Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$? Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality 
$$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?

This is as far as I got... 
$\frac{(1-b)\sqrt{a}}{(1-b)(1-a)}$  + $\frac{(1-a)\sqrt{b}}{(1-a)(1-b)}$ + $\frac{\sqrt{c}}{(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$ 
$\frac{(1-b)\sqrt{a} + (1-a)\sqrt{b}}{(1-b)(1-a)}$ + $\frac{\sqrt{c}}{(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$ 
$\frac{(1-c)(1-b)\sqrt{a}+(1-c)(1-a)\sqrt{b}+(1-b)(1-a)\sqrt{c}}{(1-a)(1-b)(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$ 
$\frac{\sqrt{a}(1-b-c+bc)+\sqrt{b}(1-a-c+ac)+\sqrt{c}(1-a-b+ab)}{(c+ab)((1-c)} $ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$ 
$\frac{\sqrt{a}(bc+a)+\sqrt{b}(ac+b)+\sqrt{c}(ab+c)}{(ab+c)(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
Can anyone explain the rest to me? Can you say that (bc+a),(ac+b),(ab+c) $\leq$ 1 ? Also how would you formally say that, I mean I know that two fractions times each other is less than 1, but I don't know how to formally state that. Thanks!
