Question 99
Let $a$, $b$ and $c$ be positive real numbers satisfying $abc = 1$. Prove that
$$
\frac{1}{a^3 + 2b^2 + 2b + 4} + \frac{1}{b^3 + 2c^2 + 2c + 4} + \frac{1}{c^3 + 2a^2 + 2a + 4} \le \frac{1}{3}.
$$
$$
\frac{1}{a^3 + 2b^2 + 2b + 4} + \frac{1}{b^3 + 2c^2 + 2c + 4} + \frac{1}{c^3 + 2a^2 + 2a + 4} \le \frac{1}{3}.
$$