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?
 A: 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 $3$.
A: $$\sqrt{b}+\sqrt{c}=\sqrt{b+c+2\sqrt{bc}}>\sqrt{b+c}>\sqrt{a},$$
which says that all denominators are positives.
Now, by C-S $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}\frac{a}{\sqrt{ab}+\sqrt{ac}-a}\geq$$
$$\geq\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sum\limits_{cyc}(\sqrt{ab}+\sqrt{ac}-a)}=\frac{\sum\limits_{cyc}(a+2\sqrt{ab})}{\sum\limits_{cyc}(2\sqrt{ab}-a)}\geq3$$
because the last inequality it's just $$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\geq0.$$
The equality occurs for $a=b=c$, which says that $3$ is a minimal value.
We can use also the Rearrangement.
Indeed, the triples $(\sqrt{a},\sqrt{b},\sqrt{c})$ and $\left(\frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}},\frac{1}{\sqrt{a}+\sqrt{c}-\sqrt{b}},\frac{1}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)$ are the same ordered.
Thus, $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\geq\frac{1}{2}\sum_{cyc}\left(\frac{\sqrt{b}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{\sqrt{b}+\sqrt{c}-\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\right)=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$ and we are done again.
Another way:
$$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-3=\sum_{cyc}\left(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-1\right)=$$
$$=\sum_{cyc}\frac{\sqrt{a}-\sqrt{b}-(\sqrt{c}-\sqrt{a})}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}(\sqrt{a}-\sqrt{b})\left(\frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-\frac{1}{\sqrt{c}+\sqrt{a}-\sqrt{b}}\right)=$$
$$=2\sum_{cyc}\frac{(\sqrt{a}-\sqrt{b})^2}{(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})}\geq0.$$
A: 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 $a=b=c$
