Prove that positive $x,y,z$ satisfy $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$ 
Prove that positive $x,y,z$ satisfy $$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$$

Actually, this is a part of my solution to another problem, which is:

If $a,b,c$ are sides of a triangle, prove that $$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \sqrt{a+b-c}+\sqrt{a-b+c}+\sqrt{-a+b+c}$$

I substituted $a=x+y$, $b=y+z$, $c=z+x$. It's often called "Ravi substitution".
Here's a similar Math.SE question.
 A: It's easy to show that
$$\sqrt{u+v}\ge{\sqrt u+\sqrt v\over\sqrt2}$$
Thus
$$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge{\sqrt x+\sqrt y\over\sqrt2}+{\sqrt y+\sqrt z\over\sqrt2}+{\sqrt z+\sqrt x\over\sqrt2}=\sqrt{2x}+\sqrt{2y}+\sqrt{2z}$$
A: \begin{align}
& 2(\sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}) \geq 2(\sqrt{2x}+\sqrt{2y}+\sqrt{2z}) \\
& \Leftrightarrow (2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y})+(2\sqrt{x+z}-\sqrt{2x}-\sqrt{2z})+(2\sqrt{y+z}-\sqrt{2y}-\sqrt{2z}) \geq 0 
\end{align}
Note that 
\begin{align}
& (2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y}) \geq 0 \\
& \Leftrightarrow 2\sqrt{x+y} \geq \sqrt{2x}+\sqrt{2y} \\
& \Leftrightarrow 4(x+y) \geq (\sqrt{2x}+\sqrt{2y})^2=2(x+y)+4\sqrt{xy} \\
& \Leftrightarrow 2(\sqrt{x}-\sqrt{y})^2 \geq 0
\end{align}
Thus $(2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y}) \geq 0$. Similarly $(2\sqrt{x+z}-\sqrt{2x}-\sqrt{2z}) \geq 0$ and $(2\sqrt{y+z}-\sqrt{2y}-\sqrt{2z}) \geq 0$ so we are done. Equality holds iff $x=y=z$.
A: Let $x\geq y\geq z$.
Hence, $(2x,2y,2z)\succ(x+y,x+z,y+z)$ and since $f(x)=\sqrt{x}$ is a concave function,
we are done by Karamata.
