Document Text Contents
Page 2
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
Page 262
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).
261
Page 522
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
.
521
Page 523
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.
522
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
Page 262
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).
261
Page 522
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
.
521
Page 523
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.
522
