ ⎣ x1 ( x 2 + x 3 ) ⎦ ⎣ x 2 ( x 3 + x 4 ) ⎦ ⎣ x n −1 ( x n + x1 ) ⎦ ⎣ x n ( x1 + x 2 ) ⎦ =

( x1 + x 2 )( x1 + x3 ) + ( x 2 + x3 )( x 2 + x 4 ) + ... + ( x n −1 + x n )( x n −1 + x1 ) + ( x n + x1 )( x n + x 2 ) x1 ( x 2 + x 3 ) x2 ( x3 + x 4 ) x n −1 ( x n + x1 ) x n ( x1 + x 2 )

≥ nn

( x1 + x3 )( x 2 + x 4 ) ... ( x n −1 + x1 )( x n + x 2 ) ≥ n n 2n

x1x 3 x 2 x 4 ... x n −1x1 x n x 2 = 2n . x1x 2 ...x n −1x n

x1x 2 ...x n −1x n

Therefore,

x12 + x 2 x 3 x 22 + x 3 x 4 x 2n −1 + x n x1 x 2n + x1x 2 + + ... + + ≥n. x1 ( x 2 + x 3 ) x 2 ( x 3 + x 4 ) x n −1 ( x n + x1 ) x n ( x1 + x 2 ) 14. (1 ≤ a, b, c ≤ 2 ) , 1 1 1⎞ + + ⎟ ≤ 10 . ⎝a b c⎠

( a + b + c) ⎜⎛ Solution.

Without loss of generality, we can assume that 1 ≤ a ≤ b ≤ c ≤ 2 , we obtain,

a b b c a c ⎛ a ⎞⎛ b ⎞ ⎛ b ⎞⎛ c ⎞ ⎜1 − ⎟ ⎜1 − ⎟ + ⎜1 − ⎟ ⎜1 − ⎟ ≥ 0 ⇔ + + + ≤ 2 + + . b c a b c a ⎝ b ⎠⎝ c ⎠ ⎝ a ⎠⎝ b ⎠ Therefore, 1 1 1⎞ ⎛a b b c⎞ a c ⎛a c⎞ + + ⎟ = 3 + ⎜ + + + ⎟ + + ≤ 5 + 2⎜ + ⎟ . ⎝a b c⎠ ⎝b c a b⎠ c a ⎝ c a⎠

( a + b + c) ⎛⎜ On the other hand,

2

1 a a ⎞⎛ 1 a ⎞ 5 a a c 5 ⎛ ⎛a⎞ ≤ ≤ 1 ⇒ ⎜ 2 − ⎟⎜ − ⎟ ≤ 0 ⇒ ⎜ ⎟ + 1 ≤ . ⇒ + ≤ . 2 c c ⎠⎝ 2 c ⎠ 2 c c a 2 ⎝ ⎝c⎠ Thus,

1 1 1⎞ + + ⎟ ≤ 10 . ⎝a b c⎠

( a + b + c) ⎜⎛ 15. ( a, b, c, d > 0 ) ,

a2 b 2 c2 d 2 a + b + c + d + + + ≥ 4 . b 2 c2 d 2 a2 abcd

Solution.

Applying the AM – GM Inequality we get

a2 b 2 c2 a6 8a 8 3. 2 + 2. 2 + 2 + 2 ≥ 8 2 2 2 = 4 , b c d bcd abcd b2 c2 d 2 b6 8b 8 3. 2 + 2. 2 + 2 + 2 ≥ 8 2 2 2 = 4 , c d a cda abcd c2 d 2 a2 c6 8c 8 3. 2 + 2. 2 + 2 + 2 ≥ 8 2 2 2 = 4 , d a b dab abcd d2 a2 b 2 d6 8d 8 3. 2 + 2. 2 + 2 + 2 ≥ 8 2 2 2 = 4 . a b c abc abcd

Adding these four inequalities, we obtain ⎛ a2 b 2 c2 d 2 ⎞ ⎛ a+ b+c+d ⎞ ⎛ a+ b+c+d ⎞ ⎛ a+ b+c+d ⎞ 6 ⎜ 2 + 2 + 2 + 2 ⎟ + 8 ≥ 8⎜ 4 = 6⎜ 4 ⎟ ⎟ + 2⎜ ⎟ c d a ⎠ abcd ⎠ abcd ⎠ ⎝ 4 abcd ⎠ ⎝ ⎝ ⎝b

4 4 abcd ⎛ a+ b+c+d ⎞ ⎛ a+ b+c+d ⎞ ≥ 6⎜ 4 + 2. = 6⎜ 4 ⎟ ⎟+8. 4 abcd ⎠ abcd abcd ⎠ ⎝ ⎝ Thus, a2 b 2 c2 d 2 a + b + c + d + + + ≥ 4 . b 2 c2 d 2 a2 abcd

16. ( 0 ≤ x ≤ 2 ) , 4x − x 3 + x + x 3 ≤ 3 4 3 .

Solution.

Applying the Cauchy – Buniakowski Inequality we get

(

)

4x − x 3 + x + x 3 =

=

17. ( x ∈

1⎡ 1 2 8x − 2x 3 + 2 x + x 3 ⎤ ≤ ⎦ 2 2⎣

1 6 4 2 . 2x ( 9 − x 242

( 2 + 4 ) ( 8x − 2x3 + x + x3 ) =

2 2 2 6 ⎛ 2x + ( 9 − x ) + ( 9 − x ) ⎞ ⎟ = 34 3 . = 4 4⎜ ⎜ ⎟ 3 2 2 ⎝ ⎠ 3

)

2 2

), 2 sin x + 15 − 10 2 cos x ≤ 6 .

Solution.

Applying the Cauchy – Buniakowski Inequality we get

2 sin x + 15 − 10 2 cosx ≤

(

18. x, y, z > 0, x 2 + y 2 + z 2 = 1, n ∈

+

(1 + 5) ( 2sin2 x + 3 − 2

)

2 cosx = 6 ⎡6 − ⎢⎣

(

)

2 2 cosx + 1 ⎤ ≤ 6 . ⎥⎦

),

2n + 1) 2n 2n + 1 ( x y z . + + ≥ 1 − x 2n 1 − y 2n 1 − z 2n 2n Solution.

We set f ( t ) = t (1 − t 2n ) , t ∈ ( 0,1) . It is easy to show that f ' ( t ) = 1 − ( 2n + 1) t 2n+1 , f ' ( t ) = 0 ⇔ t = 2n

1 2n ⎛ 1 ⎞ 2n ⇒ f ( t ) ≤ f ⎜ 2n ⎟ ⇒ t (1 − t ) ≤ 2n + 1 ( 2n + 1) 2n 2n + 1 ⎝ 2n + 1 ⎠

Since 0 < x, y,z < 1 , thus 2n + 1) 2n 2n + 1 2n + 1) 2n 2n + 1 2n + 1) 2n 2n + 1 ( ( ( 1 1 1 ≥ , ≥ , ≥ 2n 2n 2n x (1 − x 2n ) y (1 − y 2n ) z (1 − z 2n )

Adding these three inequalities, we obtain

⎛ x y z + + ⎜ 2n 2n 1− y 1 − z2n ⎝1− x

(x ≥

2

⎞ ⎞ ⎛ x2 y2 z2 ⎜ ⎟≥ = + + ⎟ 2n 2n 2n ⎠ ⎝⎜ x (1 − x ) y (1 − y ) z (1 − z ) ⎠⎟

+ y 2 + z 2 ) ( 2n + 1 ) 2 n 2n + 1 2n

2n + 1 ) 2 n 2n + 1 ( = . 2n

19. ( a, b > 0, a + b < 1) ,

a2 b2 1 9 + + +a+ b ≥ . 1− a 1− b a + b 2 Solution.

We have,

⎛ a2 ⎞ ⎛ a2 ⎞ ⎛ 1 b2 1 b2 1 1 1 ⎞ + + + a + b = + 1 + a + + 1+ b + − 2⎟ = ⎜ + + − 2⎟ ⎜ ⎟ ⎜ 1− b a + b ⎠ ⎝ 1− a 1− b a + b ⎠ ⎝ 1− a 1− b a + b ⎠ ⎝ 1− a Applying the Cauchy – Buniakowski Inequality we get

(1 + 1 + 1) 1 1 ⎞ 9 ⎛ 1 + + = . ⎜ ⎟≥ ⎝ 1 − a 1 − b a + b ⎠ 1(1 − a) + 1(1 − b) + 1. ( a + b ) 2 2

20. ( a, b, c > 0, abc = 1) ,

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

Solution.

Applying the AM – GM Inequality we get

a2 + b2 ≥ 2ab and b 2 + 1 ≥ 2b . Therefore,

a2 + 2b2 + 3 ≥ 2 ( ab + b +1) ⇒

1 1 ≤ . 2 a + 2b + 3 2 ( ab + b +1) 2

Similarly,

1 1 1 1 ≤ and 2 . ≤ 2 2 b + 2c + 3 2 ( bc + c +1) c + 2a + 3 2 ( ac + a + 1) 2

Adding these three inequalities, we obtain

1 1 1 1⎛ 1 1 1 ⎞ + 2 + 2 ≤ ⎜ + + ⎟= 2 2 2 a + 2b + 3 b + 2c + 3 c + 2a + 3 2 ⎝ ab + b + 1 bc + c + 1 ca + a + 1 ⎠ 2

1⎛ 1 ab b ⎞ 1 ab + b + 1 1 = ⎜ + + = . ⎟= 2 ⎝ ab + b + 1 b + 1 + ab 1 + ab + b ⎠ 2 ab + b + 1 2

(

)

21. x, y > 0, x 2 + y 2 = 1 ,

⎛

(1 + x ) ⎜ 1 + ⎝

1⎞ ⎛ 1⎞ ⎟ + (1 + y ) ⎜ 1 + ⎟ ≥ 4 + 3 2 . y⎠ ⎝ x⎠

Solution.

We have, ⎛

(1 + x ) ⎜ 1 + ⎝

1⎞ 1 ⎞ ⎛ 1 ⎞ ⎛x y⎞ 1⎛1 1⎞ ⎛ 1⎞ ⎛ ⎟ + (1 + y ) ⎜ 1 + ⎟ = ⎜ x + ⎟ + ⎜ y + ⎟ + ⎜ + ⎟ + ⎜ + ⎟ + 2 . y⎠ 2x ⎠ ⎝ 2y ⎠ ⎝ y x ⎠ 2 ⎝ x y ⎠ ⎝ x⎠ ⎝

Applying the AM – GM Inequality we get x+

1⎛1 1⎞ 1 1 2 1 1 x y = ≥ = 2, + ≥2. ≥ 2, y + ≥ 2, ⎜ + ⎟≥ 2 2 2x 2y y x 2⎝x y⎠ x +y xy 4 x 2 y 2

Adding these four inequalities, we obtain ⎛

(1 + x ) ⎜ 1 + ⎝

1⎞ ⎛ 1⎞ ⎟ + (1 + y ) ⎜ 1 + ⎟ ≥ 4 + 3 2 . y⎠ ⎝ x⎠

22. ( 0 < a, b, c ≤ 1) , 1 1 ≥ + (1 − a)(1 − b )(1 − c) . a+ b+c 3

Solution.

Applying the AM – GM Inequality we get 1=

(1 − b ) + (1 − c ) + (1 + b + c ) ≥ 3 3

(1 − b )(1 − c )(1 + b + c ) ⇔ (1 − b )(1 − c )(1 + b + c ) ≤ 1 .

Therefore,

(1 − a )(1 − b )(1 − c ) ≤

(

1− a 1− a 1 1 1 1 ≤ ≤ − ⇒ ≥ + (1 − a)(1 − b)(1 − c) . 1+ b + c a + b + c a + b + c 3 a+ b+c 3

)

23. x ≥ y ≥ z > 0,32 − 3x 2 = z 2 = 16 − 4y 2 ,

xy + yz + zx ≤

32 3 + 16 . 5

Solution.

