Let $x,y,z>0$ such that $x+y+z+2\sqrt{xyz}\le 1$. Prove that $$\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{\sqrt{xyz}}-2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\ge \frac{\sqrt{xy}}{y}+\frac{\sqrt{yz}}{z}+\frac{\sqrt{xz}}{x}$$
My try: Let $$\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow \left(\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}};\frac{b}{\sqrt{\left(c+b\right)\left(b+a\right)}};\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)$$
Or we need to prove $$\sum \frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{a}-2\left(\sum \frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)\ge \sum \frac{a\sqrt{b+c}}{b\sqrt{a+c}}$$
I tried to use C-S: $\sqrt{\left(a+b\right)\left(a+c\right)}\ge a+\sqrt{bc}$ and AM-GM but failed. Otherwise we have $cos^2 A+cos ^2 B+ cos ^2 C+2 cos A cos Bcos C=1$ then i let $x=cos^2 A$ but i cant continue.
The inequality can be written as:
$$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq 2(\sqrt{x}+\sqrt{y}+\sqrt{z})+\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}}$$
From the condition and using AM-GM:
$$ \begin{aligned} (1-\sqrt{x})(1+\sqrt{x})&=1-x\\ &\geq y+z+2\sqrt{xyz}\\ &\geq 2\sqrt{yz}+2\sqrt{xyz}\\ &=2\sqrt{yz}(1+\sqrt{x}) \end{aligned} $$
Therefore $1\geq \sqrt{x}+2\sqrt{yz}$ and similarly $1\geq \sqrt{y}+2\sqrt{zx}, 1\geq \sqrt{z}+2\sqrt{xy}$. Thus:
$$ \begin{aligned} \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}&\geq \frac{\sqrt{z}+2\sqrt{xy}}{\sqrt{x}}+\frac{\sqrt{x}+2\sqrt{yz}}{\sqrt{y}}+\frac{\sqrt{y}+2\sqrt{zx}}{\sqrt{z}} \\ &= 2(\sqrt{x}+\sqrt{y}+\sqrt{z})+\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}} \end{aligned} $$
Equality occurs when $x=y=z=\frac{1}{4}$.