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.
 A: Well,I think you can use the famouse Iran 96 inequality as a result to solve this problem.
the Iran 96 inequality
Let $x,y,z\geq 0$ we have
$$ \frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(x+z)^{2}}\geq \frac{9}{4(xy+yz+zx)} $$
square both side,we can rewrite the inequality into
$$ \sum_{cyc}{\frac{1}{(x+y)^{2}}}+2\sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq \frac{25}{4}$$
Now,Using this inequality as a known result,it's suffice to prove
$$ \sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq 2 $$
after simple homogenous,it's
$$ (xy+yz+xz)(x+y+z)\geq (x+y)(y+z)(z+x) $$
Or
$$ xyz\geq 0 $$
Which is obviously true,Equality occurs if and only if $x=y=1, z=0 $ or it's permutation.
The proof is complete
A: assume $z$ is the smallest.
if we fix the sum $x+y$,
$\dfrac{1}{x+y} + \dfrac{1}{z+x}+\dfrac{1}{z+y} = \dfrac{1}{x+y} +\dfrac{x+y+2z}{1+z^2}$, take it as a function of $z$, you may see, it is increasing.[you may differentiate it or just plugin $z=0$ and compare]
Thus $z=0$, it will be smallest.
The rest is easy.

Update:
After we get $z=0$, the restriction on $x,y,z$ turns out to be $xy=1$, and the goal is to minimize
$\dfrac{1}{x+y}+x+y$
Since now $x=\dfrac{1}{y}$, thus
$\dfrac{1}{x+y} +x+y = \dfrac{1}{x+\dfrac{1}{x}}+x+\dfrac{1}{x}$. Take $t = x+\dfrac{1}{x}$, we know that $t\ge 2$. then the objective function is to minimize
$$t+\dfrac{1}{t}$$
for $t\ge 2$.
You can take a derivative to see that $t+1/t$ is increasing when $t\ge 1$. So minimum will be obtained at $t = 2$. which is $x=y=1$.
A: 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!
