<s>[INST] 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 
$$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?
 [/INST] Although the function $f(x)=\sqrt{x}/(1-x)$ is not convex on (0,1), its tangent at $x=1/3$ lower bounds the
function and passes through the origin. 
That is, for $0\leq x\leq 1$, we have 
$${\sqrt{x}\over 1-x}\geq {3\sqrt{3}\over2}\, x.$$
Plugging in $a,b,c$ and adding gives
$${\sqrt{a}\over 1-a}+{\sqrt{b}\over 1-b}+{\sqrt{c}\over 1-c}\geq {3\sqrt{3}\over2}.$$ 
Added reference: Exercise 8.1 on page 131 of The Cauchy-Schwarz Master Class by J. Michael Steele asks 
you to prove that for $p\geq 1$, and positive $a,b,c$,
$${a^p\over b+c}+{b^p\over a+c}+{c^p\over a+b}\geq {1\over 2}\,3^{2-p}\,(a+b+c)^{p-1}.\tag1$$ 
He notes that for $p=1$ this reduces to  Nesbitt's inequality.
The inequality (1) fails for $0<p<p_c$, where $p_c={3\log(2)-2\log(3)\over \log(2)-\log(3)}=.29048$
 by looking at $a=b=1/2$ and $c$ close to zero.  But it holds again for $p=0$ by Jensen's inequality .
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