Question 32
If $a$, $b$, $c$ are positive real numbers, prove that:
$$
\frac{1}{ab(b+1)(c+1)} + \frac{1}{bc(c+1)(a+1)} + \frac{1}{ca(a+1)(b+1)} - \frac{3}{(1+abc)^2} \ge 0.
$$
$$
\frac{1}{ab(b+1)(c+1)} + \frac{1}{bc(c+1)(a+1)} + \frac{1}{ca(a+1)(b+1)} - \frac{3}{(1+abc)^2} \ge 0.
$$