We have y 2 =

16 − z 2 2 32 − z 2 16 − z 2 4 ≥ z2 ⇒ z ≤ . ,x = , y≥z⇒ 4 4 3 5

On the other hand, 32 − z 2 48 − 3z 2 5z 2 − 16 x − 3y = − = ≤ 0 ⇒ x ≤ y 3 ⇒ xy ≤ 3y 2 . 3 4 12 2

2

1 3 ⎛ x2 3 2 2 ⎛ x ⎞ 2⎞ We have xz ≤ 3 ⎜ .z ⎟ ≤ y + z ) and yz ≤ ( y 2 + z 2 ) . ( ⎜ +z ⎟≤ 2 ⎝ 3 ⎠ 2 ⎝ 3 ⎠ 2 Adding these two inequalities, we obtain

3 2 2 1 2 2 ⎛ 3 1 ⎞ ⎛ 16 − z2 ⎞ ⎛ 3 + 1 ⎞ 2 ( xy + yz + zx ) ≤ 3y + ( y + z ) + ( y + z ) = ⎜⎜ 3 + + ⎟⎟ ⎜ ⎟z ⎟+⎜ 2 2 20 2 ⎠ ⎝ 4 ⎠ ⎜⎝ 2 ⎟⎠ ⎝ ⎛ 3+ 3 ⎞ 2 ⎛ 3 + 3 ⎞ 16 32 3 + 16 = 2 3 3 + 1 + ⎜⎜ . ⎟⎟ z ≤ 2 3 3 + 1 + ⎜⎜ ⎟⎟ . = 5 ⎝ 8 ⎠ ⎝ 8 ⎠ 5 2

(

π ⎛ 24. ⎜ 0 < x < , n ∈ 2 ⎝

+

)

(

)

⎞ ⎟, ⎠ sin n + 2 x cosn + 2 x + ≥ 1. cosn x sin n x

Solution.

We have,

( sin

2n

x − cos2n x )( sin 2 x − cos2 x ) ≥ 0

⇔ ( tg n x − cot g n x )( sin 2 x − cos2 x ) ≥ 0 ⇔ tg n x sin 2 x + cot g n x cos2 x ≥ cot g n x sin 2 x + tg n x cos2 x . Thus, sin n + 2 x cosn + 2 x 1 n 1 + ≥ ( tg x + cot g n x )( sin 2 x + cos2 x ) ≥ .2. tg n x.cot g n x = 1 . n n cos x sin x 2 2 1 ⎛ ⎞ 25. ⎜ 0 < a, b, c, 0 ≤ x, y, z ≤ , a + b + c = x + y + z = 1⎟ , 2 ⎝ ⎠

ax + by + cz ≥ 8abc . Solution.

Without loss of generality, we can assume that a ≤ b ≤ c . We have,

4ab ≤ ( a + b ) = (1 − c ) ⇒ 8abc ≤ 2c (1 − c ) ≤ 2

2

2

1− c . 2

1 1− c ⎛ 1 ⎞ ⎛1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ a+b 0 ≤ x, y,z ≤ ⇒ cz = c ⎜ − x ⎟ + c ⎜ − y ⎟ ≥ a ⎜ − x ⎟ + b ⎜ − y ⎟ = − ( ax + by) = − ( ax + by) . 2 2 2 ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠

Thus,

ax + by + cz ≥

(

1− c ≥ 8abc . 2

)

26. x, y ≥ 0, x 3 + y3 = 1 ,

x +2 y ≤ Solution.

We set α =

1 , β = 2 5 2α . 5 1+ 2 2

6

(1 + 2 2 ) 5

5

.

Applying the AM – GM Inequality we get

x 3 + 5α

x 3 + 5α ≥ 6 6 x 3α 5 ⇒ x ≤

6 6 α5

x 3 + 5β ≥ 6 6 y3β5 ⇒ 2 y ≤

y3 + 5β 3 6 β5

,

.

Adding these two inequalities, we obtain

(

)

x +2 y ≤

x 3 + y3 + 5 ( α + β ) 6 6 α5

=

6

(1 + 2 2 ) 5

5

.

⎛ ⎞ 3x 4y 2z 27. ⎜ x, y, z > 0, + + = 2⎟, x +1 y +1 z +1 ⎝ ⎠ x3y4z2 ≤

1 . 89

Solution.

Applying the AM – GM Inequality we get

1 x 2x 4y 2z = 1− = + + = x +1 x +1 x +1 y +1 z +1

x x y y y y z z x2 y4z2 + + + + + + + ≥ 88 . 2 4 2 x +1 x +1 y +1 y +1 y +1 y +1 z +1 z +1 ( x + 1) ( y + 1) ( z + 1)

= Similarly,

1 y3 x 3z 2 1 x3y4z ≥ 88 and ≥ 88 . 3 3 2 3 4 2 1+ y 1+ z ( y + 1) ( x + 1) ( z + 1) ( x + 1) ( y + 1) ( z + 1)

Thus, 3

4

2

x 24 y32 z16 x3 y 4z2 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 9 9 8 8 ≥ = 8 ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ 24 32 16 3 4 2 ⎝ 1+ x ⎠ ⎝1+ y ⎠ ⎝ 1+ z ⎠ ( x + 1) ( y + 1) ( z + 1) ( x + 1) ( y + 1) ( z + 1) or x 3 y 4 z 2 ≤

1 . 89

28. ( a, b, c > 0 ) ,

(a

3

⎛ 1 1 1 ⎞ 3⎛ b+c c+a a+ b⎞ + b3 + c3 ) ⎜ 3 + 3 + 3 ⎟ ≥ ⎜ + + ⎟. b c ⎠ ⎝a b c ⎠ 2⎝ a

Solution.

We have,

2 ( a3 + b3 + c3 ) ≥ ab ( a + b ) + bc ( b + c ) + ca ( c + a ) and Therefore,

1 1 1 3 + 3+ 3≥ . 3 a b c abc

(a

3

29. ( x ∈

⎛ 1 1 1 ⎞ 3⎛ b+c c+a a+ b⎞ + b3 + c3 ) ⎜ 3 + 3 + 3 ⎟ ≥ ⎜ + + ⎟ b c ⎠ ⎝a b c ⎠ 2⎝ a

),

(16 cos

4

x + 3 ) + 768 ≥ 2048 cos x . 4

Solution.

Applying the AM – GM Inequality we get

(16cos

4

)

(

4

x + 3) = (16cos4 x + 1 + 1 + 1) + 768 ≥ 4 4 16cos4 x + 768 = 4

4

= 4096 cos4 x + 256 + 256 + 256 ≥ 4 4 4096 cos4 x.2563 = 2048 cos x ≥ 2048 cos x .

30. ( x ∈

), 4 1 (1 + x ) + 16x ≤ ≤ 17 . 2 4 8 1 x + ( ) 8

Solution.

If a, b are two real numbers, then

a +b 4

4

(a ≥

2

+ b2 ) 2

2

(a + b) ≥

4

8

.

Thus,

(1 + x ) + 16x 4 8

(1 + x )

2 4

4

4 ⎡(1 + x )2 ⎤ + ( −2x )4 1 + x2 ) ( 1 ⎣ ⎦ = ≥ = . 2 4 2 4 8 8 (1 + x ) (1 + x )

On the other hand, 4

(1 + x )

2 4

4 ⎡ (1 + x )2 ⎤ (1 + x )8 4 4 2 ≥⎢ and (1 + x2 ) = (1 + x2 − 2 x + 2 x ) = ⎡(1 − x ) + 2 x ⎤ ≥ 16x 4 . ⎥ ≥ ⎣ ⎦ 16 ⎢⎣ 2 ⎥⎦

Therefore,

(1 + x )

8

+ 16x 4

(1 + x ) 2

4

≤ 16 + 1 = 17

31. ( a, b, c > 0 ) ,

a 2 + b 2 b 2 + c2 c 2 + a 2 a2 + b 2 + c2 . + + ≤3 a+ b b+c c+a a+ b+c Solution.

a 2 + b 2 b 2 + c2 c2 + a 2 a2 + b 2 + c 2 + + ≤3 a+ b b+c c+a a+ b+c

⎡ a 2 + b 2 b 2 + c 2 c2 + a 2 ⎤ 2 2 2 ⇔ ( a + b + c) ⎢ + + ⎥ ≤ 3(a + b + c ) b+c c+a ⎦ ⎣ a+ b

⇔

c ( a2 + b 2 ) a+ b

⇔c − 2

⇔

+

b ( c2 + a2 ) c+a

c ( a2 + b 2 ) a+ b

+b − 2

+

a ( b 2 + c2 ) b+c

b ( c2 + a2 ) c+a

≤ a 2 + b 2 + c2

+a − 2

a ( b 2 + c2 ) b+c

≥0

ac ( c − a ) bc ( c − b ) ab ( b − a ) bc ( b − c ) ab ( a − b ) ac ( a − c ) + + + + + ≥0 a+ b a+ b c+a c+a b+c b+c

ac ( c − a ) bc ( c − b ) ab ( b − a ) ⇔ + + ≥ 0. ( a + b )( b + c ) ( a + b )( a + c ) ( a + c )( b + c ) 2

2

2

32. ( a, b, c > 0, n ∈ , n ≥ 2 ) , n

a b c n n +n +n > n −1 . b+c c+a a + b n −1

Solution.

Applying the AM – GM Inequality we get

n

a+ b ⎞ ⎛ n − 1) ( ⎜ ⎟ + 1 + 1 + ... + 1 n − 1 a + b + c c ⎠ ( a + b)( n − 1) ≤ ⎝ ( )( ) ⇒ n c ≥ n n n −1 c . n −1 = c n nc a + b n −1 a+ b+c

Similarly, n

b n n b ≥ and n −1 c + a n −1 a+ b+c

n

a n n a ≥ . n −1 b + c n −1 a+ b+c

Adding these three inequalities, we obtain n

a b c n n +n +n ≥ n −1 . b+c c+a a + b n −1

Equality holds if and only if ⎧( n −1)( a + b) = c 3 ⎪ ⎨( n −1)( c + a) = b ⇒ 2 ( n −1)( a + b + c) = a + b + c ⇒ n = (!) . 2 ⎪( n −1)( b + c) = a ⎩

Therefore, n

a b c n n n −1 . +n +n > b+c c+a a + b n −1

33. ( x, y, z ≥ 0, x + y + z = 1, n ∈ , n ≥ 2 ) ,

xn y + ynz + zn x ≤

nn

( n + 1)

n +1

.

Solution.

Without loss of generality, we can assume that x = max{x,y,z} . Since z ≤ x , thus znx ≤ zxn , znx ≤ z2xn−1 . On the other hand,

n >1⇒

n −1 1 n −1 z z≥ . ≥ ⇒ n 2 n 2

We have,

1 1 zx n z 2 x n −1 x n y + y n z + z n x ≤ x n y + x n −1yz + z n x + z n x ≤ x n y + x n −1yz + + = 2 2 2 2 z⎞ n −1 ⎞ ⎡x x x x + z ⎛ n − 1 ⎞⎤ ⎛ ⎛ z ⎟ ≤ n n ⎢ . ... . .⎜ y + z⎟ = x n −1 ( x + z ) ⎜ y + ⎟ ≤ x n −1 ( x + z ) ⎜ y + 2⎠ n ⎠ n ⎠ ⎥⎦ ⎝ ⎝ ⎣n n n n ⎝ x x+z n −1 ⎤ ⎡ n − 1) + z +y+ ( ⎢ n n n ⎥ ≤ nn ⎢ ⎥ n +1 ⎢ ⎥ ⎣ ⎦

n +1

(x + y + z) . n +1 ( n + 1)

n +1

=n

n

=

nn

( n + 1)

n +1

.

34. ( x, y, z > 0 ) ,

16xyz ( x + y + z ) ≤ 3 3 ( x + y) ( y + z ) ( z + x ) . 4

4

4

Solution.

Applying the AM – GM Inequality we get 1 3

1 3

1 3

( x + y )( y + z )( z + x ) = ( x + y + z ) xy + ( x + y + z ) xy + ( x + y + z ) xy + 1 1 1 + yz ( x + y + z ) + yz ( x + y + z ) + yz ( x + y + z ) + xz 2 + zx 2 ≥ 3 3 3

( xyz ) ( x + y + z ) 6

≥8

8

36

6

( xyz ) ( x + y + z ) 3

=8

4

3

27

.

Thus,

16xyz ( x + y + z ) ≤ 3 3 ( x + y ) ( y + z ) ( z + x ) . 4

4

4

⎛ ⎡π π⎤⎞ 35. ⎜ x, y, z ∈ ⎢ , ⎥ ⎟ , ⎣6 2⎦⎠ ⎝ 2

sinx − siny siny − sinz sinz − sinx ⎛ 1 ⎞ + + ≤ ⎜1− ⎟ . sinz sinx siny 2⎠ ⎝ Solution.

