<s>[INST] 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.
 [/INST] 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 $x=y=1$.
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