If $a+b+c=2$ where $0 \leq {a,b,c} \leq 1$ find the range of $\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}$

Question

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.

Answer 1

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: $$[8,+\infty).$$

Answer 2

First of all, you need $a,b,c < 1$, equal to $1$ will cause problems.

edit: And also for the lower bound of $0$ you will run into trouble, so for a lower bound, see the answer by Robert Z.

As the last step, let $a$ get really close to $1$ and let $b,c$ get really close to $1/2$. What will happen to your product? Can you guess an upper bound?

Once you have an idea what the range will be, you can prove it, for example, by specifying the "get really close to ..." using a proper $\epsilon$ based on the number you want to get in the product.

Answer 3

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 $g([3,+\infty))=[8,+\infty)$.