Question 40
Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that
$$
\frac{1}{a^2(b+c)} + \frac{1}{b^2(c+a)} + \frac{1}{c^2(a+b)} \ge \frac{3}{2}.
$$
$$
\frac{1}{a^2(b+c)} + \frac{1}{b^2(c+a)} + \frac{1}{c^2(a+b)} \ge \frac{3}{2}.
$$