TITLE: Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$?
QUESTION [11 upvotes]: Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality 
$$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?

REPLY [2 votes]: $\sqrt{a} = x, b=y^2, c=z^2 => x^2+y^2+z^2=1$
We have to prove $$\frac{x}{y^{2}+z^{2}}+\frac{y}{x^{2}+z^{2}}+\frac{z}{x^{2}+y^{2}}\geq \frac{3\sqrt{3}}{2}$$:
$$\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )\leq \left ( x^{2}+\frac{1}{3} \right )\left ( y^{2}+z^{2} \right )\leq \frac{\left ( x^{2}+y^{2}+z^{2}+\frac{1}{3} \right )^{2}}{4}=\frac{4}{9}$$
Do it the same for $\frac{2\sqrt{3}}{3}y\left ( z^{2}+x^{2} \right ), \frac{2\sqrt{3}}{3}z\left ( y^{2}+x^{2} \right )$
So: $$\frac{leftside}{\frac{2\sqrt{3}}{3}}=\sum \frac{x^{2}}{\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )}\geq \frac{\sum x^{2}}{\frac{4}{9}}=\frac{9}{4}$$
$$\Rightarrow leftside=\sum \frac{x}{y^{2}+z^{2}}\geq \frac{3\sqrt{3}}{2}$$