Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$. 
If $x,y,z$ are positive proper fractions satisfying $x+y+z=2$, prove that  $$\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\ge 8$$

Applying $GM \ge HM$, I get $$\left[\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\right]^{1/3}\ge \dfrac{3}{\frac 1x-1+\frac 1y-1+\frac 1z-1}\\=\dfrac{3}{\frac 1x+\frac 1y+\frac 1z-3}$$
Then how to proceed. Please help.

We begin by setting $a=1-x, b=1-y, c=1-z$, and noting that $a+b+c=1$.  The conditions of the problem imply that $a,b,c\in (0,1)$.
We now need Maclaurin's inequality: $$\frac{a+b+c}{3}\ge \sqrt{\frac{ab+bc+ac}{3}}\ge \sqrt[3]{abc}$$
Ignoring the middle part temporarily, we have $\frac{1}{3}\ge \sqrt[3]{abc}$ or $3\sqrt[3]{abc}\le 1$.  Ignoring the first part and squaring, we have $\frac{ab+bc+ac}{3}\ge (\sqrt[3]{abc})^2\ge 3abc$, where we multiplied by $3\sqrt[3]{abc}\le 1$ in the last step.  We rearrange this to get $$ab+bc+ac\ge 9abc$$
then rearrange that to get $$1-a-b-c+ab+bc+ac-abc\ge 8abc$$
The left side factors as $(1-a)(1-b)(1-c)$; divide by $abc$ and we are done.
