Prove that $\frac 1 {x+y}+\frac 1 {y+z}+\frac 1 {z+x}\geq \frac 5 2$. 
Given:

*

*$x,y,z\geq0$


*$xy+yz+zx=1$
Prove that $\displaystyle \frac 1 {x+y}+\frac 1 {y+z}+\frac 1 {z+x}\geq \frac 5 2$.

I tried using Cauchy's inequality LHS $\geq\frac 9 {2a+2b+2c}$, but failed. Please give me some ideas. Thank you.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition does not depend on $w^3$ and we need to prove that
$$\frac{\sum\limits_{cyc}(x^2+3xy)}{\prod\limits_{cyc}(x+y)}\geq\frac{5}{2}$$ or
$$\frac{9u^2+3v^2}{9uv^2-w^3}\geq\frac{5}{2}$$ or
$f(w^3)\geq0$, where $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $w^3$,
which happens in the following cases.


*

*$y=x$, $z=\frac{1-x^2}{2x}$, where $0<x\leq1$ and we get something obvious;

*$w^3=0$.
Let $z=0$. Hence, we need to prove that
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{x+y}\geq\frac{5}{2}$$ or
$$x+y+\frac{1}{x+y}\geq\frac{5}{2}$$ or
$$(x+y-2)(2(x+y)-1)\geq0,$$
which is true because $x+y\geq2\sqrt{xy}=2$ and we are done!
