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.

Two proofs of the inequality have been posted. The following is simply a comment. In the attempted solution of the OP, the inequality
$$\left(\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot \frac{z}{1-z}\right)^{1/3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}\tag{1}$$
has been proved. 
One can complete things by showing, under the hypothesis $x+y+z=2$, that the right-hand side of (1) is $\ge 2$.
However, that is not true. For instance, take $x=\frac{5}{6}$, $y=\frac{5}{6}$, and $z=\frac{2}{6}$. Then 
$$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}=\frac{18}{10}\lt 2.$$
So too much has been given away in producing (1): there is probably no direct path from it to the desired result. 
