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.

Write $(1-x)=a, (1-y)=b \text { and}  (1-z)=c$ 
$x=2-(y+z)=b+c$
$y=2-(z+x)=a+c$
$z=2-(x+y)=a+b$
Thus we have the same expression in simpler form:
$\dfrac{b+c}{a} \cdot \dfrac{a+c}{b} \cdot \dfrac{a+b}{c}$
Now we have AM-GM:
$b+c \ge 2 \sqrt{bc}$
$a+c \ge 2 \sqrt{ac}$
$b+a \ge 2 \sqrt{ba}$
$\dfrac{b+c}{a} \cdot \dfrac{a+c}{b} \cdot \dfrac{a+b}{c} \ge \dfrac{2^3 abc}{abc} =8$, Done.
