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+ (y + z)2 + (z + x)2)(x+ y + z) + 4(x+ y)(y + z)(z + x)
3
=
4(x3 + y3 + z3 + 3x2y + 3xy2 + 3y2z + 3yz2 + 3z2x+ 3zx2 + 5xyz)
3
=
4((x+ y + z)3 − xyz)
3
=
4(26
27
(x+ y + z)3 + (x+y+z
3
)3 − xyz)
3
≥ 4(
26
27(x+ y + z)
3)
3
=
13
3
.
Solution 6 (2, DDucLam). Using the familiar Inequality (equivalent to Schur)
abc ≥ (b+ c− a)(c + a − b)(a+ b− c) ⇒ abc ≥ 4
3
(ab+ bc+ ca)− 3.
Therefore
P ≥ a2 + b2 + c2 + 16
9
(ab+ bc+ ca) − 4
= (a+ b+ c)2 − 2
9
(ab+ bc+ ca)− 4 ≥ 5− 2
27
(a+ b+ c)2 = 4 +
1
3
.
Equality holds when a = b = c = 1.
Solution 7 (3, pi3.14). With the conventional denotion in triangle, we have
abc = 4pRr , a2 + b2 + c2 = 2p2 − 8Rr − 2r2.
Therefore
a2 + b2 + c2 +
4
3
abc =
9
2
− 2r2.
Moreover,
p ≥ 3
√
3r ⇒ r2 ≤ 1
6
.
Thus
a2 + b2 + c2 +
4
3
abc ≥ 41
3
.
14 www.batdangthuc.net
∇
Problem 5 (7, China Northern Mathematical Olympiad 2007). Let a, b, c be positive
real numbers such that abc = 1. Prove that
ak
a+ b
+
bk
b+ c
+
ck
c+ a
≥ 3
2
.
for any positive integer k ≥ 2.
Solution 8 (Secrets In Inequalities, hungkhtn). We have
ak
a + b
+
bk
b+ c
+
ck
c+ a
≥ 3
2
⇔ ak−1 + bk−1 + ck−1 ≥ 3
2
+
ak−1b
a+ b
+
bk−1c
b + c
+
ck−1a
c+ a
By AM-GM Inequality, we have
a+ b ≥ 2
√
ab, b+ c ≥ 2
√
bc, c+ a ≥ 2√ca.
So, it remains to prove that
ak−
3
2 b
1
2 + bk−
3
2 c
1
2 + ck−
3
2 a
1
2 + 3 ≤ 2 (ak−1 + bk−1 + ck−1) .
This follows directly by AM-GM inequality, since
ak−1 + bk−1 + ck−1 ≥ 3 3
√
ak−1bk−1ck−1 = 3
and
(2k − 3)ak−1 + bk−1 ≥ (2k − 2)ak−32 b 12
(2k − 3)bk−1 + ck−1 ≥ (2k − 2)bk−32 c 12
(2k − 3)ck−1 + ak−1 ≥ (2k − 2)ck−32 a 12
Adding up these inequalities, we have the desired result.
∇
Problem 6 (8, Revised by NguyenDungTN). Let a, b, c > 0 such that a + b + c = 1.
Prove that:
a2
b
+
b2
c
+
c2
a
≥ 3(a2 + b2 + c2).
www.batdangthuc.net 15
Solution 9. By Cauchy-Schwarz Inequality:
a2
b
+
b2
c
+
c2
a
≥ (a
2 + b2 + c2)2
a2b + b2c + c2a
.
It remains to prove that
(a2 + b2 + c2)2
a2b+ b2c + c2a
≥ 3(a2 + b2 + c2)
⇔ (a2 + b2 + c2)(a+ b+ c) ≥ 3(a2b+ b2c+ c2a)
⇔ a3 + b3 + c3 + ab2 + bc2 + ca2 ≥ 2(a2b+ b2c+ c2a)
⇔ a(a − b)2 + b(b− c)2 + c(c − a)2 ≥ 0.
So we are done!