⎡1 ⎤ We set a = sin x, b = sin y, c = sin z and observe that a, b, c ∈ ⎢ ,1⎥ . ⎣2 ⎦

The inequality is equivalent to

( a − b )( b − c )( c − a ) ≤ ⎛ 1 − 1 ⎞ . a−b b−c c−a ⎛ 1 ⎞ + + ≤ ⎜1 − ⇔ ⎟ ⎜ ⎟ c a b abc 2⎠ 2⎠ ⎝ ⎝ 2

Without loss of generality, we can assume that

2

1 ≤ a ≤ b ≤ c ≤ 1. 2

a b 1 We set u = , v = and observe that ≤ u ≤ v ≤ 1 . c c 2

We will prove that

( v − u )(1 − u )(1 − v ) ≤ ⎛ 1 − uv

⎜ ⎝

2

1 ⎞ ⎟ . 2⎠

We have 1 ⎞⎛ 1 ⎞ ⎛ v − ⎟⎜1 − ⎟ (1 − v ) 2 ⎜ ( v − u)(1 − u )(1 − v ) ≤ ⎝ 2 ⎠⎝ 2 ⎠ 1 1 1 1 ⎛ 1 ⎞ = 1 + − v − ≤ 1 + − 2 v. = ⎜1 − ⎟ . 1 uv 2 2v 2 2v 2 ⎝ ⎠ v 2 36. ( x, y, z, t ∈ , ( x + y )( z + t ) + xy + 88 = 0 ) ,

x 2 + 9y 2 + 6z 2 + 24t 2 ≥ 352 . Solution.

Since ( x + y )( z + t ) + xy + 88 = 0 , thus 4 ( x + y )( z + t ) + 4xy + 352 = 0 . We have

x2 + 9y2 + 6z2 + 24t 2 + 4 ( x + y )( z + t ) + 4xy = x2 + 9y2 + 6z2 + 24t 2 + 4xz + 4xt + 4yz + 4yt + 4xy = = x 2 + 4x ( z + y + t ) + 4 ( z + y + t ) + 4y 2 − 4yz + z 2 + z 2 − 8zt + 16t 2 + y 2 − 4yt + 4t 2 = 2

2 2 2 2 = ⎡ x + 2 ( z + y + t ) ⎤ + ( 2y − z ) + ( z − 4t ) + ( y − 2t ) ≥ 0 . ⎣ ⎦

Therefore, x 2 + 9y 2 + 6z 2 + 24t 2 ≥ 352 . 37. ( x, y, z > 0, xyz = 1) , x2 y2 z2 + + ≥ 3. x + y + y3z y + z + z3 x z + x + x 3 y

Solution.

We set x2 y2 z2 A= + + . x + y + y3z y + z + z3 x z + x + x 3 y

We have

A=

x3 y3 z3 x3 y3 z3 + + = + + . x2 + yz + y3zx y2 + zy + z3xy z2 + xz + x3yz x2 + yx + y2 y2 + zy + z2 z2 + xz + x2

Applying the Cauchy – Buniakowski Inequality we get

A. ⎡⎣ x ( x 2 + xy + y 2 ) + y ( y 2 + yz + z 2 ) + z ( z 2 + xz + x 2 ) ⎤⎦ ≥ ( x 2 + y 2 + z 2 )

2

⇔ A ( x + y + z ) ( x2 + y2 + z2 ) ≥ ( x2 + y2 + z2 ) . 2

Thus,

A≥

x 2 + y 2 + z 2 x + y + z 3 3 xyz ≥ ≥ = 3. x+y+z 3 3

38. ( x, y, z > 0, xyz = 1) , x2y2 y2z2 z2 x2 + + ≤1. x2 y2 + x 7 + y7 y2z2 + y7 + z 7 z2 x2 + z 7 + x 7

Solution.

If x, y are positive real numbers, then x 7 + y 7 ≥ x 3 y3 ( x + y ) . Thus, x 2 y2 x 2 y2 1 1 z z . ≤ 2 2 3 3 = = = = 2 2 7 7 x y + x y ( x + y ) 1 + xy ( x + y ) xyz + xy ( x + y ) xyz ( x + y + z ) x + y + z x y +x +y

Similarly, y2z2 x z2 x2 y ≤ and 2 2 7 ≤ . 2 2 7 7 7 y z +y +z x+y+z z x +z +x x+y+z

Adding these three inequalities, we obtain x2 y2 y2z2 z2 x2 x+y+z + + ≤ = 1. 2 2 7 7 2 2 7 7 2 2 7 7 x y +x +y y z +y +z z x +z +x x+y+z

39. ( a, b, c > 0, a + b + c = 1) ,

a b abc 3 3 + + ≤ 1+ . a + bc b + ca c + ab 4 Solution.

We set

T= We have

a b abc + + . a + bc b + ca c + ab

ab a b abc 1 1 T= + + = + + c . a + bc b + ca c + ab 1 + bc 1 + ca 1 + ab a b c We set

bc A ca B = tan2 , = tan2 , 0 < A, B < π . Since a 2 b 2

ab ca bc ab ca bc . + . + . c b a c b a

1= a+ b+c = thus,

A B tan ab 2 2 = cot g ⎛ A + B ⎞ . = ⎜ ⎟ A B c ⎝ 2 ⎠ tan + tan 2 2 1 − tan

Therefore,

A+B π < and 2 2

ab C = tg , ( C = π − ( A + B) ∈ ( 0, π ) ) . c 2

We obtain

T=

1 1 + tan2

+

1

+

tan

C 2

A B C 1 + tan2 1 + tan2 2 2 2

= cos2

A B sin C 1 + cos2 + = 1 + ( cos A + cos B + sin C) . 2 2 2 2

On the other hand,

A+B A−B π cos A + cos B + sin C + sin = 2 cos .cos + 2 sin 3 2 2 A+B ≤ 2 cos + 2.cos 2

π π C− 3 .cos 3 2 2

C+

π π A + B+ C− 3 ≤ 4 cos 3 = 4 cos π = 2 3 . 2 4 6

C−

Thus,

1⎛ 3⎞ 3 3 T ≤ 1 + ⎜⎜ 2. 3 − . ⎟⎟ = 1 + 2⎝ 2 ⎠ 4 40. ( x, y, z > 0, x + y + z = 2007 ) ,

x 20 y 20 z 20 + 11 + 11 ≥ 3.669 9 . 11 y z x Solution.

Applying the AM – GM Inequality we get

x20 x20 x20 20 + 11y + 8.669 = 11 8 + y + y + ... + y + 669 + 669 + ... + 669 ≥ 20 11 .y11.6698 = 20x . 11 8 8 y 669 y 669 y .669 8 11

Similarly,

y 20 z 20 11z 8.669 20y + + ≥ and + 11x + 8.669 ≥ 20z . z11 .6698 x11 .6698 Adding these three inequalities, we obtain x 20 y 20 z 20 + 11 + 11 ≥ ⎡⎣ 9 ( x + y + z ) − 3.8.669 ⎤⎦ .6698 = 3.669 9 11 y z x

41. ( a, b, c > 0 ) ,

5b3 − a3 5c3 − b3 5a3 − c3 + + ≤ a+ b + c. ab + 3b 2 bc + 3c2 ca + 3a2 Solution.

Since a3 + b3 ≥ ab( a + b) , thus a3 − 5b3 ≥ ab( a + b) − 6b3 = b( a2 + ab − 6b2 ) = b( a + 3b)( a − 2b)

⇒ 5b3 − a3 ≤ ( ab + 3b2 ) ( 2b − a ) ⇒

5b3 − a3 ≤ 2b − a . ab + 3b 2

Similarly,

5c3 − b3 5a3 − c3 2c b ≤ − and ≤ 2a − c . bc + 3c2 ca + 3a2 Adding these three inequalities, we obtain

5b3 − a3 5c3 − b3 5a3 − c3 + + ≤ a+ b + c. ab + 3b 2 bc + 3c2 ca + 3a2

(

)

42. a, b > 0, x, y,z, t ∈ ,a ( x 2 + y 2 ) + b ( z 2 + t 2 ) = 1 ,

( x + z )( y + t ) ≤

a+ b . 2ab

Solution.

We have

a ( x2 + y2 ) + b ( z2 + t 2 ) = 1 ⇔

x2 z2 y2 t 2 1 + + + = . b a b a ab

Applying the Cauchy – Buniakowski Inequality we get

⎛ x2 z2 ⎞ ⎛ y2 t 2 ⎞ 2 2 ( b + a) ⎜ + ⎟ ≥ ( x + z) and ( b + a ) ⎜ + ⎟ ≥ ( y + t ) . ⎝ b a⎠ ⎝b a⎠ Adding these two inequalities, we obtain

⎛ x2 z2 y2 t2 ⎞ b + a + + + ⎟= ⎝ b a b a ⎠ ab

( x + z) + ( y + t ) ≤ ( b + a) ⎜ 2

2

⇒ ( x + z )( y + t )

(

(x + z) + (y + t) ≤ 2

2

2

≤

a+ b 2ab

)

43. x, y, z > −1, x 3 + y3 + z3 ≥ x 2 + y 2 + z 2 ,

x 5 + y5 + z5 ≥ x2 + y2 + z2 . Solution.

Since x > −1 , thus

( x − 1) ( x + 2 ) ≥ 0 ⇔ x3 − 3x + 2 ≥ 0 ⇔ x 5 ≥ 3x 3 − 2x 2 . 2

Similarly,

y 5 ≥ 3y3 − 2y 2 and z 5 ≥ 3z3 − 2z 2 . Adding these three inequalities, we obtain

x5 + y5 + z5 ≥ 3( x3 + y3 + z3 ) − 2( x2 + y2 + z2 ) = ( x3 + y3 + z3 ) + 2 ⎡⎣( x3 + y3 + z3 ) − ( x2 + y2 + z2 )⎤⎦ ≥ x3 + y3 + z3 . 44. ( α, β, γ ∈ , sin α + sin β + sin γ ≥ 2 ) ,

cos α + cos β+ cos γ ≤ 5 . Solution.

Applying the Cauchy – Buniakowski Inequality we get

cos α + cos β+ cos γ ≤ 3. cos2 α + cos2 β+ cos2 γ = 3. 3 − ( sin2 α + sin2 β+ sin2 γ ) ≤ 3.

( sin α + sin β + sin γ ) 3− 3

2

≤ 3. 3 −

4 = 5. 3

45. ( a, b, c > 0,a + b + c = 6 ) ,

a2 +

1 1 1 3 17 + b2 + + c2 + ≥ . b+c c+a a+ b 2

Solution. First solution.

Applying the Cauchy – Buniakowski Inequality we get 2

⎛ ⎞ 2 1 16a 1 a a a 1 1 ⎜a a a 1 ⎟ 1⎛ 1 ⎞ 2 = + = + + ... + + ≥ + + ... + + = ⎜ 4a + a + ⎟ . b + c 16 b + c 16 16 16 b + c 17 ⎜⎜ 4 4 4 b + c ⎟⎟ 17 ⎝ b+c ⎠ 16 ⎝ ⎠ 16 2

2

2

⇒ a2 + Similarly,

2

1 1 ⎛ 1 ⎞ ≥ 4a + ⎜ ⎟. b+c 17 ⎝ b+c ⎠

⇒ b2 +

1 1 ⎛ 1 ⎞ 4b + ≥ ⎜ ⎟ and c+a 17 ⎝ c+a ⎠

c2 +

1 ≥ a+b

1 17

⎛ ⎜ 4c + ⎝

1 a+b

⎞ ⎟. ⎠

Adding these three inequalities, we obtain

a2 +

1 1 1 1 ⎡ 1 1 ⎞⎤ ⎛ 1 + b2 + + c2 + ≥ + + ⎢4( a + b + c) + ⎜ ⎟⎥ ≥ b+c c+a a + b 17 ⎣ ⎝ a + b b + c c + a ⎠⎦

⎛ 1 ⎜ 24 + ≥ 17 ⎜⎜ ⎝

⎛ ⎞ ⎜ 3 ⎟ ≥ 1 ⎜ 24 + 3 a+ b ( )( b + c)( c + a ) ⎟⎟⎠ 17 ⎜⎜ ⎝

⎞ ⎟ 3 ⎟ = 3 17 . ( a + b ) + ( b + c) + ( c + a ) ⎟⎟ 2 3 ⎠

Second solution.

