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?

$$\sqrt{b}+\sqrt{c}=\sqrt{b+c+2\sqrt{bc}}>\sqrt{b+c}>\sqrt{a},$$
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:
$$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-3=\sum_{cyc}\left(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-1\right)=$$
$$=\sum_{cyc}\frac{\sqrt{a}-\sqrt{b}-(\sqrt{c}-\sqrt{a})}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}(\sqrt{a}-\sqrt{b})\left(\frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}}-\frac{1}{\sqrt{c}+\sqrt{a}-\sqrt{b}}\right)=$$
$$=2\sum_{cyc}\frac{(\sqrt{a}-\sqrt{b})^2}{(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})}\geq0.$$