Solution 10 (2, By Zaizai).
a2
b
+
b2
c
+
c2
a
≥ 3(a2 + b2 + c2)
⇔
∑(a2
b
− 2a+ b
)
≥ 3(a2 + b2 + c2)− (a+ b+ c)2
⇔
∑ (a− b)2
b
≥ (a − b)2 + (b − c)2 + (c− a)2
⇔
∑
(a− b)2
(
1
b
− 1
)
≥ 0
⇔
∑ (a− b)2(a + c)
b
≥ 0.
This ends the solution, too.
∇
Problem 7 (9, Romania Junior Balkan Team Selection Tests 2007). . Let a, b, c be three
positive reals such that
1
a + b+ 1
+
1
b+ c+ 1
+
1
c+ a + 1
≥ 1.
Show that
a + b+ c ≥ ab+ bc+ ca.
Solution 11 (Mathlinks, Reposted by NguyenDungTN). By Cauchy-Schwarz Inequality,
we have
(a + b+ 1)(a+ b+ c2) ≥ (a+ b+ c)2.
16 www.batdangthuc.net
Therefore
1
a + b+ 1
≤ c
2 + a+ b
(a+ b+ c)2
,
or
1
a+ b+ 1
+
1
b+ c+ 1
+
1
c+ a+ 1
≤ a
2 + b2 + c2 + 2(a+ b+ c)
(a + b+ c)2
⇒ a2 + b2 + c2 + 2(a+ b+ c) ≥ (a+ b+ c)2
⇒ a+ b+ c ≥ ab+ bc+ ca.
Solution 12 (DDucLam). Assume that a+ b+ c = ab+ bc+ ca, we have to prove that
1
a+ b+ 1
+
1
b+ c + 1
+
1
c+ a+ 1
≤ 1
⇔ a + b
a+ b+ 1
+
b+ c
b+ c+ 1
+
c+ a
c+ a+ 1
≥ 2
By Cauchy-Schwarz Inequality,
LHS ≥ (a+ b+ b+ c+ c + a)
2∑
cyc(a+ b)(a+ b+ 1)
= 2.
We are done
Comment. This second very beautiful solution uses Contradiction method. If you can't
understand the principal of this method, have a look at Sang Tao Bat Dang Thuc, or Secrets
In Inequalities, written by Pham Kim Hung.
∇
Problem 8 (10, Romanian JBTST V 2007). Let x, y, z be non-negative real numbers.
Prove that
x3 + y3 + z3
3
≥ xyz + 3
4
|(x− y)(y − z)(z − x)|.
Solution 13 (vandhkh). We have
x3 + y3 + z3
3
≥ xyz + 3
4
|(x− y)(y − z)(z − x)|
⇔ x
3 + y3 + z3
3
− xyz ≥ 3
4
|(x− y)(y − z)(z − x)|
⇔ ((x−y)2+(y−z)2 +(z−x)2(((x+y)+(y+ z)+(z +x)) ≥ 9|(x−y)(y−z)(z −x)|.
Notice that
x+ y ≥ |x− y|; y + z ≥ |y − z|; z + x ≥ |z − x|,
and by AM-GM Inequality,
((x− y)2 + (y − z)2 + (z − x)2)(|x− y|+ |y − z|+ |z − x|) ≥ 9|(x− y)(y − z)(z − x)|.
So we are done. Equality holds for x = y = z.
www.batdangthuc.net 17
Solution 14 (Secrets In Inequalities, hungkhtn). The inequality is equivalent to
(x+ y + z)
∑
(x − y)2 ≥ 9
2
|(x− y)(y − z)(z − x)|.
By the entirely mixing variable method, it is enough to prove when z = 0
x3 + y3 ≥ 9
4
|xy(x − y)|.
This last inequality can be checked easily.
∇
Problem 9 (11, Yugoslavia National Olympiad 2007). Let k be a given natural number.
Prove that for any positive numbers x, y, z with the sum 1, the following inequality holds
xk+2
xk+1 + yk + zk
+
yk+2
yk+1 + zk + xk
+
zk+2
zk+1 + xk + yk
≥ 1
7
.
When does equality occur?
Solution 15 (NguyenDungTN). We can assume that x ≥ y ≥ z. By this assumption, easy
to refer that
xk+1
xk+1 + yk + zk
≥ y
k+1
yk+1 + zk + xk
≥ z
k+1
zk+1 + xk + yk
;
zk+1 + yk + xk ≥ yk+1 + xk + zk ≥ xk+1 + zk + yk ;
and
xk ≥ yk ≥ zk.
