<s>[INST] For real $a$, $b$, $c$ in $(0,1)$, if $a+b+c=2$, then $\frac{a}{1-a}\times\frac{b}{1-b}\times\frac{c}{1-c}\geq 8$ 
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$ 
  Prove that
  $$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$

I tried by substituting $x$ for$(1-a)$ and similarly for others and used AM-GM inequality, but couldn't get to the desired result.
 [/INST] $$a+b-c=2-2c>0,$$ which says that $a$, $b$ and $c$ are sides-lengths of a triangle.
Now, let $a=y+z$, $b=x+z$ and $c=x+y.$
Thus, $x$, $y$ and $z$ are positives, $x+y+z=1$ and by AM-GM 
$$\prod_{cyc}\frac{a}{1-a}=\prod_{cyc}\frac{y+z}{x}\geq\prod_{cyc}\frac{2\sqrt{yz}}{x}=8.$$
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