$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ 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.

We have that $$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)}$$
$$ = \left[\frac{xy}{z^2(x + y)} + \frac{1}{z}\right] + \left[\frac{yz}{x^2(y + z)} + \frac{1}{x}\right] + \left[\frac{zx}{y^2(z + x)} + \frac{1}{y}\right] - \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$
$$ = \frac{xy + yz + zx}{z^2(x + y)} + \frac{xy + yz + zx}{x^2(y + z)} + \frac{xy + yz + zx}{y^2(z + x)} - \frac{xy + yz + zx}{xyz}$$
$$ = (xy + yz + zx)\left[\frac{1}{z^2(x + y)} + \frac{1}{x^2(y + z)} + \frac{1}{y^2(z + x)} - 2\right]$$
Now we have to prove that $$\frac{1}{z^2(x + y)} + \frac{1}{x^2(y + z)} + \frac{1}{y^2(z + x)} \ge 3$$
It is evident that
$$\frac{1}{z^2(x + y)} + \frac{1}{x^2(y + z)} + \frac{1}{y^2(z + x)} = \frac{2xyz}{z^2(x + y)} + \frac{2xyz}{x^2(y + z)} + \frac{2xyz}{y^2(z + x)}$$
$$2\left(\frac{xy}{zx + yz} + \frac{yz}{xy + zx} + \frac{zx}{yz + xy}\right)$$
$$ = 2\left[(xy + yz + zx)\left(\frac{1}{xy + yz} + \frac{1}{yz + zx} + \frac{1}{zx + xy}\right) - 3\right]$$
$$ \ge 2\left[(xy + yz + zx)\frac{9}{2(xy + yz + zx)} - 3\right] = 3$$
The equality sign occurs when $x = y = z = \sqrt[3]{\dfrac{1}{2}}$.
