Title [Colectiv] Inequalities Methods and Olympiad Probl(BookZZ.org) 2.6 MB 523
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Document Text Contents
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Contents

1 Part I 3

1.1 Squares are positive . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Some special inequalities and identities for two numbers . . . . 10

1.3 Some special inequalities and identities for more numbers . . . 13

1.4 The mathematical induction . . . . . . . . . . . . . . . . . . . . 17

1.5 The AM-GM inequality . . . . . . . . . . . . . . . . . . . . . . 25

1.6 The quadratic trinomial . . . . . . . . . . . . . . . . . . . . . . 29

1.7 Cauchy-Schwartz Inequality . . . . . . . . . . . . . . . . . . . . 41

1.8 Young Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.9 Advanced techniques with Cauchy-Buniakowski-Schwarz and
Holder Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.10 The principle of extremality and monotonicity . . . . . . . . . . 84

1.11 Breaking the inequality . . . . . . . . . . . . . . . . . . . . . . 86

1.12 Separating the squares . . . . . . . . . . . . . . . . . . . . . . . 92

1.13 The Dual Principle . . . . . . . . . . . . . . . . . . . . . . . . . 101

1.14 Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

1.15 Homogenization and dehomogenization . . . . . . . . . . . . . . 109

1.16 Unimonotonic sequences . . . . . . . . . . . . . . . . . . . . . . 111

1.17 Working backwards . . . . . . . . . . . . . . . . . . . . . . . . . 128

1.18 Mixing variables . . . . . . . . . . . . . . . . . . . . . . . . . . 129

1.19 Limits in inequalities . . . . . . . . . . . . . . . . . . . . . . . . 133

1.20 Derivatives in Inequalities . . . . . . . . . . . . . . . . . . . . . 135

1.21 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

1.22 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 141

1.23 The shrinking principle and Karamata’s Inequality . . . . . . . 144

1.24 Schur’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 148

1.25 The generalized Means . . . . . . . . . . . . . . . . . . . . . . . 149

1.26 Inequalities between the symmetric sums . . . . . . . . . . . . . 150

1.27 The pqr technique . . . . . . . . . . . . . . . . . . . . . . . . . 151

1.28 The tangent line technique and its extensions . . . . . . . . . . 159

1.29 Using identities to prove inequalities . . . . . . . . . . . . . . . 178

2 Problems 185

3 Solutions 225

1

Page 261

using the facts that θi−1 < θi and that sinx < x for x > 0, so that

n∑
i=1

sin θi − sin θi−1
cos θi−1

<

n∑
i=1

(θi − θi−1) = θn − θ0 <
π

2
,

as claimed.
?F?

96.6. (a) Find the minimum value of xx for x a positive real number.

(b) If x and y are positive real numbers, show that xy + yx > 1.

(France 1996)

First solution: (a) Since xx = ex log x and ex is an increasing function of x, it
su�ces to determine the minimum of x log x. This is easily done by setting its

derivative 1+log x to zero, yielding x =
1

e
. The second derivative

1

x
is positive

for x > 0, so the function is everywhere convex, and the unique extremum is
indeed a global minimum. Hence xx has minimum value e−

1
e .

(b) If x ≥ 1, then xy ≥ 1 for y > 0, so we may assume 0 < x, y < 1. Without
loss of generality, assume x ≤ y, now note that the function f(x) = xy + yx
has derivative f ′(x) = xy log x + yx−1. Since yx ≥ xx ≥ xy for x ≤ y and
1

x
≥ − log x, we see that f ′(x) > 0 for 0 ≤ x ≤ y and so the minimum of

f occurs with x = 0, in which case f(x) = 1, since x > 0, we have strict
inequality.

Second solution: We present another way to prove (b). Firstly, we will

prove that ab ≥
a

a+ b− ab
for any a, b ∈ (0, 1). Indeed, from the Bernoull’s

Inequality, it follows that a1−b = [1+(a−1)]1−b ≤ 1+(a−1)(1−b) = a+b−ab,
and thus the conclusion follows. Now, if x or y is at least 1, we are done.
Otherwise, let 0 < x, y < 1. In this case, we apply the above observation and
�nd that

xy + yx ≥
x

x+ y − xy
+

y

x+ y − xy
>

x

x+ y
+

y

x+ y
= 1.

?F?

96.7. Let a, b, c be positive real numbers such that abc = 1. Prove that

ab

a5 + b5 + ab
+

bc

b5 + c5 + bc
+

ca

c5 + a5 + ca
≤ 1.

(IMO Shortlist 1996)

Solution: For any positive numbers x, y, we have (x3− y2)(x2− y2) ≥ 0, and
hence x5 + y5 ≥ x2y2(x+ y). According to this inequality, we have

ab

a5 + b5 + ab
=

a2b2c

a5 + b5 + a2b2c

a2b2c

a2b2 (a+ b) + a2b2c
=

c

a+ b+ c
.

260

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Adding this to the two analogous inequalities, we get the desired result. Note
that the equality holds if and ony if a = b = c = 1.

?F?

96.8. Let a and b be positive real numbers with a+ b = 1. Prove that

a2

a+ 1
+

b2

b+ 1

1

3
.

(Hungary 1996)

First solution: Using the condition a + b = 1, we can reduce the given
inequality to homogeneous one

1

3

a2

(a+ b)[a+ (a+ b)]
+

b2

(a+ b)[b+ (a+ b)]
,

or
a2b+ ab2 ≤ a3 + b3,

which follows from

(a3 + b3)− (a2b+ ab2) = (a− b)2(a+ b) ≥ 0.

The equality holds if and only if a = b =
1

2
.

