<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] By Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.
$$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}\geq 8$$
where the equality holds for $a=b=c=2/3$. 
Moreover, for $t\geq 3$, by letting $b(t)=c(t)=1-1/t\in (0,1)$ and $a(t)=2/t\in (0,1)$, then $a(t)+b(t)+c(t)=2$ and
$$g(t):=\frac{a(t)}{1-a(t)}\cdot\frac{b(t)}{1-b(t)}\cdot\frac{c(t)}{1-c(t)}= \frac{2(t-1)^2}{t-2}.$$
Hence $g(3)=8$ and $\lim_{t\to +\infty} g(t)=+\infty$.
Finally, by continuity, it follows that the required range is $g([3,+\infty))=[8,+\infty)$.
</s>