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.
 A: 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.
A: 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. 
A: 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.
A: Note that $x$, $y$ and $z$ are sides of a triangle. Let $p$ be half the perimiter (i.e., $p=\left(x+y+z\right)/2=1$). Then you'd like to show that $xyz\geq 8(p-x)(p-y)(p-z)$. From geometry we know that $xyz = 4RS$ where $R$ and $S$ denote the radius of the circumcircle the area of the corresponding triangle. Furthermore, $p(p-x)(p-y)(p-z)=S^2$ by Heron's formula. Therefore, we have to show that $$4RS\geq 8S^2/p$$ or equivalently $$R\geq 2S/p.$$ Considering the fact that $S=pr$ where $r$ is the inradius of the triangle, the last inequality is equivalent to $R\geq 2r$ which is a well-known geometric inequality (Euler's inequality).
