Topics in Inequalities - Theorems and Techniques Hojoo ... - Index of

Proof. Case 1. b 1 ≥ a 2 : It follows from a 1 ≥ a 1 +a 2 −b 1 and from a 1 ≥ b 1 that a 1 ≥ max(a 1 +a 2 −b 1 , b 1 ) so
that max(a 1 , a 2 ) = a 1 ≥ max(a 1 +a 2 −b 1 , b 1 ). From a 1 +a 2 −b 1 ≥ b 1 +a 3 −b 1 = a 3 and a 1 +a 2 −b 1 ≥ b 2 ≥ b 3 ,
we have max(a 1 + a 2 − b 1 , a 3 ) ≥ max(b 2 , b 3 ). Apply the theorem 8 twice to obtain
∑
x a 1
y a 2
z a 3
= ∑
z a 3
(x a 1
y a 2
+ x a 2
y a 1
)
sym
≥
cyclic
∑
z a3 (x a1+a2−b1 y b1 + x b1 y a1+a2−b1 )
cyclic
= ∑
x b 1
(y a 1+a 2 −b 1
z a 3
+ y a 3
z a 1+a 2 −b 1
)
≥
cyclic
∑
x b 1
(y b 2
z b 3
+ y b 3
z b 2
)
cyclic
= ∑ sym
x b1 y b2 z b3 .
Case 2. b 1 ≤ a 2 : It follows from 3b 1 ≥ b 1 + b 2 + b 3 = a 1 + a 2 + a 3 ≥ b 1 + a 2 + a 3 that b 1 ≥ a 2 + a 3 − b 1
and that a 1 ≥ a 2 ≥ b 1 ≥ a 2 + a 3 − b 1 . Therefore, we have max(a 2 , a 3 ) ≥ max(b 1 , a 2 + a 3 − b 1 ) and
max(a 1 , a 2 + a 3 − b 1 ) ≥ max(b 2 , b 3 ). Apply the theorem 8 twice to obtain
∑
x a 1
y a 2
z a 3
= ∑
x a 1
(y a 2
z a 3
+ y a 3
z a 2
)
sym
≥
cyclic
∑
x a1 (y b1 z a2+a3−b1 + y a2+a3−b1 z b1 )
cyclic
= ∑
y b1 (x a1 z a2+a3−b1 + x a2+a3−b1 z a1 )
≥
cyclic
∑
y b 1
(x b 2
z b 3
+ x b 3
z b 2
)
cyclic
= ∑ sym
x b1 y b2 z b3 .
Remark 3.2.2. The equality holds if and only if x = y = z. However, if we allow x = 0 or y = 0 or z = 0,
then one may easily check that the equality holds when a 1 , a 2 , a 3 > 0 and b 1 , b 2 , b 3 > 0 if and only if
x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0.
We can use Muirhead’s theorem to prove Nesbitt’s inequality.
(Nesbitt) For all positive real numbers a, b, c, we have
a
b + c +
b
c + a +
c
a + b ≥ 3 2 .
Proof 6. Clearing the denominators of the inequality, it becomes
2 ∑
cyclic
a(a + b)(a + c) ≥ 3(a + b)(b + c)(c + a) or ∑ sym
a 3 ≥ ∑ sym
(IMO 1995) Let a, b, c be positive numbers such that abc = 1. Prove that
1
a 3 (b + c) + 1
b 3 (c + a) + 1
c 3 (a + b) ≥ 3 2 .
a 2 b.
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