Inequality with the constraint $xy=x+y$ I have to prove that if $x$ and $y$ are two positive numbers such that $xy=x+y$, the following inequality
$$\frac{x}{y^2+4}+\frac{y}{x^2+4}\geq \frac{1}{2}$$
holds. Any help is appreciated, thank you in advance.

By renaming $x$ as $\frac{1}{u}$ and $y$ as $\frac{1}{v}$ we have to prove that $u,v>0$ and $u+v=1$ grant
$$ \frac{u^2}{v+4u^2 v}+\frac{v^2}{u+4uv^2}\geq \frac{1}{2} $$
that is the same as showing that for any $x\in(0,1)$ the inequality
$$ \frac{x^2}{(1-x)(1+4x^2)}+\frac{(1-x)^2}{x(1+4(1-x)^2)} \geq \frac{1}{2} $$
holds. That is almost trivial since $f(x)=\frac{x^2}{(1-x)(1+4x^2)}$ is a convex function on $(0,1)$, hence
$$ f(x)+f(1-x) \geq 2\cdot f\left(\frac{1}{2}\right) = \frac{1}{2} $$
as wanted.
