$x$, $y$ 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.
On
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 $x = y = z = \sqrt[3]{\dfrac{1}{2}}$.
On
$$\frac{xy}{z^2(x+y)}+\frac{yz}{x^2(z+y)}+\frac{xz}{y^2(x+z)}$$ $$=\frac{2x^2y^2}{z(x+y)}+\frac{2y^2z^2}{x(z+y)}+\frac{2x^2z^2}{y(x+z)}$$ $$\geq 2\frac{(xy+yz+xz)^2}{2(xy+yz+zx)}=xy+yz+xz$$
Using Titu's Lemma which is a variant of the Cauchy-Schwarz inequality.
Let $x=\frac{1}{a},$ $y=\frac{1}{b}$ and $z=\frac{1}{c}.$
Thus, $abc=2$ and by C-S we obtain: $$\sum_{cyc}\frac{xy}{z^2(x+y)}=\sum_{cyc}\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=xy+xz+yz.$$