For real $a$, $b$, $c$ in $(0,1)$, if $a+b+c=2$, then $\frac{a}{1-a}\times\frac{b}{1-b}\times\frac{c}{1-c}\geq 8$ 
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$ 
  Prove that
  $$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$

I tried by substituting $x$ for$(1-a)$ and similarly for others and used AM-GM inequality, but couldn't get to the desired result.
 A: For the edited question

Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$ 
  Prove that
  $$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$

The inequality is equivalent with
$$abc \geq 8(1-a)(1-b)(1-c)$$
or 
$$ abc \geq 8\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right)$$
or
$$abc \geq (a+b-c)(b+c-a)(c+a-b)$$
Notice that each bracket on the right hand side is positive, so using AM-GM we have
$$\sqrt{(a+b-c)(b+c-a)} \leq \frac{1}{2}[(a+b-c)+(b+c-a)]=b$$
Now multiply with the other two similar inequalities to arrive at
$$(a+b-c)(b+c-a)(c+a-b) \leq abc$$
Equality occurs when $a=b=c=\dfrac{2}{3}$.
This part of the answer is for initial problem which was

Let$ a,b,c$ be three real numbers such that $0<a,b,c<1$and $$a + b + c= 2$$ 
Prove that:
$$\frac a{1-a} + \frac b{1-b} + \frac c {1-c}≥8$$
I tried by substituting $x$ for$(1-a)$ and similarly for others and
  used AM-GM inequality,but couldn't get to the desired reasult.

The correct lower bound is $6$. From Cauchy-Schwarz:
$$(a+b+c)^2 \leq 3(a^2+b^2+c^2)\Rightarrow a^2+b^2+c^2 \geq \frac{4}{3}$$
So, using Cauchy-Schwarz again:
$$LHS = \sum \frac{a^2}{a-a^2} \geq \frac{(a+b+c)^2}{a+b+c-(a^2+b^2+c^2)} \geq \frac{4}{2-\frac{4}{3}} = 6$$
Equality occurs when $a=b=c=\dfrac{2}{3}$.
A: $$a+b-c=2-2c>0,$$ which says that $a$, $b$ and $c$ are sides-lengths of a triangle.
Now, let $a=y+z$, $b=x+z$ and $c=x+y.$
Thus, $x$, $y$ and $z$ are positives, $x+y+z=1$ and by AM-GM 
$$\prod_{cyc}\frac{a}{1-a}=\prod_{cyc}\frac{y+z}{x}\geq\prod_{cyc}\frac{2\sqrt{yz}}{x}=8.$$