Applying the Minkowski Inequality and the AM – GM Inequality we get

1 1 1 + b2 + + c2 + ≥ a + b+c c+a a+ b 2

⎛ ≥ 36 + ⎜ ⎜⎜ ⎝

⎛ ⎞ ⎜ 3 ⎟ ≥ 36 + ⎜ ⎟ ⎜ 3 a+ b ( )( b + c)( c + a ) ⎠⎟ ⎜ ⎝ 2

(

2

1 1 ⎞ ⎛ 1 + + ( a + b + c) + ⎜ ⎟ ≥ b+c c+a ⎠ ⎝ a+ b 2

2

⎞ ⎟ 3 ⎟ = 3 17 . 2 ( a + b ) + ( b + c ) + ( c + a) ⎟⎟ 3 ⎠

)

46. x, y, z ∈ , x 2 + 2y 2 + 2x 2 z 2 + y 2 z 2 + 3x 2 y 2 z 2 = 9 ,

xyz ≥ −1 . Solution.

Applying the AM – GM Inequality we get

9 = x 2 + y 2 + y 2 + x 2 z 2 + x 2 z 2 + x 2 z 2 + y 2 z 2 + x 2 y 2 z 2 + x 2 y 2 z 2 + x 2 y 2 z 2 ≥ 9 9 x12 y12 z12 . Thus, 1 ≥ xyz ⇒ xyz ≥ −1

47. ( x, y, z, t > 0, xyzt = 1) ,

1 1 1 1 4 + 3 + 3 + 3 ≥ . x ( yz + zt + ty ) y ( xz + zt + tx ) z ( xt + ty + yx ) t ( xy + yz + zx ) 3 3

Solution.

1 1 1 1 We set a = , b = , c = , d = and observe that abcd = 1 . The inequality can rewrite x y z t a2 b2 c2 d2 4 + + + ≥ . b+c+d c+d +a d +a+ b a+ b+ c 3 Applying the Cauchy – Buniakowski Inequality we get

( a + b + c + d ) = a + b + c + d ≥ 4 4 abcd = 4 . a2 b2 c2 d2 + + + ≥ b + c + d c + d + a d + a + b a + b + c 3(a + b + c + d ) 3 3 3 2

48. ( k, n ∈

+

,a1 , a2 , ..., ak > 0,a1 + a2 + ... + ak ≥ k ) , a1n + a2n + ... + ank ≤1. a1n +1 + a2n +1 + ... + ank +1

Solution.

If a is a positive real numbers then

(a

n

− 1) ( a − 1) ≥ 0 ⇔ an +1 − an ≥ a − 1 .

Therefore,

(a

n +1 1

+ a2n +1 + ... + ank +1 ) − ( a1n + a2n + ... + ank ) ≥ ( a1 + a2 + ... + a k ) − k ≥ 0 .

49. ( a, b, c > 0, abc + a + c = b ) ,

2 2 3 10 . − 2 + 2 ≤ a +1 b +1 c +1 3 2

Solution.

We have abc + a + c = b ⇔ ac +

a c + = 1. b b

Since a, b, c > 0 , thus there exist A, B, C ∈ ( 0, π ) such that A + B + C = π and a = tan

A 1 B C , = tan , c = tan . 2 b 2 2

Therefore,

2 2 3 C C A−B 1 A−B 1 10 − 2 + 2 = −3sin 2 + 2 sin cos + 3 ≤ cos2 +3≤ +3= . a +1 b +1 c +1 2 2 2 3 3 3 3 2

50. ( x, y ∈ , x 2 + y 2 − 2x − 2y ≤ 0 ) ,

2x + y ≤ 10 + 3 . Solution.

Applying the Cauchy – Buniakowski Inequality we get 2x + y − 3 = 2 ( x − 1) + ( y − 1) ≤ 5.

⎛ 51. ⎜ 0 ≤ x, y, z ≤ 1, x + y + z = ⎝

( x − 1) + ( y − 1) 2

3⎞ ⎟, 2⎠

x 2 + y2 + z2 ≤ Solution.

2

5 . 4

≤ 5. 2 = 10 or 2x + y ≤ 10 + 3 .

⎛ 1 ⎞ It is easy to show that ( x, y, z ) ≺ ⎜1, , 0 ⎟ and f ( t ) = t 2 is convex on [ 0, +∞ ] . ⎝ 2 ⎠ Applying the Karamata Inequality we get

5 ⎛1⎞ f ( x ) + f ( y ) + f ( z ) ≤ f ( 0 ) + f ⎜ ⎟ + f (1) or x 2 + y 2 + z 2 ≤ . 4 ⎝2⎠ 52. ( a, b, c > 0 ) ,

⎛ a+ b b+c c+a c a b ⎞ + + ≥ 2 ⎜⎜ + + ⎟. c a b b+c a + c ⎠⎟ ⎝ a+b Solution.

If x, y are two positive real numbers, then

x + y ≤ 2 ( x + y),

1 1 4 + ≥ . x y x+y

Thus,

a+ b b+c c+a 1 ⎛ a b⎞ 1 ⎛ b c⎞ 1 ⎛ c c⎞ + + ≥ + + + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟ c a b 2⎝ c c⎠ 2⎝ a a⎠ 2⎝ b b ⎟⎠ = ≥

a⎛ 1 1 ⎞ b⎛ 1 1 ⎞ c⎛ 1 1 ⎞ 2 2a 2 2b 2 2c + + + + + ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟≥ 2⎝ b c⎠ 2⎝ a c⎠ 2⎝ a b⎠ b+ c a+ c a+ b 2 2a 2 ( b + c)

+

2 2b 2 ( a + c)

⎛ a b c ⎞ = 2 ⎜⎜ + + ⎟⎟ . b + c a + c a + b 2 (a + b) ⎝ ⎠ 2 2c

+

53. ( a, b, c > 0, ab + bc + ca = 1) ,

a + b + c + abc ≥

10 3 . 9

Solution.

We have

(a + b + c)

2

≥ 3 ( ab + bc + ca ) = 3 ⇒ a + b + c ≥ 3 .

Applying the AM – GM Inequality we get 3

⎡ a + b + c ⎞⎤ ⎛ 3⎞ 1 ≤ − (1 − a )(1 − b )(1 − c ) ≤ ⎢1 − ⎛⎜ ⎜ ⎟ ⎟⎥ ⎜ 3 3 ⎟⎠ ⎠⎦ ⎝ ⎣ ⎝ ⎛ 3⎞ ⇔ 1 + ab + bc + ca − a − b − c − abc ≤ ⎜⎜ 1 − ⎟ 3 ⎟⎠ ⎝ ⇔ a + b + c + abc ≥ 54. ( 0 ≤ a1 , a 2 , ..., a n ≤ 1, n ≥ 2 ) ,

10 3 . 9

3

3

a1 a2 an + + ... + ≤ n −1 . a2 a3 ...an + 1 a1a3 ...an + 1 a1a2 ...an −1 + 1 Solution.

Since 0 ≤ ai ≤ 1 thus a1a2 ...ai −1ai +1 ...an + 1 ≥ a1a2 ...an + 1 . Therefore, n a1 a2 an ai a + a 2 +... + an . + + ... + ≤∑ = 1 a2 a3 ...an + 1 a1a3 ...an + 1 a1a2 ...an −1 + 1 i =1 a1a2 ...an + 1 a1a2 ...an + 1

It is easy show that: If 0 ≤ a, b ≤ 1 then 0 ≤ (1− a)(1− b) or a + b ≤ 1+ ab . Thus,

a1 + a2

≤ 1 + a1a2

a3 + a1a2

≤ 1 + a1a2 a3

........................................... an + a1a2 ...an −1 ≤ 1 + a1a2 ...an Adding these n inequalities, we obtain a1 + a2 + ... + an ≤ ( n − 1) + a1a2 ...an ≤ ( n − 1)(1 + a1a2 ...an ) .

Therefore,

a1 a2 an + + ... + ≤ n −1 . a2 a3 ...an + 1 a1a3 ...an + 1 a1a2 ...an −1 + 1 55. ( a, b, c > 0, a + b + c = 3) , 3 ( a 2 + b 2 + c 2 ) + 4abc ≥ 13 .

Solution.

Since a, b, c > 0 and a + b + c = 3 thus,

abc ≥ ( a + b − c)( b + c − a)( c + a − b) ⇒ abc ≥ ( 3 − 2a )( 3 − 2b )( 3 − 2c ) ⇒ abc ≥ 27 − 18 ( a + b + c ) + 12 ( ab + bc + ca ) − 8abc ⇒ abc ≥ −27 + 12 ( ab + bc + ca ) − 8abc ⇒ 3abc ≥ 4 ( ab + bc + ca ) − 9 .

We have 3 ( a2 + b 2 + c2 ) + 4abc ≥ 3 ( a2 + b 2 + c2 ) − 12 + = 3 ( a + b + c ) − ( ab + bc + ca ) − 12 + 2

16 ( ab + bc + ca ) = 3

16 ( ab + bc + ca ) = 3

= 27 − 12 −

2 2 2 ( ab + bc + ca ) ≥ 15 − ( a + b + c ) = 15 − 2 = 13 . 3 9

56. ( a, b, c > 0,a2 b2 + b2 c2 + c2 a2 ≥ a2 b2 c2 ) ,

a2 b 2 b 2 c2 c2 a2 3 . + + ≥ 3 2 2 3 2 2 3 2 2 c (a + b ) a ( b + c ) b (c + a ) 2 Solution.

We set a =

1 1 1 , b = , c = and observe that x, y,z > 0, x 2 + y 2 + z 2 ≥ 1 . x y z

The inequality can rewrite

x3 y3 z3 3 + + ≥ . 2 2 2 2 2 2 y +z z +x x +y 2 Applying the Cauchy – Buniakowski Inequality we get

(x

2

+ y2 + z

)

2 2

⎛ x x ⎞ y y z z =⎜ . x y2 + z2 + . y x2 + z2 + . z y2 + x2 ⎟ ⎜ y2 + z2 ⎟ x2 + z2 y2 + x2 ⎝ ⎠

2

⎛ x3 y3 z3 ⎞ ⎡x y2 + z2 ) + y ( z2 + x2 ) + z ( x 2 + y2 )⎤ . ≤⎜ 2 2 + 2 + 2 2 2 ⎟⎣ ( ⎦ + + + y z z x x y ⎝ ⎠

On the other hand, 2

⎡x ( y2 + z2 ) + y( z2 + x2 ) + z ( x2 + y2 ) ⎤ = ⎡x y2 + z2 . y2 + z2 + y x2 + z2 . x2 + z2 + z x2 + y2 . x2 + y2 ⎤ ≤ ⎣ ⎦ ⎣ ⎦ 2

≤ ⎡⎣ x 2 ( y 2 + z 2 ) + y 2 ( z 2 + x 2 ) + z 2 ( x 2 + y 2 ) ⎤⎦ ( y 2 + z 2 + z 2 + x 2 + x 2 + y 2 ) = = 4 ( x 2 y 2 + y 2 z 2 + z 2 x 2 )( x 2 + y 2 + z 2 ) ≤

3 4 2 x + y2 + z2 ) . ( 3

Thus,

(x

2

2 ⎛ x3 y3 z3 + y2 + z2 ) ≤ ⎜ 2 + + 2 z2 + x2 x2 + y2 ⎝y +z

⇒

⎞ 4 2 2 2 3 ⎟. (x + y + z ) ⎠ 3

x3 y3 z3 3 3 + + ≥ . . x 2 + y2 + z 2 ≥ 2 2 2 2 2 2 y +z z +x x +y 2 2

57. ( a, b, c > 0, a + b + c = 6 ) , 1 ⎞⎛ 1 ⎞⎛ 1 ⎞ 729 ⎛ . ⎜1 + 3 ⎟⎜ 1 + 3 ⎟⎜1 + 3 ⎟ ≥ ⎝ a ⎠⎝ b ⎠⎝ c ⎠ 512

Solution.

