Question 111
Let $a$, $b$, $c$ be positive numbers with $abc \ge 1$. Prove that
$$
\frac{1}{a^3 + 2b^3 + 6} + \frac{1}{b^3 + 2c^3 + 6} + \frac{1}{c^3 + 2a^3 + 6} \le \frac{1}{3}.
$$
$$
\frac{1}{a^3 + 2b^3 + 6} + \frac{1}{b^3 + 2c^3 + 6} + \frac{1}{c^3 + 2a^3 + 6} \le \frac{1}{3}.
$$