Proving a sharp inequality After experimenting, I've come to the conclusion that if $x\geq y\geq z\geq 0:$
$$\sum_{x,y,z}\frac{x}{\sqrt{x+y}}\geq \sum_{x,y,z}\frac{y}{\sqrt{x+y}}$$
(the sums are cyclic)
Does anyone know how to prove this? I've tried variants of Cauchy Schwarz, and the rearrangement inequality, but these are not powerful enough.

The condition $x^2+y^2+z^2=1$ is redundant.
Let $a=\sqrt{y+z},b=\sqrt{z+x},c=\sqrt{x+y}$ we have $c\ge b \ge a \ge 0$ and $$x=\frac{b^2 + c^2 - a^2}{2},y=\frac{c^2 + a^2 - b^2}{2},z=\frac{a^2 + b^2 - c^2}{2}.$$
Replacing into the original inequality, it becomes
$$\sum_{a,b,c} \frac{b^2 + c^2 - a^2}{2c} \ge \sum_{a,b,c} \frac{c^2 + a^2 - b^2}{2c},$$
which is true because
\begin{align}
LHS-RHS &= \sum_{a,b,c} \frac{b^2 - a^2}{c} \\
&= \frac{b^2 - a^2}{c} + \frac{c^2 - b^2}{a} +\frac{(a^2-b^2) +(b^2 - c^2)}{b}\\
&= (b^2 - a^2)\left(\frac{1}{c}-\frac{1}{b}\right) + (c^2 - b^2)\left(\frac{1}{a}-\frac{1}{b}\right) \\
&= \frac{(b-a)(c-b)}{b} \left(-\frac{b+a}{c}+\frac{b+c}{a}\right) \\
&= \frac{(b-a)(c-b)(c-a)(a+b+c)}{abc}\\
&\ge 0.
\end{align}
