Inequality: Find Min $S=\frac{a}{\sqrt{1-a}}+\frac{b}{\sqrt{1-b}}$

Question

Inequality: Find Min a,b>0, a+b=1. $S=\frac{a}{\sqrt{1-a}}+\frac{b}{\sqrt{1-b}}$

Answer 1

By Holder's Inequality, $$LHS^2\cdot(2ab)=\left(\frac{a}{\sqrt b}+\frac{b}{\sqrt a}\right)^2(ab+ba)\ge(a+b)^3=1$$

We also know $2ab\le\frac12(a+b)^2=\frac12$. Using this in the above, we get $$LHS\ge\sqrt2$$

As equality is obtained when $a=b=\frac12$, this is indeed then minimum.

Answer 2

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 $a=b=\frac{1}{2}$.

Answer 3

It hasn't "Min S" if you don't have restrictions for $a$ and $b$, since $$f(x)=\frac{x}{\sqrt{1-x}}$$ is an increasing function.

Note that $$f'(x)=\frac{\sqrt{1-x}+ x\frac{1}{2\sqrt{1-x}}} {1-x}=\frac{\frac{1}{2}(1-x)+\frac12} {(1-x)\sqrt{1-x}}>0$$