# $\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ 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: $$\sum_{cyc}\frac{xy}{z^2(x+y)}=\sum_{cyc}\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=xy+xz+yz.$$

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

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

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 $$x = y = z = \sqrt[3]{\dfrac{1}{2}}$$.