Question 129
Let $a$, $b$, $c$ be real numbers with sum $a + b + c = 6$. Prove that
$$
\frac{a^4 + 1}{b^2 + 1} + \frac{b^4 + 1}{c^2 + 1} + \frac{c^4 + 1}{a^2 + 1} \ge \frac{51}{5}.
$$
$$
\frac{a^4 + 1}{b^2 + 1} + \frac{b^4 + 1}{c^2 + 1} + \frac{c^4 + 1}{a^2 + 1} \ge \frac{51}{5}.
$$