Applying the AM – GM Inequality we get 1 ⎞⎛ 1 ⎞⎛ 1⎞ 1 1 ⎞ 1 ⎛ ⎛ 1 1 1⎞ ⎛ 1 ⎜1 + 3 ⎟ ⎜1 + 3 ⎟ ⎜1 + 3 ⎟ = 1 + ⎜ 3 + 3 + 3 ⎟ + ⎜ 3 3 + 3 3 + 3 3 ⎟ + 3 3 3 ≥ b ⎠⎝ c ⎠ b c ⎠ ⎝a b bc ca ⎠ abc ⎝ a ⎠⎝ ⎝a

⎛ ⎜ 3 3 3 1 1 ⎞ ⎜ 1 ⎛ ≥ 1+ + 2 2 2 + 3 3 3 = ⎜1 + ≥ ⎟ ⎜1 + 3 abc a b c abc abc ⎠ ⎝ ⎛a+ b+c⎞ ⎜ ⎜ ⎟ 3 ⎝ ⎠ ⎝

3

⎞ ⎟ ⎟ = 729 . ⎟ 512 ⎟ ⎠

58. ( a, b, c ≥ 0,a + b + c = 1) ,

a2 + 1 b 2 + 1 c2 + 1 7 + + ≤ . b 2 + 1 c2 + 1 a2 + 1 2 Solution.

b 2 + 1 c2 + 1 1 2 . + 2 ≤ ( b + c) + 1 + 2 2 c +1 a +1 a +1

We assume that a = max {a, b, c} . We will prove

b 2 + 1 c2 + 1 1 2 + 2 ≤ ( b + c) + 1 + 2 2 c +1 a +1 a +1

( b + c) c2 + ( b + c) + c2 − b2 c2 ⇔ 2 ≤ a +1 c2 + 1 2

2

b + c ) c2 + 2bc + 2c2 ( c2 . ⇔ 2 ≤ a +1 c2 + 1 2

Thus,

a2 + 1 b 2 + 1 c2 + 1 1 2 + 2 + 2 ≤ a2 + ( b + c ) + 2 +2. 2 b +1 c +1 a +1 a +1 We have to prove that

a2 + ( b + c ) +

1 3 ≤ . ( a + 1) 2

a2 + ( b + c ) +

1 3 ≤ ( a + 1) 2

2

2

We have 2

⇔ ⎛ 59. ⎜ a, b, c > 0, k ≥ ⎝

2

(1 − a ) (1 − 3a − 4a2 ) 2 (1 + a

2

)

1⎞ ⎛ ≤ 0 ⎜ since 1 ≥ a ≥ ⎟ . 3⎠ ⎝

2⎞ ⎟, 3⎠ k

k

k

3 ⎛ a ⎞ ⎛ b ⎞ ⎛ c ⎞ ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ≥ k. 2 ⎝ b+c⎠ ⎝ c+a⎠ ⎝ a+ b⎠ Solution.

Applying the AM – GM Inequality we get

2a + ( b + c ) + ( b + c ) 2 ≥ ( a + b + c) = 3 3

2

3

( 2a )( b + c )

3 a ⎛ a ⎞3 ⇒⎜ . ⎟ ≥3 . 4 a+ b+c ⎝ b+c⎠

2

Similarly, 2

2

3 b 3 c ⎛ b ⎞3 ⎛ c ⎞3 ≥ and . . ⎜ ⎟ ⎜ ⎟ ≥3 . 3 4 a+ b+c 4 a+ b+c ⎝ c+a⎠ ⎝ c+a⎠

Adding these three inequalities, we obtain k

k

k

3 ⎛ 2⎞ ⎛ a ⎞ ⎛ b ⎞ ⎛ c ⎞ ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ≥ k ⎜k = ⎟. 2 ⎝ 3⎠ ⎝ b+c⎠ ⎝ c+a⎠ ⎝ a+ b⎠ If k >

2 3 , then we set m = k, m > 1 . 3 2 m

Applying the well – known Inequality:

xm + ym + zm ⎛ x + y + z ⎞ ≥⎜ ⎟ , we set 3 3 ⎝ ⎠ 2

2

2

⎛ a ⎞3 ⎛ b ⎞3 ⎛ c ⎞3 x=⎜ ⎟ , y=⎜ ⎟ , z=⎜ ⎟ . ⎝ b+c⎠ ⎝ c+a⎠ ⎝ a+ b⎠

We obtain k

k

k

3 ⎛ a ⎞ ⎛ b ⎞ ⎛ c ⎞ ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ≥ k. 2 ⎝ b+c⎠ ⎝ c+a⎠ ⎝ a+ b⎠ 60. ( x, y > 0, x + y = 1) ,

1 1 + ≥ 4+2 3. 3 x + y xy 3

Solution.

Since x + y = 1 , thus 1 = ( x + y ) = x 3 + y 3 + 3xy ( x + y ) = x 3 + y3 + 3xy . 3

Applying the AM – GM Inequality we get

1 1 x3 + y3 + 3xy x3 + y3 + 3xy 3xy x3 + y3 3xy x3 + y3 + = + = 4 + + ≥ 4 + 2 = 4+2 3 . x3 + y3 xy x3 + y3 xy x3 + y3 xy x3 + y3 xy 61. ( a, b, c,d ∈ ,a2 + b2 = c + d = 4 ) , ac + bd + cd ≤ 4 + 4 2 .

Solution.

Applying the AM – GM Inequality we get 1⎛ c2 ⎞ 1⎛ d2 ⎞ 2 ac ≤ ⎜ 2a2 + , bd 2b ≤ + ⎟ ⎜ ⎟. 2⎝ 2⎝ 2⎠ 2⎠

Therefore,

( c + d ) + cd ⎛ 1 − 1 ⎞ ≤ 1 2 1 ac + bd + cd ≤ a + b2 ) + c2 + d 2 ) + cd = 2 2 + ( ( ⎜ ⎟ 2 2 2 2 2 2⎠ ⎝ 2

(

≤ 2 2+4 2

)

(c + d ) + 4

2

1 ⎞ ⎛ ⎜1 − ⎟= 4+4 2 . 2⎠ ⎝

62. ( x, y, z > 0, x = max {x, y, z} ) ,

x y z + 1+ + 3 1+ ≥ 1+ 2 + 3 2 . y x x Solution.

Applying the AM – GM Inequality we get

⎛x y z⎞ x y 3 y ⎜⎜ + 1 + + 3 1 + ⎟⎟ ≥ + 2 4 + 2 6 = x x⎠ y z z ⎝y =

1 ⎛x y z⎞ ⎛ 1 ⎞x ⎛3 6 ⎞6 z . +⎜ 2 − ⎜⎜ + 4 4 + 6 6 ⎟⎟ + ⎜ 1 − ⎟ ⎟ z x ⎠ ⎝ 2 2⎠y ⎝ 2 2⎝y 2 2⎠ x

On the other hand,

1 ⎛x y z ⎞ 11 x y z 11 11 = , ⎜⎜ + 4 4 + 6 6 ⎟⎟ ≥ z x⎠ 2 2 yzx 2 2 2 2⎝y x z 1 6 ≥ 1 ≥ and − >0> 32− . y x 2 2 2 2 Adding these three inequalities, we obtain

x y z 11 ⎛ 1 ⎞ ⎛3 6 ⎞ + 1+ + 3 1+ ≥ + ⎜1 − +⎜ 2 − = 1+ 2 + 3 2 . ⎟ ⎟ y x x 2 2 ⎝ 2 2⎠ ⎝ 2 2⎠ 63. ( a > 0, x, y,z ∈ , xy + yz + zx = 1) ,

a ( x2 + y2 ) + z2 ≥

−1 + 1 + 8a . 2

Solution.

Let α ∈ ( 0,a ) . Applying the AM – GM Inequality we get αx 2 +

z2 α z2 α ≥2 xz, αy 2 + ≥ 2 yz, 2 2 2 2

α 2 α x + y2 ) ≥ 2 xy . ( 2 2

Adding these three inequalities, we obtain

⎛ α⎞ 2 α 2 2 . ⎜⎜ α + ⎟⎟ ( x + y ) + z ≥ 2 2⎠ 2 ⎝ We will find α ∈ ( 0,a ) such that α +

α =a⇔ 2

α −1 + 1 + 8a = , we get 2 4

a ( x2 + y2 ) + z2 ≥

−1 + 1 + 8a . 2

64. ( a, b, c > 1) ,

alogb c + b logc a + cloga b ≥ 3 3 abc . Solution.

Since a, b, c > 1, thus log b c > 0, log c b > 0 . Applying the AM – GM Inequality we get 2 log b c.logc b

alogb c + blogc a = alogb c + blogc b.logb a = alogb c + alogc b ≥ 2 alogb calogc b = 2 alogb c+ logc b ≥ 2 a

= 2a .

Similarly,

blogc a + cloga b ≥ 2b and cloga b + alogb c ≥ 2c . Adding these three inequalities, we obtain

a log b c + b log c a + c log a b ≥ a + b + c ≥ 3 3 abc . 65. ( x, y ≥ 0, x 2 + y 2 = 1) ,

xy + max {x, y} ≤

3 3 . 4

Solution.

Without loss of generality, we can assume that x = max {x, y} . Applying the AM – GM Inequality we get

( xy + max{x,y})

2

4

1 1 ⎛ 3 − 3y +1+ y +1+ y +1+ y ⎞ 27 3 = ⎡⎣x ( y +1) ⎤⎦ = (1− y ) (1+ y) = ( 3 − 3y)(1+ y) ≤ ⎜ ⎟ = . 3 3⎝ 4 ⎠ 16 2

2

2

Therefore,

xy + max {x, y} ≤

3 3 . 4

66. ( a, b, c > 0, a + b + c = 1) ,

a6 b3 c6 1 + + ≥ . 3 3 3 3 3 3 b + c c + a a + b 18 Solution.

Applying the Cauchy – Buniakowski Inequality we get 3 a3 + b3 + c3 ) ( a6 b3 c6 a3 + b3 + c3 1 ⎛ a + b + c ⎞ 1 + + ≥ = ≥ .3. ⎜ ⎟ = . 3 3 3 3 3 3 3 3 3 b + c c + a a + b 2 (a + b + c ) 2 2 ⎝ 3 ⎠ 18 2

67. ( x, y,z > 0, x + y + z = 1) ,

xy yz zx 3 + + ≤ . z + xy x + yz y + zx 2 Solution.

Applying the AM – GM Inequality we get

xy xy = = z + xy z( x + y + z) + xy

xy 1⎛ x y ⎞ ≤ ⎜ + . ( x + z)( y + z) 2 ⎝ x + z y + z ⎟⎠

Similarly,

yz 1⎛ y z ⎞ ≤ ⎜ + ⎟ and x + yz 2 ⎝ x + y x + z ⎠

zx 1⎛ z x ⎞ ≤ ⎜ + ⎟. y + zx 2 ⎝ y + z x + y ⎠

Adding these three inequalities, we obtain

xy yz zx 1⎛x+y y+z z+x⎞ 3 + + ≤ ⎜ + + ⎟= . z + xy x + yz y + zx 2 ⎝ x + y y + z z + x ⎠ 2 68. ( a, b, c > 0, a + b + c = 1) ,

ab a2 + b2 + 2c2

+

bc b2 + c2 + 2a2

+

ca c2 + a2 + 2b2

≤

1 . 2

Solution.

Applying the Cauchy – Buniakowski Inequality we get

( a + b + c + c) ⇒

ab a + b + 2c 2

2

2

≤

2

≤ 4 ( a2 + b 2 + 2c2 )

2ab 2ab ⎛ 1 1 ⎞ 1 ⎛ ab ab ⎞ ≤ + + . ⎜ ⎟= ⎜ ( a + c ) + ( a + c ) 4 ⎝ c + a c + b ⎠ 2 ⎝ c + a c + b ⎟⎠

Similarly, bc

1 ⎛ bc bc ⎞ ≤ ⎜ + ⎟ and b 2 + c2 + 2a2 2 ⎝ a + b a + c ⎠

ca

1 ⎛ ca ca ⎞ ≤ ⎜ + ⎟. c2 + a2 + 2b 2 2 ⎝ b + a b + c ⎠

Adding these three inequalities, we obtain

ab a + b + 2c 2

≤ 69.

