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 ?

As shown by Michael Rozenberg, the proof given is not correct. Here is another AGM approach, using
$$
\sum_{k=1}^n\frac1{p_k}=1\implies\sum_{k=1}^nx_k\ge\prod_{k=1}^n(p_kx_k)^{1/p_k}\quad\text{where}\quad x_k,p_k\gt0\tag1
$$
Suppose $x+y+z=1$, then $(1)$ says that
$$
\begin{align}
&(1-x)(1-y)(1-z)\\
&=1-(x+y+z)+(xy+yz+zx)-xyz\\
&=(x+y+z)(xy+yz+zx)-xyz\\
&=x^2y+xy^2+2xyz+y^2z+yz^2+zx^2+z^2x\\
&\ge\left(8x^2y\right)^{1/8}\left(8xy^2\right)^{1/8}\left(8xyz\right)^{1/4}\left(8y^2z\right)^{1/8}\left(8yz^2\right)^{1/8}\left(8zx^2\right)^{1/8}\left(8z^2x\right)^{1/8}\\
&=8xyz\tag2
\end{align}
$$
Let $a=1-x$, $b=1-y$, $c=1-z$, then we get that if $a+b+c=2$, then $(2)$ is equivalent to
$$
abc\ge8(1-a)(1-b)(1-c)\tag3
$$
Thus, if $a+b+c=2$, then
$$
\frac{a}{1-a}\frac{b}{1-b}\frac{c}{1-c}\ge8\tag4
$$
and $a=b=c=\frac23$ shows that $(4)$ is sharp.
