Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$

Question

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?

Answer 1

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 $3$.

Answer 2

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