Second solution: By the Cauchy Schwarz Inequality, we have that

a2

a+ 1
+

b2

b+ 1

(a+ b)
2

a+ 1 + b+ 1
=

1

3
,

as desired.
?F?

96.9. Prove the following inequality for positive real numbers x, y, z,

(xy + yz + zx)

[
1

(y + z)2
+

1

(z + x)2
+

1

(x+ y)2

]

9

4
.

(Iran 1996)

First solution: Without loss of generality, we may assume that x ≥ y ≥ z.
Denote

P (x, y, z) =
1

(y + z)2
+

1

(z + x)2
+

1

(x+ y)2

9

4(xy + yz + zx)
.

We have to show that P (x, y, z) ≥ 0. Note that

P (t, t, z) =
z(z − t)2

2t2(2z + t)(z + t)2
≥ 0

for any positive reals t ≥ z > 0. Therefore, a key to solve this problem is to
�nd a suitable number t ≥ z such that

P (x, y, z) ≥ P (t, t, z).

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8 Fie a1, a2, ..., an numere reale astfel �̂ncât |ai| ≤ 2 pentru i = 1, 2, ..., n �si
a31 + a

3
2 + ...+ a

3
n = 0. Ar�atat�i c�a

a1 + a2 + ...+ an ≤
2n

3
.

9 Pentru n ≥ 3, �e x1, x2, · · · , xn ∈ [−1, 1] astfel �̂ncât

x51 + x
5
2 + ...+ x

5
n = 0.

Demonstrat�i c�a

x1 + x2 + ...+ xn ≤
8n

15
.

10 Fie x1, x2, · · · , xn numere reale astfel �̂ncât x1+x2+...+xn = na. Ar�atat�i
c�a

n∑
k=1

(xk − a)2 ≤
1

2
(
n∑
k=1

|xk − a|)2.

11 Dac�a x, y, z ≥ 0, atunci

x(y + z − x)2 + y(z + x− y)2 + z(x+ y − z)2 ≥ 3xyz.

12 Dac�a x, y, z sunt numere reale pozitive, atunci

1

(x+ y)2
+

1

(y + z)2
+

1

(z + x)2

9

4(xy + yz + zx)
.

13 (Ĉ�rtoaje, Ji Chen) Dac�a a, b, c �si x, y, z sunt numere reale nenegative,
atunci

2

(a+ b)(x+ y)
+

2

(b+ c)(y + z)
+

2

(c+ a)(z + x)

9

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

14 Fie x, y, z numere pozitive astfel �̂ncât xyz = 1. Demonstrat�i c�a

x

x3 + y + z
+

y

y3 + z + x
+

z

z3 + x+ y
≤ 1.

15 Dac�a a, b, c sunt numere reale pozitive astfel �̂ncât a+ b+ c = 3, atunci

6(
a

b
+
b

c
+
c

a
+ 3 ≥ 7(a2 + b2 + c2).

16 Dac�a a, b, c,d sunt numere reale nenegative astfel �̂ncât a+ b+ c+d = 4,
atunci

a2bc+ b2cd+ c2da+ d2ab ≤ 4.

17 Dac�a a, b, c sunt numere reale nenegative, cel mult unul egal cu 0, atunci

1

b2 − bc+ c2
+

1

c2 − ca+ a2
+

1

a2 − ab+ b2

12

(a+ b+ c)2
.

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18 Dac�a a, b, c sunt numere reale nenegative, cel mult unul egal cu 0, atunci

ab− bc+ ca
b2 + c2

+
bc− ca+ ab
c2 + a2

+
ca− ab+ bc
a2 + b2

3

2
.

19 (Ĉ�rtoaje) Dac�a a ≥ b ≥ c > 0, atunci

3(a− c)2

4(a+ b+ c)
≤ a+ b+ c− 3 3

abc ≤

4(a− c)2

a+ b+ c
.

20 Fie a, b, c numere reale pozitive astfel �̂ncât abc = 1. Ar�atat�i c�a

3 + a

(1 + a)2
+

3 + b

(1 + b)2
+

3 + c

(1 + c)2
≥ 3.

21 Dac�a a, b, c, d sunt numere reale pozitive astfel �̂ncât a + b + c + d = 4,
atunci

1

ab
+

1

bc
+

1

cd
+

1

da
≥ a2 + b2 + c2 + d2.

22 Dac�a 0 < a ≤ b ≤ c ≤ 0, atunci

a

b
+
b

c
+
c

a

2a

b+ c
+

2b

c+ a
+

2c

a+ b
.

23 Dac�a a, b, c, d 6= 1
3

sunt numere reale pozitive care satisfac relat�ia

abcd = 1,

atunci

1

(3a− 1)2
+

1

(3b− 1)2
+

1

(3c− 1)2
+

1

(3d− 1)2
≥ 1.

24 Dac�a a, b, c sunt numere reale pozitive care satisfac relat�ia ab+bc+ca = 3,
atunci

bc+ 2

a2 + 2
+
ca+ 2

b2 + 2
+
ab+ 2

c2 + 2
≥ 3.

25 Fie a, b, c numere reale nenegative, dintre care cel mult unul egal cu 0.
Demonstrat�i c�a

a(b+ c)

b2 + bc+ c2
+

b(c+ a)

c2 + ca+ a2
+

c(a+ b)

a2 + ab+ b2
≥ 2.

26 Fie a, b, c, d, e numere reale astfel �̂ncât a+b+c+d+e = 0. Demonstrat�i
c�a

a2 + b2 + c2 + d2 + e2 ≥ 3(ab+ bc+ cd+ de+ ea).

27 Dac�a a1, a2, ...an ∈ [1, 2] , atunci
n∑
i=1

3

ai + 2ai+1

n∑
i=1

2

ai + ai+1
,

unde an+1 = a1.

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