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Problems of Ukrainian Authors

Algebra Difficulty 4.7 AIME Prove it Ukraine

For positive numbers $a$ , $b$ with $a + b = ab$ prove inequality:
$\frac{a}{b^2 + 4} + \frac{b}{a^2 + 4} \ge \frac{1}{2}.$

Solution

$ab = a + b \ge 2\sqrt{ab} \Rightarrow ab \ge 4$ , then
$\begin{aligned} \frac{a}{b^2+4} + \frac{b}{a^2+4} &\ge \frac{a}{b^2+ab} + \frac{b}{a^2+ab} = \\ &= \frac{a}{b(a+b)} + \frac{b}{a(a+b)} = \frac{a^2+b^2}{(a+b)^2} \ge \frac{1}{2}. \end{aligned}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.