<s>[INST] Show that $\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$. 
Given $x,y,z\geq0$ and $xy+yz+zx=1$. Show that $\displaystyle\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.

I've tried many things but all failed. The only thing I know is that the equality holds when $x=y=1, z=0$. 
Please help. Thank you.
 [/INST] $z=\dfrac{1-xy}{x+y} \ge 0 \to xy\le 1$
LHS$=\sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x+y}{1+x^2}}+\sqrt{\dfrac{x+y}{1+y^2}}=f(x,y)$
$f'_{x}=\dfrac{1-x^2-2xy}{(x^2+1) \sqrt{x^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(y^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<1>
$f'_{y}=\dfrac{1-y^2-2xy}{(y^2+1) \sqrt{y^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(x^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<2>
<1>$*(1+y^2)$-<2>$*(1+x^2)$:
$(y^2-x^2)\left(2(x+y)(1-xy)+\dfrac{(1+x^2)(1+y^2)}{\sqrt{1+x^2}{1+y^2}}\right)=0$
so we have $x=y$ only. put in <1>, we get: $4x-4x^3=(x^2+1)^{\frac{3}{2}}$,ie:
$15x^6-35x^4+13x^2-1=0 \to (3x^2-1)(5x^4-10x^2+1)=0$, we get two roots:
$x_1=\dfrac{1}{\sqrt{3}}, x_2=\sqrt{1-\dfrac{2}{\sqrt{5}}}$
$f(x_1)=2.79,f(x_2)=2.77$,but we note $f(x,y)$ when $x ,y \to \infty, f(x,y) \to 0$,so we have to check the  bound, which is $xy=1$. 
$f(x,y)=\sqrt{x}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{\dfrac{1}{x}+x}}$
let $m=\sqrt{x}+\dfrac{1}{\sqrt{x}} \ge2$
$f(m)=m+\dfrac{1}{\sqrt{m^2-2}}$, when $m\ge 2$, $f(m)$ is mono increasing, so the minimum will be got when $m=2 \to x=y=1, f(1)=2+\dfrac{1}{\sqrt{2}}< 2.77$, so the  minimum is $2+\dfrac{1}{\sqrt{2}}$.
To prove $f(m)=m+\dfrac{1}{\sqrt{m^2-2}}$ is mono increasing function, 
$f'(m)=1-\dfrac{m}{(m^2-2)^{3/2}}=1-\dfrac{1}{\sqrt{m^2-2}\left(m-\dfrac{2}{m}\right)}>0$ when $m\ge 2$.
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