Question

If abc = 1, show that $\frac{1}{1+a+b^{-1}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+c^{-1}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+c+a^{-1}}}=1$.

Answer 1

Consider the left hand side:
$$\begin{gathered} \frac{1}{1+a+b^{-1}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+c^{-1}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+c+a^{-1}}} \\ =\frac{1}{1+a+{\displaystyle \frac{1}{b}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+\frac{1}{c}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+c+\frac{1}{a}}} \\ =\frac{1}{{\displaystyle \frac{b+ab+1}{b}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+ab}}+\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{{\displaystyle 1}}{ab}+\frac{1}{a}}}\left(abc=1\right) \\ =\frac{b}{{\displaystyle b+ab+1}}+\frac{{\displaystyle 1}}{{\displaystyle 1+b+ab}}+\frac{{\displaystyle ab}}{{\displaystyle ab+1+b}} \\ =\frac{b+1+ab}{{\displaystyle b+ab+1}} \\ =1 \end{gathered}$$
Hence proved.