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 $3$.