2

2

+

bc b + c + 2a 2

2

2

+

ca c + a2 + 2b 2 2

≤

1 ⎡⎛ ac + bc ⎞ ⎛ ab + ac ⎞ ⎛ ab + bc ⎞ ⎤ 1 1 ⎜ ⎟+⎜ ⎟+⎜ ⎟⎥ = ( a + b + c) = . ⎢ 2 ⎣⎝ a + b ⎠ ⎝ b + c ⎠ ⎝ c + a ⎠ ⎦ 2 2

( a, b > 0, a + b = 1) , a b + ≥ 2. b a

Solution.

Applying the Cauchy – Buniakowski Inequality we get

2 ab ≤ a + b = ( a + b ) = 2

(

a. 4 a3 + 4 b. 4 b3

4

) ≤( 2

)(

)

a+ b a a+b b ≤

(

)

(

)

≤ 2 (a + b) a a + b b = 2 a a + b b . Thus,

a b + ≥ 2. b a 70. ( x, y > 0, x 2 + y 2 = 1) ,

x+y+

1 1 + ≥3 2 . x y

Solution.

Applying the AM – GM Inequality we get x2 + y2 1 1 = ⇒ ≥2. xy ≤ 2 2 xy

Thus,

x+y+

1 1 1 1 1 1 1 1 + =x+ + +y+ + ≥ 66 4 ≥ 6 6 3 = 3. 2 x y 2x 2x 2y 2y 2 xy 2

71. ( x ≥ 0 ) ,

2 2 + x ≤ x + 9. x +1 Solution.

Applying the Cauchy – Buniakowski Inequality we get 2 2 x + 1 + x x + 1 = 1.2 2 x + 1 + x. x + 1 ≤

≤

(

12 +

( x ) ) ((2 2 ) + ( 2

2

x +1

) )= 2

x + 1. x + 9 .

Thus,

2 2 + x ≤ x + 9. x +1 72. ( a > b ≥ 0 ) ,

2a +

32

( a − b )( 2b + 3)

2

≥ 5.

Solution.

Applying the AM – GM Inequality we get 2a +

32

( a − b )( 2b + 3)

2

32 ⎛ 2b + 3 ⎞ ⎛ 2b + 3 ⎞ = 2(a − b) + ⎜ −3≥ ⎟+⎜ ⎟+ 2 ⎝ 2 ⎠ ⎝ 2 ⎠ ( a − b )( 2b + 3)

2. ( a − b ) . ( 2b + 3) .32 2

≥ 44

4. ( a − b ) . ( 2b + 3)

2

− 3 = 4 4 16 − 3 = 5 .

73. ( a, b > 0, a + b ≥ 4 ) ,

6 10 2a + 3b + + ≥ 18 . a b Solution.

Applying the AM – GM Inequality we get 6 10 4⎞ ⎛1 1⎞ 4 4 ⎛ = 18 2a + 3b + + = 2 ( a + b ) + ⎜ b + ⎟ + 6 ⎜ + ⎟ ≥ 2.4 + 2 b. + 6. a b b⎠ ⎝a b⎠ b a+b ⎝ 74. ( a, b, c ≥ 0, a + b + c = 3) , 5

2a + b + 5 2b + c + 5 2c + a ≤ 3 5 3 .

Solution.

Applying the AM – GM Inequality we get 5

2a + b =

1 5

34

5

( 2a + b ) .3.3.3.3 ≤

1 5

34

.

( 2a + b ) + 3 + 3 + 3 + 3 5

.

Similarly, 5

2b + c ≤

1 5

34

.

( 2b + c ) + 3 + 3 + 3 + 3 5

and

5

2c + a ≤

1 5

34

.

( 2c + a ) + 3 + 3 + 3 + 3 5

.

Adding these three inequalities, we obtain 5

2a + b + 5 2b + c + 5 2c + a ≤

3 ( a + b + c ) + 3.4.3 1 45 = . = 35 3 . 5 4 5 4 5 3 3 5 1

.

75. ( x, y, z > 0 ) ,

( x + y + z)

6

≥ 432xy2z3 .

Solution.

Applying the AM – GM Inequality we get 6 ⎛ 6x + 3y + 3y + 2z + 2z + 2z ⎞ ⎡ 6 ( x + y + z ) ⎤ 432.xy z = 6x.3y.3y.2z.2z.2z ≤ ⎜ ⎥ = ( x + y + z) . ⎟ =⎢ 6 6 ⎝ ⎠ ⎣ ⎦ 6

2 3

76. ( 0 ≤ a ≤ 1) , 13. a 2 − a 4 + 9. a 2 + a 4 ≤ 16 .

Solution.

Applying the Cauchy – Buniakowski Inequality we get

6

(

1− a

and

2

)

4a ≤

(

9a .4 1 + a 2

(

2

)

)

4 1 − a2 + a2

2

2

≤

(

=

9a 2 + 4 1 + a 2 4

4 − 3a 2 52 − 39a 2 ⇒ 13 a 2 − a 4 ≤ 2 4

) = 13a

2

2

+4

⇒ 9 a2 + a4 ≤

39a 2 + 12 . 4

Adding these two inequalities, we obtain 13. a 2 − a 4 + 9. a 2 + a 4 ≤

52 − 39a 2 39a 2 + 12 + = 16 . 4 4

77. ( a, b, c, d > 0 ) ,

3a ⎞ ⎛ 3b ⎞ ⎛ 3c ⎞ ⎛ 3d ⎞ 28561 ⎛ . ⎜ 2 + ⎟⎜ 2 + ⎟⎜ 2 + ⎟⎜ 2 + ⎟ ≥ 5b ⎠ ⎝ 5c ⎠ ⎝ 5d ⎠ ⎝ 5a ⎠ 625 ⎝ Solution.

Applying the AM – GM Inequality we get 2+

3a 1 1 1 a a a a3 = + + ... + + + + ≥ 13.13 13 3 . 5b 5 5 5 5b 5b 5b 5 .b 10

Similarly, 2+

3b b3 3c c3 3d d3 ≥ 13.13 13 3 , 2 + ≥ 13.13 13 3 , 2 + ≥ 13.13 13 3 . 5c 5a 5a 5 .c 5 .a 5 .a

Thus,

3a ⎞ ⎛ 3b ⎞ ⎛ 3c ⎞ ⎛ 3d ⎞ a 3 .b3 .c3 .d 3 134 28561 ⎛ 4 . 2 2 2 2 13 . + + + + ≥ = = 13 ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ 4 13 4 3 3 3 3 5b ⎠ ⎝ 5c ⎠ ⎝ 5d ⎠ ⎝ 5a ⎠ 5 625 ⎝ 5 .a .b .c .d

( )

78. ( a, b, c, d > 0, a + b + c + d ≤ 1) ,

⎛ 1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞ 4 ⎜1 + + ⎟ ⎜1 + + ⎟ ⎜1 + + ⎟ ⎜ 1 + + ⎟ ≥ 9 . ⎝ a b ⎠⎝ b c ⎠⎝ c d ⎠⎝ d a ⎠ Solution.

Applying the AM – GM Inequality we get 1 1 1 1 1 1 1 1 1 ⎛ 1 1⎞ . + + + + + + + ≥ 99 ⎜1 + + ⎟ = 1 + 4 4a 4a 4a 4a 4b 4b 4b 4b ⎝ a b⎠ ( 4a.4b )

Similarly, 1 1 1 ⎛ 1 1⎞ ⎛ 1 1⎞ ⎛ 1 1⎞ , ⎜1 + + ⎟ ≥ 9 9 , ⎜1 + + ⎟ ≥ 9 9 . ⎜1 + + ⎟ ≥ 9 9 4 4 4 ⎝ b c⎠ ( 4b.4c ) ⎝ c d ⎠ ( 4c.4d ) ⎝ d a ⎠ ( 4d.4a )

Thus,

1 1 ⎛ 1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞ 4 ≥ 94. ≥ 94 . ⎜1 + + ⎟⎜1 + + ⎟⎜1 + + ⎟⎜1 + + ⎟ ≥ 9 9 8 4 ⎝ a b ⎠⎝ b c ⎠⎝ c d ⎠⎝ d a ⎠ ( 4a.4b.4c.4d ) 9 ⎛ 4( a + b + c + d) ⎞ .⎜ ⎟ 4 ⎝ ⎠ 79. ( a, b, c, d > 0, a + b + c + d ≤ 1) ,

⎛ 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞ 4 ⎜1 + + ⎟⎜1 + + ⎟⎜1 + + ⎟⎜ 1 + + ⎟ ≥ 13 . a b b c c d d a ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ Solution.

Applying the AM – GM Inequality we get 1 1 1 1 1 1 1 1 ⎛ 2 1⎞ + + ... + + + + + ≥ 1313 . ⎜1 + + ⎟ = 1 + 8 4 4a 4a 4a 4b 4b 4b 4b ⎝ a b⎠ ( 4a ) ( 4b ) 8

Similarly, 1 1 1 ⎛ 2 1⎞ ⎛ 2 1⎞ ⎛ 2 1⎞ , ⎜1 + + ⎟ ≥ 1313 , ⎜1 + + ⎟ ≥ 1313 . ⎜1 + + ⎟ ≥ 1313 8 4 8 4 8 4 ⎝ b c⎠ ( 4b ) ( 4c ) ⎝ c d ⎠ ( 4c ) ( 4d ) ⎝ d a ⎠ ( 4d ) ( 4a )

⎛ 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞ ⎜1 + + ⎟⎜1 + + ⎟⎜1 + + ⎟⎜ 1 + + ⎟ ≥ ⎝ a b ⎠⎝ b c ⎠⎝ c d ⎠⎝ d a ⎠ ≥ 134 13

1

( 4a.4b.4c.4d )

12

≥ 134. 13

1 ⎛ 4 (a + b + c + d) ⎞ .⎜ ⎟ 4 ⎝ ⎠

48

≥ 134 .

80. ( a, b, c, d > 0, abcd ≥ 16 ) ,

1 ⎞⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞ 625 ⎛ . ⎜ a + ⎟⎜ b + ⎟⎜ c + ⎟⎜ d + ⎟ ≥ b ⎠⎝ c ⎠⎝ d ⎠⎝ a ⎠ 16 ⎝ Solution.

Applying the AM – GM Inequality we get a+

1 1 1 1 1 a . =a+ + + + ≥ 55 4 b 4b 4b 4b 4b ( 4b )

Similarly, b+

1 b 1 c 1 d , c + ≥ 55 , d + ≥ 55 . ≥ 55 4 4 4 c a a ( 4c ) ( 4a ) ( 4a )

Thus, 3 1 ⎞⎛ 1 ⎞⎛ 1 ⎞⎛ 1⎞ 54 625 ⎛ 4 5 ( abcd ) 4 5 ( abcd ) 4 5 16 =5 . ≥5 . = = ⎜ a + ⎟⎜ b + ⎟⎜ c + ⎟⎜ d + ⎟ ≥ S ≥ 5 b ⎠⎝ c ⎠⎝ d ⎠⎝ a⎠ 416 a.b.c.d 168 168 24 16 ⎝ 4

81. ( a, b, c, d > 0, abcd ≥ 16 ) ,

3

2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞ 2401 ⎛ . ⎜ a + + ⎟⎜ b + + ⎟⎜ c + + ⎟⎜ d + + ⎟ ≥ b c ⎠⎝ c d ⎠⎝ d a ⎠⎝ a b ⎠ 16 ⎝ Solution.

Applying the AM – GM Inequality we get 2 1⎞ a a a a 1 1 1 a4 ⎛ 7 a 7 . + + = + + + + + + ≥ ⎜ ⎟ b c⎠ 4 4 4 4 b b c b 2 c1 ⎝

Similarly, 2 1⎞ b4 ⎛ 2 1⎞ c4 ⎛ 2 1⎞ d4 ⎛ 7 7 7 b + + ≥ 7 c + + ≥ 7 d + + ≥ 7 , , . ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ c d⎠ d a⎠ a b⎠ c 2 d1 ⎝ d 2 a1 ⎝ a 2 b1 ⎝

Thus,

( abcd ) ≥ 74 = 2401 . 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1 ⎞⎛ 2 1⎞ ⎛ 4 7 + + + + + + + + ≥ a b c d 7 ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ 3 b c ⎠⎝ c d ⎠⎝ d a ⎠⎝ a b⎠ 24 16 ⎝ 416 ( abcd ) 4

82. ( a, b > 0, a + b ≤ 1) ,

1 1 1 + 2 + 2 ≥ 20 . 3 a + b a b ab 3

Solution.

