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

Question

$x$, $y$ 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.

Answer 1

$$\frac{xy}{z^2(x+y)}+\frac{yz}{x^2(z+y)}+\frac{xz}{y^2(x+z)}$$ $$=\frac{2x^2y^2}{z(x+y)}+\frac{2y^2z^2}{x(z+y)}+\frac{2x^2z^2}{y(x+z)}$$ $$\geq 2\frac{(xy+yz+xz)^2}{2(xy+yz+zx)}=xy+yz+xz$$

Using Titu's Lemma which is a variant of the Cauchy-Schwarz inequality.