By Chebyshev Inequality, we have
xk+2
xk+1 + yk + zk
+
yk+2
yk+1 + zk + xk
+
zk+2
zk+1 + xk + yk
≥ x+ y + z
3
(
xk+1
xk+1 + yk + zk
+
yk+1
yk+1 + zk + xk
+
zk+1
zk+1 + xk + yk
)
=
1
3
(
xk+1
xk+1 + yk + zk
+
yk+1
yk+1 + zk + xk
+
zk+1
zk+1 + xk + yk
)∑
cyc(x
k+1 + yk + zk)∑
cyc(xk+1 + yk + zk)
=
1
3
(∑
cyc
(
xk+1
xk+1 + yk + zk
∑
cyc
(xk+1 + yk + zk)
1∑
cyc(xk+1 + yk + zk)
))
≥ 1
3
(xk+1+yk+1+zk+1).
1∑
cyc(xk+1 + yk + zk)
=
xk+1 + yk+1 + zk+1
xk+1 + yk+1 + zk+1 + 2(xk + yk + zk)
18 www.batdangthuc.net
Also by Chebyshev Inequality,
3(xk+1 + yk+1 + zk+1) ≥ 3x+ y + z
3
(xk + yk + zk) = xk + yk + zk.
Thus
xk+1 + yk+1 + zk+1
xk+1 + yk+1 + zk+1 + 2(xk + yk + zk)
≥ x
k+1 + yk+1 + zk+1
xk+1 + yk+1 + zk+1 + 6(xk+1 + yk+1 + zk+1)
=
1
7
.
So we are done. Equality holds for a = b = c =
1
3
.
∇
Problem 10 (Macedonia Team Selection Test 2007). Let a, b, c be positive real numbers.
Prove that
1 +
3
ab+ bc+ ca
≥ 6
a + b+ c
.
Solution 16 (VoDanh). The inequality is equivalent to
a+ b+ c+
3(a+ b+ c)
ab+ bc+ ca
≥ 6.
By AM-GM Inequality,
a+ b+ c+
3(a+ b+ c)
ab+ bc+ ca
≥ 2
√
3(a+ b+ c)2
ab+ bc+ ca
.
It is obvious that (a+ b+ c)2 ≥ 3(ab+ bc+ ca), so we are done!
∇
Problem 11 (14, Italian National Olympiad 2007). a). For each n ≥ 2, find the maximum
constant cn such that:
1
a1 + 1
+
1
a2 + 1
+ . . .+
1
an + 1
≥ cn,
for all positive reals a1, a2, . . . , an such that a1a2 · · ·an = 1.
∇
b). For each n ≥ 2, find the maximum constant dn such that
1
2a1 + 1
+
1
2a2 + 1
+ . . .+
1
2an + 1
≥ dn,
for all positive reals a1, a2, . . . , an such that a1a2 · · ·an = 1.
www.batdangthuc.net 19
Solution 17 (Mathlinks, reposted by NguyenDungTN). a). Let
a1 = n−1, ak =
1

∀k 6= 1,
then let  → 0, we easily get cn ≤ 1. We will prove the inequality with this value of cn.
Without loss of generality, assume that a1 ≤ a2 ≤ · · · ≤ an. Since a1a2 ≤ 1, we have
n∑
k=1
1
ak + 1
≥ 1
a1 + 1
+
1
a2 + 1
=
1
a1 + 1
+
a1
a2 + a1a2
≥ 1
a1 + 1
+
a1
a1 + 1
= 1.
This ends the proof.
b). Consider n = 2, it is easy to get d2 = 23 . Indeed, let a1 = a, a2 =
1
a . The inequality
becomes
1
2a+ 1
+
a
a+ 2
≥ 2
3
⇔ 3(a+ 2) + 3a(2a+ 1) ≥ 2(2a+ 1)(a+ 2)
⇔ (a − 1)2 ≥ 0.
When n ≥ 3, similar to (a), we will show that dn = 1. Indeed, without loss of generality,
we may assume that
a1 ≤ a2 ≤ · · · ≤ an ⇒ a1a2a3 ≤ 1.
