$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$. 
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$

Before you complain, this problem is adapted from a recent competition. I have put my solution down below, there might be more practical and correct answers. In that case, please post them.

Let $x=\frac{1}{a},$ $y=\frac{1}{b}$ and $z=\frac{1}{c}.$
Thus, $abc=2$ and by C-S we obtain:
$$\sum_{cyc}\frac{xy}{z^2(x+y)}=\sum_{cyc}\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=xy+xz+yz.$$
