Question:
If $a b c=1$, show that $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$
Solution:
Consider the left hand side:
$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$
$=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$
$=\frac{1}{\frac{b+a b+1}{b}}+\frac{1}{1+b+a b}+\frac{1}{1+\frac{1}{a b}+\frac{1}{a}} \quad(a b c=1)$
$=\frac{b}{b+a b+1}+\frac{1}{1+b+a b}+\frac{a b}{a b+1+b}$
$=\frac{b+1+a b}{b+a b+1}$
$=1$
Hence proved.