<s>[INST] If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{a+b+c+15}}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$
It seems nice enough.
I proved this inequality by Holder, but it quits very ugly.
Maybe there is something nice? Thank you!
 [/INST] Put the following substitution :

$\frac{a}{b+c}=x$$\quad$$\frac{b}{a+c}=y$$\quad$$\frac{c}{b+a}=z$

Remark that :
$$a+b+c=((\frac{1}{x}+1)(\frac{1}{y}+1)(\frac{1}{z}+1))^{\frac{1}{3}}$$
And :
$$abc=1 \iff -2=-\frac{1}{xyz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
So we get the inequality related to this Post that you have proved by your own with brio . 
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