Inequality with the constraint $xy=x+y$

Question

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.

Answer 1

The inequality is not true for $0<x,y<1.$ $x,y$ must be greater than $1.$

Answer 2

Since $\frac{(x+y)^2}{4}\geq xy=x+y$, we obtain $x+y\geq4$.

Thus, by C-S $\frac{x}{y^2+4}+\frac{y}{x^2+4}\geq\frac{(x+y)^2}{xy^2+x^2y+4(x+y)}=\frac{x+y}{xy+4}=\frac{1}{1+\frac{4}{x+y}}\geq\frac{1}{2}$.

Answer 3

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.