TITLE: Show that $\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
QUESTION [18 upvotes]: 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.

REPLY [4 votes]: Without loss of generality, we assume that $x\ge y\ge z$.

Lemma 1:
  $$\dfrac{1}{\sqrt{y^2+1}}+\dfrac{1}{\sqrt{z^2+1}}\ge 1+\dfrac{1}{\sqrt{(y+z)^2+1}}$$

Proof: this inequality is equivalent to
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}+\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge 1+\dfrac{1}{(y+z)^2+1}+\dfrac{2}{\sqrt{(y+z)^2+1}}.$$
Notice that
$$(y+z)^2+1-(y^2+1)(z^2+1)=yz(2-yz)\ge 0,$$
hence
$$\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge\dfrac{2}{\sqrt{(y+z)^2+1}}.$$
Therefore, it suffices to prove that
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\ge1+\dfrac{1}{(y+z)^2+1}.$$
And this is equivalent to
$$\dfrac{yz[2-2yz-yz(y+z)^2]}{(y^2+1)(z^2+1)[(y+z)^2+1]}\ge 0.$$
Above is true because
$$2-2yz-yz(y+z)^2=2x(y+z)-yz(y+z)^2=(y+z)[2x-yz(y+z)]\ge (y+z)[2x-x^2(y+z)]=x(y+z)(2-xy-xz)\ge 0.$$


Lemma 2:
  $$\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}\ge\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}.$$

Proof:
\begin{align}\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{(y+z)(x+y)}}+\dfrac{1}{\sqrt{(z+x)(z+y)}}\right)\\
&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{\sqrt{1+z^2}}\right)
\end{align}
and apply Lemma 1, then done!

Thus the original inequality is equivalent to
$$\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}\ge 2+\dfrac{1}{\sqrt{2}}.$$
Letting $$t=\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}\ge 2,$$ the inequality becomes
$$t+\dfrac{1}{\sqrt{t^2-2}}\ge 2+\dfrac{1}{\sqrt{2}}.$$
This can be shown by the following: 
\begin{align}
&t+\dfrac{1}{\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}}
\\
=&t+\dfrac{2\sqrt{2}}{2\sqrt{2}\cdot\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}}
\\
\ge& t+\dfrac{2\sqrt{2}}{2+t^2-2}-2-\dfrac{1}{\sqrt{2}}
\\
=&t+\dfrac{2\sqrt{2}}{t^2}-2-\dfrac{1}{\sqrt{2}}
\\
=&\dfrac{(t-2)(\sqrt{2}t^2-t-2)}{\sqrt{2}t^2}\ge 0.
\end{align}