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?

When $a = b = c$, $S = 3$. Next it will be proved that $S \geqslant 3$ for all possible $a, b, c$.

Denote $\displaystyle u = \frac{\sqrt{a}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, $\displaystyle v = \frac{\sqrt{b}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, $\displaystyle w = \frac{\sqrt{c}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, then $\sum u = 1$. Since$$ \sqrt{a} < \sqrt{b + c} < \sqrt{b} + \sqrt{c}, $$ then $\displaystyle 0 < u < \frac{1}{2}$. Analogously, $\displaystyle 0 < v, w < \frac{1}{2}$. It suffices to prove$$ S = \sum \frac{u}{v + w - u} = \sum \frac{u}{1 - 2u} \geqslant 3. $$

Define $\displaystyle f(x) = \frac{x}{1 - 2x} \ (0 < x < \frac{1}{2})$. Because $\displaystyle f''(x) = \frac{4}{(1 - 2x)^3} > 0$, by Jensen's inequality,$$ S = \sum f(u) \geqslant 3 f\left(\frac{1}{3} \sum u\right) = 3f\left(\frac{1}{3}\right) = 3. $$ Therefore the minimum of $S$ is $3$.

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 $a=b=c$