Question 10
Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 6$. Prove that
$$
\frac{1}{a^2 b + 16} + \frac{1}{b^2 c + 16} + \frac{1}{c^2 a + 16} \geq \frac{1}{8}
$$
$$
\frac{1}{a^2 b + 16} + \frac{1}{b^2 c + 16} + \frac{1}{c^2 a + 16} \geq \frac{1}{8}
$$