Inequality Solution Correctness Let $a,b,c \in \Re $ and $ 0 < a < 1 ,  0 < b < 1 ,  0 < c < 1 $ & $ \sum_{cyc} a = 2$
Prove that 
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
My solution 
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
or 
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$ 
From $ \sum_{cyc} a = 2$ we can conclude $\prod_{cyc}a \le \frac{8}{27} $ , thus getting a maximum bound on $\prod_{cyc}a$ .
Going back to 
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$ 
We can say $$\prod_{cyc}(1-a) \le \left(\frac{\sum_{cyc}(1-a)}{3}\right )^3=\left(\frac{1}{3}\right )^3=\frac{1}{27}.$$
Which is true from $AM-GM$ inequality . 
Is this solution correct ?

You need to prove that:
$$\prod_{cyc}(1-a)\leq\frac{1}{8}abc.$$
You proved that $$\prod_{cyc}(1-a)\leq\frac{1}{27}.$$
Thus, it's enough to prove that $$\frac{1}{27}\leq\frac{1}{8}abc,$$ which is wrong, which says that your solution is wrong.
Since $$a+b-c=2-2c>0,$$ we see that $a$, $b$ and $c$ are sides-lengths of a triangle.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x+y+z=1$ and $x$, $y$ and $z$ are positives and after homogenization we need to prove that:
$$(x+y)(x+z)(y+z)\geq8xyz,$$ which is true by AM-GM.
