TITLE: Inequality: Find Min $S=\frac{a}{\sqrt{1-a}}+\frac{b}{\sqrt{1-b}}$
QUESTION [5 upvotes]: Inequality: 
Find Min a,b>0, a+b=1. $S=\frac{a}{\sqrt{1-a}}+\frac{b}{\sqrt{1-b}}$

REPLY [2 votes]: For $x>0$, the function $f(x)=x^{-1/2}$ is convex, so using $a+b=1$, we have
$$
af(b)+bf(a)\geq f(ab+ba)=f(2ab)=\frac{1}{\sqrt{2ab}}\geq\frac{\sqrt{2}}{a+b}=\sqrt{2}.
$$
The second inequality above uses $(a+b)^2\geq 4ab$. All equalities above are achieved when $a=b=\frac{1}{2}$.