Let
x = 9
√
a2a3
a21
, y = 9
√
a1a3
a22
, z = 9
√
a1a2
a23
then a1 ≤ 1x3 , a2 ≤ 1y3 , a3 ≤ 1z3 , xyz = 1. Thus
n∑
k=1
1
ak + 1
≥
3∑
k=1
1
ak + 1
≥ x
3
x3 + 2
+
y3
y3 + 2
+
z3
z3 + 2
=
x2
x2 + 2yz
+
y2
y2 + 2xz
+
z2
z2 + 2xy
≥ x
2
x2 + y2 + z2
+
y2
x2 + y2 + z2
+
z2
x2 + y2 + z2
= 1.
This ends the proof.
∇
Problem 12 (15, France Team Selection Test 2007). . Let a, b, c, d be positive reals such
that a+ b+ c+ d = 1. Prove that:
6(a3 + b3 + c3 + d3) ≥ a2 + b2 + c2 + d2 + 1
8
.
20 www.batdangthuc.net
Solution 18 (NguyenDungTN). By AM-GM Inequality
2a3 +
1
43
≥ 3a
2
4
a2 +
1
42
≥ a
2
.
Therefore
6(a3 + b3 + c3 + d3) +
3
16
≥ 9(a
2 + b2 + c2 + d2)
4
5(a2 + b2 + c2 + d2)
4
+
5
16
≥ 5(a+ b+ c+ d
8
=
5
8
Adding up two of them, we get
6(a3 + b3 + c3 + d3) ≥ a2 + b2 + c2 + d2 + 1
8
Solution 19 (Zaizai). We known that
6a3 ≥ a2 + 5a
8
− 1
8
⇔ (4a− 1)
2(3a+ 1)
8
≥ 0
Adding up four similar inequalities, we are done!
∇
Problem 13 (16, Revised by NguyenDungTN). Suppose a, b and c are positive real
numbers. Prove that
a+ b+ c
3
≤
√
a2 + b2 + c2
3
≤ 1
3
(
bc
a
+
ca
b
+
ab
c
)
.
Solution 20. The left-hand inequality is just Cauchy-Schwarz Inequality. We will prove the
right one. Let
bc
a
= x,
ca
b
= y,
ab
c
= z.
The inequality becomes √
xy + yz + zx
3
≤ x+ y + z
3
.
Squaring both sides, the inequality becomes
(x+ y + z)2 ≥ 3(xy + yz + zx)⇔ (x− y)2 + (y − z)2 + (z − x)2 ≥ 0,
which is obviously true.
∇
Problem 14 (17, Vietnam Team Selection Test 2007). Given a triangle ABC. Find the
minimum of: ∑ (cos2(A2 )(cos2(B2 )
cos2(C2 )
www.batdangthuc.net 21
Solution 21 (pi3.14). We have
T =
∑ (cos2(A2 )(cos2(B2 )
(cos2(C2 )
=
∑ (1 + cosA)(1 + cosB)
2(1 + cosC)
.
Let a = tanA
2
; b = tanB
2
; c = tanC
2
. We have ab+ bc+ ca = 1. So
T =
∑ (1 + a2)
(1 + b2)(1 + c2)
=
∑ 1
(1+b2)(1+c2)
1+a2
=
∑ 1
(ab+bc+ca+b2)(ab+bc+ca+c2)
(ab+bc+ca+a2)
=
∑ 1
(a+b)(c+b)(a+c)(b+c)
(b+a)(b+c)
=
∑ 1
(b+ c)2
By Iran96 Inequality, we have
1
(b+ c)2
+
1
(c+ a)2
+
1
(a+ b)2
≥ 9
4(ab+ bc+ ca)
.
Thus F ≥ 9
4
Equality holds when ABC is equilateral.
∇
Problem 15 (18, Greece National Olympiad 2007). . Let a, b, c be sides of a triangle,
show that
(b+ c− a)4
a(a+ b− c) +
(c+ a− b)4
b(b+ c− a) +
(b + c− a)4
a(c+ a− b) ≥ ab+ bc+ ca.
Solution 22 (NguyenDungTN). Since a, b, c are three sides of a triangle, we can substitut