If $x,y,z>0$ are distinct and $x+y+z=1$ what is the minimum of $\left((1+x)(1+y)(1+z)\right)/\left((1-x)(1-y)(1-z)\right)$? If $x,y,z>0$ are not equal and positive and if $x+y+z=1$ the expression 
$$\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$$
is greater than what quantity?

we prove
$$\dfrac{1+x}{1-x}\cdot\dfrac{1+y}{1-y}\cdot\dfrac{1+z}{1-z}\ge 8,x+y+z=1$$
$$\Longleftrightarrow \sum \ln{\left(\dfrac{1+x}{1-x}\right)}\ge 3\ln{2}$$
and  we have
$$\ln{\left(\dfrac{1+x}{1-x}\right)}\ge\dfrac{9}{4}x-\dfrac{3}{4}+\ln{2}$$
poof:let
$$G(x)=\ln{\left(\dfrac{1+x}{1-x}\right)}-\left(\dfrac{9}{4}x-\dfrac{3}{4}+\ln{2}\right)$$
$$\Longrightarrow G'(x)=\dfrac{(3x+1)(3x-1)}{4(1-x^2)}$$
since $0<x<1$, then we have
$$G(x)\ge G(\dfrac{1}{3})=0$$
so
$$\sum \ln{\left(\dfrac{1+x}{1-x}\right)}\ge\dfrac{9}{4}(x+y+z)-\dfrac{3}{4}\cdot 3-3\cdot\ln{2}=3\ln{2}$$
by done!
