<s>[INST] If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$ 
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010

My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1-c<1.$$
Thus by AM-GM inequality we have $$\frac{(1-a)+(1-b)+(1-c)}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{(1-a)(1-b)(1-c)}\ge 27.$$
Also by AM-GM inequality we have $$\frac{a+b+c}{3}\ge (abc)^\frac{1}{3}\\ \implies \frac{2}{3}\ge (abc)^\frac{1}{3}\\\implies abc\le \frac{8}{27}\\\implies \frac{1}{abc}\ge \frac{27}{8}.$$
This implies that, we have $$\frac{1}{abc(1-a)(1-b)(1-c)}\ge \frac{27^2}{8}\\ \implies \frac{(abc)^2}{abc(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}\\\implies \frac{abc}{(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}.$$
How to proceed after this?
 [/INST] Your inequality is the same as
$$\sqrt{(1 - a)(1 - b)}\sqrt{(1 - b)(1 - c)}\sqrt{(1 - a)(1 - c)}
\leq {abc \over 8}$$
By the AM-GM inequality you have
$$\sqrt{(1 - a)(1 - b)} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$\sqrt{(1 - b)(1 - c)} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$\sqrt{(1 - a)(1 - c)} \leq {1 - a + 1 - c \over 2} = {b \over 2}$$
Multiplying these together gives the desired inequality. 
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