Question 73
Let $a$, $b$, $c$, $d$ be nonnegative reals such that $a + b + c + d = 4$. Prove the inequality
$$
\frac{a}{a^3+8} + \frac{b}{b^3+8} + \frac{c}{c^3+8} + \frac{d}{d^3+8} \leq \frac{4}{9}.
$$
$$
\frac{a}{a^3+8} + \frac{b}{b^3+8} + \frac{c}{c^3+8} + \frac{d}{d^3+8} \leq \frac{4}{9}.
$$