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Algebra Difficulty 7.7 National Olympiad, round 2 Prove it Baltic Way

Let $a$ , $b$ , $c$ , $d$ be nonnegative reals such that $a + b + c + d = 4$ . Prove the inequality
$\frac{a}{a^3+8} + \frac{b}{b^3+8} + \frac{c}{c^3+8} + \frac{d}{d^3+8} \leq \frac{4}{9}.$

Solution

By the means inequality we have $a^3 + 2 = a^3 + 1 + 1 \ge 3\sqrt[3]{a^3 \cdot 1 \cdot 1} = 3a$ . Therefore it is sufficient to prove the inequality
$\frac{a}{3a+6} + \frac{b}{3b+6} + \frac{c}{3c+6} + \frac{d}{3d+6} \le \frac{4}{9}.$
We can write the last inequality in the form
$\frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} + \frac{1}{d+2} \ge \frac{4}{3}.$
Now it follows by the harmonic and arithmetic means inequality:
$\frac{1}{4} \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} + \frac{1}{d+2} \right) \ge \frac{4}{(a+2)(b+2)(c+2)(d+2)} = \frac{4}{4+2+2+2+2} = \frac{1}{3}.$

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