Question

0< a,b,c <1

if a+b+c=2

prove that

(abc)/( (1-a)(1-b)(1-c) ) >=8

Answer 1

I'm not sure if this is helpful, but since \(a,b,c\in(0,1)\), you could rewrite
\[\frac{abc}{(1-a)(1-b)(1-c)}\]
as the product of infinite series:
\[\sum_{n=1}^\infty a^n\cdot \sum_{n=1}^\infty b^n \cdot \sum_{n=1}^\infty c^n\]
But like I said, I don't know where this would take you, nor do I see how to use the fact that \(a+b+c=2\).

Answer 2

Now do this,
(1-a) + (1-b) + (1-c) = 1
and 0< (1-a) <1, 0<(1-b)<1 and 0<(1-c)<1

Almost done... just divide and notice that the inequality sign will change...