<s>[INST] If $a+b+c=2$ where $0 \leq {a,b,c} \leq 1$ find the range of $\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}$ 
If $a+b+c=2$ where $0 < {a,b,c} <1$ find the range of $$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}.$$

I have tried using AM-GM but I was not able to solve it . I also assumed the expression equal to $k$ and tried to find the range of $k$ but I derived some absurd results. Please help me to solve it and guide me to the correct approach.
 [/INST] Let $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z.$
Thus, $x$, $y$ and $z$ are positives, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y},$ $c=\frac{z}{1+z}$ and the condition gives: 
$$\sum_{cyc}\frac{x}{1+x}=2.$$
Id est, by AM-GM we obtain:
$$\prod_{cyc}\frac{x}{1+x}=\prod_{cyc}\left(2-\frac{y}{1+y}-\frac{z}{1+z}\right)=\prod_{cyc}\left(\frac{1}{1+y}+\frac{1}{1+z}\right)\geq$$
$$\geq\prod_{cyc}\frac{2}{\sqrt{(1+y)(1+z)}}=\frac{8}{\prod\limits_{cyc}(1+x)},$$
which gives $$xyz\geq8$$ or
$$\prod_{cyc}\frac{a}{1-a}\geq8.$$
The equality occurs for $a=b=c=\frac{2}{3},$ which says that $8$ is a minimal value.
Also, since $\prod\limits_{cyc}\frac{a}{1-a}$ is a continuous and for $a\rightarrow1^-$ this expression is closed to $+\infty,$ we got the answer:
$$[8,+\infty).$$
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