Prove the inequality $\sum\limits_{\text{cyc}} \frac{x}{y+z+\sqrt{x}}≥ \frac{(x+y+z)^2}{2(xy+yz+zx)+x\sqrt{x}+y\sqrt{y}+z\sqrt{z}}$ How do I show the following?
$$
\frac{x}{y+z+\sqrt{x}}+\frac{y}{z+x+\sqrt{y}}+\frac{z}{x+y+\sqrt{z}}≥ \frac{(x+y+z)^2}{2(xy+yz+zx)+x\sqrt{x}+y\sqrt{y}+z\sqrt{z}}
$$
Any hint would do; I don't require a complete solution.

It is just the Cauchy-Schwarz inequality in the form:
$$ \frac{a^2}{A}+\frac{b^2}{B}+\frac{c^2}{C}\geq\frac{(a+b+c)^2}{A+B+C}$$
for $a=x,b=y,c=z$ and $A=xy+xz+x\sqrt{x},B=xy+zy+y\sqrt{y},C=xz+yz+z\sqrt{z}$.