Applying the AM – GM Inequality we get 1 1 1 1 1 ⎛1 1⎞ 1 1 4 + 2 + 2 = + ⎜ + ⎟≥ 2 + . ≥ 3 2 2 2 a +b a b ab ( a + b ) a − ab + b ab ⎝ a b ⎠ a − ab + b ab a + b

(

3

≥

≥

(

)

(

)

1 4 1 1 1 1 1 + = 2 + + + + ≥ 2 2 ab ab ab ab ab a − ab + b a − ab + b

)

2

(1 + 1 + 1 + 1)

(

2

a − ab + b + ab + ab + ab 2

2

)

+

1

⎛a+b⎞ ⎜ ⎟ ⎝ 2 ⎠

2

≥ 16 + 4 = 20 .

83. ( a, b, c > 0, a + b + c ≤ 1) ,

1 1 1 1 1 1 81 + 2 2+ 2 + + + ≥ . 2 2 a +b b +c c +a ab bc ca 2 2

Solution.

Applying the AM – GM Inequality and the Cauchy – Buniakowski Inequality we get 1 1 1 1 1 1 9 9 + 2 2+ 2 + + + ≥ + = 2 2 2 2 2 a +b b +c c +a ab bc ca 2 a + b + c ab + bc + ca 2

(

)

9⎛ 1 2 1 1 1 ⎞ 9⎛ ⎞ = ⎜ 2 + + + ⎟= ⎜ 2 ⎟≥ 2 2 2 2 2 ⎝ a + b + c ab + bc + ca ⎠ 2 ⎝ a + b + c ab + bc + ca ab + bc + ca ⎠

(1 + 1 + 1) 9 9 9 81 ≥ . 2 = . ≥ . 2 2 2 2 a + b + c + 2 ( ab + bc + ca ) 2 ( a + b + c ) 2 2

84. ( a, b, c > 0, a + b + c = 3) , 5

( 2a + b )( a + c ) a + 5 ( 2b + c )( b + a ) b + 5 ( 2c + a )( c + b ) c ≤ 3 5 6 .

Solution.

Applying the AM – GM Inequality we get

5

3 2

( 2a + b ) . ( a + c ) .3a.3.3 ≤

( 2a + b ) + ⎜⎛

3 3 ⎞ a + c ⎟ + 3a + 3 + 3 2 ⎠ ⎝2 . 5

Similarly,

5

5

3 2

( 2b + c ) . ( b + a ) .3b.3.3 ≤ 3 2

( 2c + a ) . ( c + b ) .3c.3.3 ≤

( 2b + c ) + ⎜⎛

3 3 ⎞ b + a ⎟ + 3b + 3 + 3 2 ⎠ ⎝2 , 5

( 2c + a ) + ⎜⎛

3 3 ⎞ c + b ⎟ + 3c + 3 + 3 2 ⎠ ⎝2 . 5

Adding these two inequalities, we obtain 5

( 2a + b )( a + c ) a + 5 ( 2b + c )( b + a ) b + 5 ( 2c + a )( c + b ) c ≤ 5

(

(

)

(

)

9 ( a + b + c ) + 18 = 35 6 . 3 3 5 .3 2 1

.

)

85. a, b, c > 0, a 2 + a + 2 ( b + 1) c 2 + 3c = 64 , 2

a 3 b 4 c5 ≤ 1 . Solution.

Applying the AM – GM Inequality we get

⎛ a2 + a + 2 ⎞ ⎛b b b b 1 1 1 1⎞ a b c = 4 . a .a.1.1 ⎜ . . . . . . . ⎟ c 2 .c.c.c ≤ 48. ⎜ ⎟ 4 ⎝4 4 4 4 4 4 4 4⎠ ⎝ ⎠ 3 4 5

8

(

2

)

(

)

= ⎛ 86. ⎜ a, b, c > 0, a + b + c ≤ ⎝

(

)

1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞ ⎛ ⎜ 3 + + ⎟⎜ 3 + + ⎟⎜ 3 + + ⎟ ≥ 343 . a b ⎠⎝ b c ⎠⎝ c a⎠ ⎝

Applying the AM – GM Inequality we get

⎛ b +1 ⎞ .⎜ ⎟ ⎝ 8 ⎠

(

1 2 . a 2 + a + 2 ( b + 2 ) c2 + 3c 8 8

3⎞ ⎟, 2⎠

Solution.

4

8

)

4

⎛ c 2 + 3c ⎞ .⎜ ⎟ = ⎝ 4 ⎠

2

=

1 .644 = 1 . 8 8

1 1⎞ 1 1 1 1 1 ⎛ + + + + 1 + 1 ≥ 7. 7 4 2 2 . ⎜3 + + ⎟ = 1+ a b⎠ 2a 2a 2b 2b 2 .a .b ⎝ Similarly, 1 1⎞ 1 1 1⎞ 1 ⎛ ⎛ ⎜ 3 + + ⎟ ≥ 7. 7 4 2 2 and ⎜ 3 + + ⎟ ≥ 7. 7 4 2 2 . b c⎠ c a⎠ 2 .b .c 2 .c .a ⎝ ⎝ Thus, 1 ⎛ 1 1 ⎞⎛ 1 1 ⎞⎛ 1 1 ⎞ 3 ≥ 73. ⎜ 3 + + ⎟⎜ 3 + + ⎟⎜ 3 + + ⎟ ≥ 7 . 7 12 4 a b b c c a 7 ⎝ ⎠⎝ ⎠⎝ ⎠ 2 . ( abc )

1 12

⎛a+b+c⎞ 212. ⎜ ⎟ 3 ⎠ ⎝

1

≥ 73.

12

7

⎛3⎞ ⎜ ⎟ 212. ⎜ 2 ⎟ ⎜2⎟ ⎝ ⎠

3⎞ ⎛ 87. ⎜ a, b, c, m, n, p > 0, a + b + c ≤ 1, m + n + p ≤ ⎟ , 2⎠ ⎝

⎛ 2 1 ⎞⎛ 2 1 ⎞ ⎛ 2 1 ⎞ 3 ⎜1 + + ⎟⎜1 + + ⎟ ⎜ 1 + + ⎟ ≥ 9 . ⎝ a m ⎠⎝ b n ⎠ ⎝ c p ⎠ Solution.

Applying the AM – GM Inequality we get 1 1 1 1 1 1 ⎛ 2 1⎞ + + ... + + + ≥ 99 . ⎜1 + + ⎟ = 1 + 6 2 3a 3a 3a 2m 2m ⎝ a m⎠ 3a 2m ( ) ( ) 6

Similarly, ⎛ 2 1⎞ 1 1 ⎛ 2 1⎞ and ⎜1 + + ⎟ ≥ 9 9 . ⎜1 + + ⎟ ≥ 9 9 6 2 6 2 c p⎠ ⎝ b n⎠ ⎝ ( 3b ) ( 2n ) ( 3c ) ( 2p )

Thus, 1 ⎛ 2 1 ⎞⎛ 2 1 ⎞ ⎛ 2 1 ⎞ 3 ⎜1 + + ⎟⎜ 1 + + ⎟ ⎜1 + + ⎟ ≥ 9 9 6 2 ⎝ a m ⎠⎝ b n ⎠ ⎝ c p ⎠ ( 3a.3b.3c ) . ( 2m.2n.2p )

1

≥ 93. 9

88. ( x, y, z ∈

⎛ 3(a + b + c + d ) .⎜ ⎜ 3 ⎝

3

⎞ ⎟ ⎟ ⎠

6

⎛ 2(m + n + p) .⎜ ⎜ 3 ⎝

),

(

)(

)(

)

27 x 2 + 3 y 2 + 3 z 2 + 3 ≥ 4 ( 3xy + 3yz + 3zx ) . 2

Solution.

We have

(

)(

)(

)

27 x 2 + 3 y 2 + 3 z 2 + 3 ≥ 4 ( 3xy + 3yz + 3zx )

2

3

⎞ ⎟ ⎟ ⎠

2

≥ 93 .

= 73 .

(

) (

)

⇔ 27 ⎣⎡ x 2 y 2 z 2 + 3 x 2 y 2 + y 2 z 2 + z 2 x 2 + 9 x 2 + y 2 + z 2 + 27 ⎦⎤ ≥ ≥ 4 ⎡9 ( xy + yz + zx ) + 6xyz ( xy + yz + zx ) + x 2 y 2 z 2 ⎤ . ⎣ ⎦ 2

We set a = x + y + z, b = xy + yz + zx, c = xyz . The inequality is equivalent to

(

)

(

)

243 a 2 − 2b + 81 b 2 − 2ca + 23c2 + 27 2 ≥ 36b 2 + 24bc

⇔ 243a 2 + 45b 2 + 23c 2 − 24bc − 162ca − 486b + 27 2 ≥ 0

(

)

(

)

⇔ 11( 3a − c ) + 12 ( b − c ) + ( b − 27 ) + 144 a 2 − 3b + 32 b 2 − 3ca ≥ 0 . 2

2

2

Since a 2 − 3b =

1⎡ 2 2 2 ( x − y ) + ( y − z ) + ( z − x ) ⎤⎦ ≥ 0 ⎣ 2

and b 2 − 3ca =

1⎡ 2 2 2 xy − yz ) + ( yz − zx ) + ( zx − xy ) ⎤ ≥ 0 ( ⎦ 2⎣

thus, the last inequality is true. 89. ( a, b, c > 0 ) ,

a 3 + b 3 + c3 a 2 + b 2 b 2 + c 2 c 2 + a 2 9 + 2 + + ≥ . 2abc c + ab a 2 + bc b 2 + ac 2 Solution.

We have a 3 + b 3 + c3 a 2 + b 2 b 2 + c 2 c 2 + a 2 + 2 + + = 2abc c + ab a 2 + bc b 2 + ac

=

a2 b2 c2 a 2 + b2 b2 + c2 c2 + a 2 + + + + + = 2bc 2ac 2ab c2 + ab a 2 + bc b 2 + ac

⎛ a 2 + bc b 2 + c 2 ⎞ ⎛ b 2 + ac c 2 + a 2 ⎞ ⎛ c 2 + ab a 2 + b 2 ⎞ 3 =⎜ + 2 + 2 + 2 ⎟+⎜ ⎟+⎜ ⎟− ≥ + + + 2bc a bc 2ac b ac 2ab c ab ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 ⎛ a 2 + bc 2bc ⎞ ⎛ b 2 + ac 2ca ⎞ ⎛ c 2 + ab 2ab ⎞ 3 ≥⎜ + 2 + 2 + 2 ⎟+⎜ ⎟+⎜ ⎟− ≥ a + bc ⎠ ⎝ 2ac b + ac ⎠ ⎝ 2ab c + ab ⎠ 2 ⎝ 2bc

≥ 2+2+2−

3 9 = 2 2

90. ( a, b > 0, ab = 1) ,

a3 b3 + ≥1. 1+ b 1+ a Solution.

Applying the AM – GM Inequality we get

(

)

2 2 4 4 a3 b3 a 3 + b3 + a 4 + b 4 ( a + b ) a − ab + b + a + b + = = ≥ 1+ b 1+ a 1 + a + b + ab a+b+2

≥

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

=

ab ( a + b + 2ab ) = 1. a+b+2

91. ( x, y, z ≥ 0, x + y + z = 2 ) ,

(

)

(

)

2 x 3 + y3 + z 3 ≤ 2 + x 4 + y 4 + z 4 . Solution.

We have

(

) (

)

2 x 3 + y3 + z 3 − x 4 + y 4 + z 4 = x 2 ( 2 − x ) + y3 ( 2 − y ) + z 3 ( 2 − z ) =

(

)

(

)

(

)

= x 3 ( y + z ) + y3 ( z + x ) + z 3 ( x + y ) = xy x 2 + y 2 + yz y 2 + z 2 + zx z 2 + x 2 ≤

(

)

≤ ( xy + yz + zx ) x 2 + y 2 + z 2 =

(

)

1 2 ( xy + yz + zx ) x 2 + y 2 + z 2 ≤ 2

(

2 2 2 1 ⎡ 2 ( xy + yz + zx ) + x + y + z ≤ ⎢ 2⎢ 2 ⎣

) ⎤⎥ ⎥⎦

2

=

1 4 x + y + z) = 2 . 3 ( 2

92. ( a, b, c, α > 0 ) , α

α

α

⎛ 2 1 ⎞ ⎛ 2 1 ⎞ ⎛ 2 1⎞ α ⎜ a + ⎟ + ⎜ b + ⎟ + ⎜ c + ⎟ ≥ 3.2 . ab ⎠ ⎝ bc ⎠ ⎝ ca ⎠ ⎝ Solution.

