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Problem 1390

National olympiad, first round

Algebra Difficulty 6.9 Prove it

Example 8.2.3. Let $a, b, c$ be positive real numbers satisfying $a b c=1$ . Prove that
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$
(Bulgarian MO 1998)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

SOLUTION. Denote $S = \sum_{cyc} a$ , $P = \sum_{cyc} ab$ , $Q = abc$ . By some calculations, we get that
$\begin{array}{c} \mathrm{LHS} = \sum_{cyc} \frac{1}{S+1-a} = \frac{S^2 + 4S + 3 + P}{S^2 + 2S + PS + P} \\ \mathrm{RHS} = \sum_{cyc} \frac{1}{2+a} = \frac{12 + 4S + P}{9 + 4S + 2P} \end{array}$

So it suffices to prove that
$\frac{S^2 + 4S + 3 + P}{S^2 + 2S + PS + P} \leq \frac{12 + 4S + P}{9 + 4S + 2P}$
which is reduced to
$(3P - 5)S^2 + (S - 1)P^2 + 6PS \geq 24S + 3P + 27$

Because $abc = 1$ , we deduce $S, P \geq 3$ , therefore
$\begin{aligned} \text{LHS} \geq 4S^2 + 2P^2 + 6PS & \geq 12S + 6(P-1)S + 6S + 2P^2 \\ & \geq 24S + 3P + (P^2 + 6S) \geq \text{RHS} \end{aligned}$

Equality holds for $S = P = 3$ or equivalently to $a = b = c = 1$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.