An algebra problem by Aly Ahmed

Since $f(x) = \dfrac 1{1-\sqrt x}$ is convex in the interval $(0,1]$ , by Jensen's inequality we have:

$\begin{aligned} \frac {f(a)+f(b)}2 & \ge f \left(\frac {a+b}2\right) \\ \frac 1{1-\sqrt a} + \frac 1{1-\sqrt b} & \ge \frac 2{1-\sqrt{\frac{a+b}2}} = \frac 2{1-\sqrt{\frac 12}} = \frac {2\sqrt 2}{\sqrt 2-1} = \boxed{4+2\sqrt 2} \end{aligned}$

$a+b=1\implies \sqrt {ab}\leq \frac{1}{2}$ .

$\dfrac {1}{1-\sqrt a}+\dfrac{1}{1-\sqrt b}=\dfrac{2-\sqrt a-\sqrt b }{1-\sqrt a-\sqrt b+\sqrt {ab}}$

$\geq \dfrac{2-\sqrt a-\sqrt b}{\frac{3}{2}-\sqrt a-\sqrt b}=1+\dfrac{1}{3-2(\sqrt a+\sqrt b) }$

$\sqrt a+\sqrt b\geq 2\sqrt[4] {ab}=\sqrt 2$ .

So, $\dfrac{1}{1-\sqrt a}+\dfrac{1}{1-\sqrt b }\geq 1+\dfrac{1}{3-2\sqrt 2}=1+ 3+2\sqrt 2=4+2\sqrt 2$ .