If $x+y+z=2$ prove that $\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\ge2+\frac{1}{\sqrt{2}}$ Let $x,y,z$ be non-negative reals whose sum is $2$. Prove that
$\frac{1}{\sqrt{x^2+y^2}}+\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{z^2+x^2}}\ge2+\frac{1}{\sqrt{2}}$
I have tried bounding them up (assuming that $a\le b\le c$), many inequalities (AM-GM, QM-AM, Cauchy) but nothing has worked. I know that equality is achieved when one of $x,y,z$ is $0$ and the other two are $1$ but that doesn't help.

Without loss of generality, assume that $x \geq y \geq z$. It is easy to verify that using $(x + \frac{z}{2}, y + \frac{z}{2}, 0)$ to replace $(x, y, z)$ would make the LHS smaller, i.e., 
$$
\frac{1}{\sqrt{x^2 + y^2}} + 
\frac{1}{\sqrt{y^2 + z^2}} + 
\frac{1}{\sqrt{x^2 + z^2}} 
\geq
\frac{1}{\sqrt{\left(x + \frac{z}{2}\right)^2 + \left(y + \frac{z}{2}\right)^2}} + 
\frac{1}{y + \frac{z}{2}} + 
\frac{1}{x + \frac{z}{2}}. 
$$
This is obvious as $\left( y + \frac{z}{2}\right)^2 \geq (y + z)^2, \left( x + \frac{z}{2}\right)^2 \geq (x + z)^2$. 
As a consequence,  we only need to prove the case where $z = 0$, i.e., for nonnegative $x$ and $y$ with $x + y = 2$,
$$
\frac{1}{\sqrt{x^2 + y^2}} + \frac{1}{x} + \frac{1}{y} \geq 2 + \frac{1}{\sqrt{2}}.\tag{*}
$$
Note that the LHS in (*) could be written as 
$$
\begin{align}
\frac{1}{\sqrt{(x+y)^2 - 2xy}} + \frac{x+y}{xy} 
&= \frac{1}{\sqrt{4 - 2xy}} + \frac{2}{xy}\\
&= 2 + \frac{1}{\sqrt{2}} + (1 - xy) \left[\frac{2}{xy} - \frac{1}{\sqrt{2}(2 - xy + \sqrt{2 - xy})}\right]\\
&\geq 2 + \frac{1}{\sqrt{2}} + (1 - xy)\left(2 - \frac{1}{2\sqrt{2}}\right)\\
&\geq 2 + \frac{1}{\sqrt{2}},
\end{align}
$$
where we have used the fact that $xy \leq \left(\frac{x + y}{2}\right)^2 = 1$.
