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 ?
 A: 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.
A: Much of your argument is correct. I was concerned that the way you have written it made it look as if the argument is circular. However @MichaelRosenberg has pointed out a much more important concern. The problem occurs with $\prod_{cyc}a \le \frac{8}{27}$ which is true but the inequality is the wrong way round for the rest of your proof.
For the part of your proof which is fine, it would be simpler to use the AM-GM inequality as you have done but say $$\prod_{cyc}(1-a) \le \left(\frac{\sum_{cyc}(1-a)}{3}\right )^3=\left(\frac{1}{3}\right )^3=\frac{1}{27}.$$
A: 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.
