Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum \frac{1}{\frac{P}{\sqrt{a}}-2}$$
Let $x=\frac{P}{\sqrt{a}}$, $y=\frac{P}{\sqrt{b}}$,$z=\frac{P}{\sqrt{c}}$
Then we have $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
By $AM \ge HM$
$$\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Hence
$$x+y+z \ge 9$$
Any way to proceed further?

I did the following:
Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}$, $y=\sqrt{a}+\sqrt{c}-\sqrt{b}$ and $z=\sqrt{a}+\sqrt{b}-\sqrt{c}$. Then $\frac{x+y}{2}=\sqrt{c}, \frac{x+z}{2}=\sqrt{b}$ and $\frac{y+z}{2}=\sqrt{a}$. Rewriting the original expression, we get $$\frac{1}{2}\left(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\right)$$
It is easy to see that this expression is greater than or equal to $3$, as $\frac{x}{y}+\frac{y}{x}\geq 2$ by AM-GM inequality. 
This minimum value is achieved when $a=b=c$
