Inequalities 3

Applying AM-GM inequality, we have:

$1+a=a+b+c+a\ge2\sqrt{(a+b)(a+c)}=\sqrt{(1-c)(1-b)}$

Similarly, we have:

$1+b\ge2\sqrt{(1-a)(1-c)}$

$1+c\ge2\sqrt{(1-a)(1-b))}$

Thus, $(1+a)(1+b)(1+c)\ge8(1-a)(1-b)(1-c)$

Or $\dfrac{(1+a)(1+b)(1+c)}{(1-a)(1-b)(1-c)}\ge8$ .

The equality holds if and only if $a=b=c=\dfrac{1}{3}$ .

So, the answer is $\boxed{8}$ .

The answer is 8.

let $f(x)=\frac{1+x}{1-x}$ . Then, since $f(x)$ is a convex function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ this implies that $f(a)+f(b)+f(c) \geq 6$ but we want minimum value of $f(a) \times f(b) \times f(c)$ . So, applying A.M.-G.M. inequality on $f(a), f(b), f(c)$ , we have $f(a) \times f(b) \times f(c) \geq \boxed 8$

Applying AM GM inequality,

(1+a)+(1+b)+(1+c)/3)^3>_ (1+a)(1+b)(1+c)

therefore, (4/3)^3>_(1+a)(1+b)(1+c)

similarly, (2/3)^3>_(1-a)(1-b)(1-c)

after solving we get, 2^3>_((1+a)(1+b)(1+c))/((1-a)(1-b)(1-c))