Applying the AM – GM Inequality we get α

α

α

α α α ⎛ 2 1 ⎞ ⎛ 2 1 ⎞ ⎛ 2 1 ⎞ ⎛ a⎞ ⎛ b⎞ ⎛ c⎞ ⎟ +⎜2 ⎟ +⎜2 ⎟ ≥ ⎜ a + ⎟ + ⎜ b + ⎟ + ⎜ c + ⎟ ≥ ⎜⎜ 2 ab ⎠ ⎝ bc ⎠ ⎝ ca ⎠ ⎝ b ⎟⎠ ⎜⎝ c ⎟⎠ ⎜⎝ a ⎟⎠ ⎝

α

α

α

α

⎛ a⎞ ⎛ b⎞ ⎛ c⎞ ⎛ 3 a b c⎞ . . ⎟⎟ = 3.2α . ≥ 3 3 ⎜⎜ 2 ⎟⎟ ⎜⎜ 2 ⎟⎟ ⎜⎜ 2 ⎟⎟ = 3 3 ⎜⎜ 2 b c a b c a⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 93. ( a, b, c > 0, abc = 1) ,

( a + b )( b + c )( c + a ) ≥ 2 (1 + a + b + c ) . Solution.

We have

( a + b )( b + c )( c + a ) ≥ 2 (1 + a + b + c ) ⇔ a 2 b + ab 2 + a 2 c + ac2 + bc2 + b 2 c + 2abc ≥ 2 + 2 ( a + b + c )

(

) (

) (

)

⇔ a 2 b + a 2c + 1 + b 2a + b2c + 1 + c2a + c2 b + 1 ≥ 3 + 2 ( a + b + c ) .

Applying the AM – GM Inequality we get

( a b + a c + 1) ≥ 3 2

3

a 4 bc = 3a ,

( b c + b a + 1) ≥ 3

3

b 4 ca = 3b ,

( c a + c b + 1) ≥ 3

3

c4 ab = 3c .

2

2

2

2

2

Adding these three inequalities, we obtain

( a b + a c +1) + ( b a + b c + 1) + ( c a + c b + 1) ≥ 3( a + b + c) ≥ 3 2

2

2

2

2

2

3

abc + 2 ( a + b + c ) = 3 + 2 ( a + b + c ) .

94. ( x, y, z > 0 ) ,

⎛1 1 1⎞ x z y + + ⎟+ + + ≥ x + y+ z + 6. ⎝x y z⎠ z y x

( xyz + 1) ⎜ Solution.

Applying the AM – GM Inequality we get ⎛ 1 1 1⎞ x z y ⎛ z⎞ ⎛ y⎞ ⎛ x⎞ 1 1 1 + + ⎟ + + + = ⎜ yz + ⎟ + ⎜ xy + ⎟ + ⎜ xz + ⎟ + + + ≥ y⎠ ⎝ x⎠ ⎝ z⎠ x y z ⎝x y z⎠ z y x ⎝

( xyz + 1) ⎜

≥ 2x + 2y + 2z +

1 1 1 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ + + = x + y+ z+⎜x + ⎟+⎜y+ ⎟+⎜z+ ⎟ ≥ x + y+ z+ 6. x y z x⎠ ⎝ y⎠ ⎝ z⎠ ⎝

95. ( 0 ≤ a, b, c, d ≤ 1) ,

a b c d + + + ≤3. bcd + 1 cda + 1 dab + 1 abc + 1 Solution.

Since 0 ≤ a, b, c, d ≤ 1 thus

(1 − a )(1 − b ) + (1 − c )(1 − d ) + (1 − ab )(1 − cd ) ≥ 0 ⇒ 3 − ( a + b + c + d ) + abcd ≥ 0 ⇒ a + b + c + d ≤ 3 + abcd .

Therefore, a b c d a + b + c + d 3 + abcd 2abcd + + + ≤ ≤ = 3− ≤ 3. bcd + 1 cda + 1 dab + 1 abc + 1 abcd + 1 1 + abcd 1 + abcd 96. ( a, b, c > 0, ab + bc + ca = 1) ,

a8

(a

2

+b

+

) (b

2 2

b8 2

+c

+

) (c

2 2

Solution.

Applying the Cauchy – Buniakowski Inequality we get

c8 2

+a

)

2 2

≥

1 . 12

2

a8

(a

2

+ b2

b8

+

) (b 2

(

2

+ c2

c8

+

) (c 2

)

2 2 2 ⎡ 2 1 ⎢ a +b +c ≥ . 3 ⎢ 2 a 2 + b2 + c2 ⎣

(

2

)

+ a2 2

) (

2

⎛ a4 b4 c4 ⎞ + + ⎜ 2 ⎟ a + b 2 b 2 + c2 c2 + a 2 ⎠ ≥⎝ ≥ 3

⎤ a 2 + b2 + c2 ⎥ = ⎥ 12 ⎦

) ≥ ( ab + bc + ca ) 2

12

2

=

1 . 12

97. ( a, b, c < 0, a + b + c ≤ 3) , a 3 b 3 c3 ⎛ 1 1 1 ⎞ + 2 + 2 + 27 ⎜ + + ⎟ ≥ 84 . 2 b c a ⎝ ab bc ca ⎠

Solution.

Applying the AM – GM Inequality we get 3 ≥ a + b + c ≥ 3 3 abc ⇒ abc ≤ 1 . Thus, 3 a 3 b 3 c3 b3 c3 1 1 1 ⎛ 1 1 1⎞ a ⎛ 1 1 1⎞ 27 + + + + + = + + 2 + + + + 26 ⎜ + + ⎟ ≥ ⎜ ⎟ 2 2 2 2 2 b c a c a ab bc ca ⎝ ab bc ca ⎠ b ⎝ ab bc ca ⎠

a 3 b 3 c3 1 1 1 1 1 1 1 1 ≥ 6. 2 . 2 . 2 . . . + 26.3. 3 . . = 6 6 + 26.3 3 ≥ 6 + 26.3 = 84 . 2 b c a ab bc ca ab bc ca abc ( abc ) 6

1 1 1⎞ ⎛ ⎛ 1 1 1⎞ 98. ⎜ a, b, c > 0, 6 ⎜ 2 + 2 + 2 ⎟ ≤ 1 + + + ⎟ , b c ⎠ a b c⎠ ⎝a ⎝ 1 1 1 1 + + ≤ . 10a + b + c a + 10b + c a + b + 10c 12 Solution.

We have 2

2

2

⎛1 1⎞ ⎛ 1 1⎞ ⎛1 1⎞ ⎛ 1 1 1⎞ ⎛1 1 1⎞ ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ ≥ 0 ⇒ 6⎜ 2 + 2 + 2 ⎟ ≥ 4⎜ + + ⎟ − 2 . b c ⎠ ⎝ a 3⎠ ⎝ b 3⎠ ⎝ c 3⎠ ⎝a ⎝a b c⎠ 1 1 1 ⎛ 1 1 1⎞ Since 6 ⎜ 2 + 2 + 2 ⎟ ≤ 1 + + + , thus b c ⎠ a b c ⎝a 1 1 1 1 1 1 ⎛1 1 1⎞ 1+ + + ≥ 4⎜ + + ⎟ − 2 ⇒ + + ≤ 1⇒ a + b + c ≥ 9 . a b c a b c ⎝a b c⎠ If x, y are two positive real numbers then

1 1⎛1 1⎞ ≤ ⎜ + ⎟. x+y 4⎝ x y⎠

Therefore, 1 1 1⎛ 1 1 ⎞ = ≤ ⎜ + ⎟= 10a + b + c 6a + ( 4a + b + c ) 4 ⎝ 6a 4a + b + c ⎠

⎞ 1⎡ 1 1⎛ 1 1⎛ 1 1 1 1 1 ⎞⎤ = ⎜⎜ + + . ⎟⎟ ≤ ⎢ + ⎜ + ⎟⎥ = 4 ⎝ 6a 3a + ( a + b + c ) ⎠ 4 ⎣ 6a 4 ⎝ 3a a + b + c ⎠ ⎦ 16a 16 ( a + b + c ) Similarly, 1 1 1 1 1 1 and . ≤ + ≤ + a + 10b + c 16b 16 ( a + b + c ) a + b + 10c 16c 16 ( a + b + c )

Adding these three inequalities, we obtain 1 1 1 1 ⎛1 1 1⎞ 3 1 3 1 + + ≤ ⎜ + + ⎟+ ≤ + = . 10a + b + c a + 10b + c a + b + 10c 16 ⎝ a b c ⎠ 16 ( a + b + c ) 16 16.9 12 99. ( a, b, c, d, e, f ∈ , ab + bc + cd + de + ef = 1) ,

a 2 + b 2 + c2 + d 2 + e2 + f 2 ≥

1 π 2 cos 7

.

Solution.

Let αi > 0,i = 1, 2,...,5 . Applying the AM – GM Inequality we get α1a 2 +

1 2 b ≥ 2ab , α1

α2b2 +

1 2 c ≥ 2bc , α2

α 3c 2 +

1 2 d ≥ 2cd , α3

α 4d 2 +

1 2 e ≥ 2de , α4

α5e2 +

1 2 f ≥ 2ef . α5

Adding these five inequalities, we obtain

⎛1 ⎞ ⎛1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 α1a 2 + ⎜ + α2 ⎟ b2 + ⎜ + α3 ⎟ c2 + ⎜ + α4 ⎟ d2 + ⎜ + α5 ⎟ e2 + f 2 ≥ 2 ( ab + bc + cd + de + ef ) = 2 . α5 ⎝ α1 ⎠ ⎝ α2 ⎠ ⎝ α4 ⎠ ⎝ α3 ⎠ We set αi =

sin ( i + 1) sin

iπ 7

π 7 ,i = 1, 2,...,5 .

It is easy to show that α1 = Therefore,

1 1 1 1 1 π + α2 = + α3 = + α4 = + α5 = = 2 cos . 7 α1 α2 α3 α4 α5

a 2 + b 2 + c2 + d 2 + e2 + f 2 ≥

1 2 cos

π 7

.

100. ( a, b, c > 0, a + b + c = 1) , a b c 27 . + 3 + 3 2 ≤ 2 2 a + a + 1 b + b + 1 c + c + 1 31 3

Solution.

Applying the AM – GM Inequality we get 1 1 ⎞ ⎛ 2 1 ⎞ 22 1 1 1 22 22 ⎛ a3 + a2 +1 = ⎜ a3 + + ≥ 3 3 a3. . + 2 a 2. + =a+ . ⎟+ ⎜a + ⎟+ 27 27 ⎠ ⎝ 9 ⎠ 27 27 27 9 27 27 ⎝ Therefore,

a ≤ a + a2 + 1 3

a 22 = 1− . 22 27a + 22 a+ 27

Similarly, c 22 b 22 . and 3 2 ≤ 1− ≤ 1− 2 c + c +1 27c + 22 b + b +1 27b + 22 3

Adding these three inequalities, we obtain

a b c 1 1 1 ⎛ ⎞ + 3 + 3 2 ≤ 3 − 22 ⎜ + + ⎟. 2 2 a + a +1 b + b +1 c + c +1 ⎝ 27a + 22 27b + 22 27c + 22 ⎠ 3

On the other hand, using the AM – GM Inequality we get 1 1 1 ⎛ ⎞ ⎡⎣( 27a + 22 ) + ( 27b + 22 ) + ( 27c + 22 ) ⎤⎦ ⎜ + + ⎟≥9 ⎝ 27a + 22 27b + 22 27c + 22 ⎠

1 1 ⎞ 9 3 ⎛ 1 ⇒⎜ + + = . ⎟≥ ⎝ 27a + 22 27b + 22 27c + 22 ⎠ 27 ( a + b + c) + 66 31 Therefore,

a b c 3 27 + 3 + 3 2 ≤ 3 − 22. = . 2 2 a + a +1 b + b +1 c + c +1 31 31 3

*****