Question

If $abc=1$, Prove that: $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

Answer 1

$\frac{1}{1+a+b^{-1}}=\frac{1}{1+a+\frac{1}{b}}=\frac{b}{1+ab+b}$......First term

Now, $abc=1\Rightarrow ab=\frac{1}{c}$

So, First term becomes $=\frac{b}{1+\frac{1}{c}+b}=\frac{bc}{1+c+bc}$

Second term $=\frac{1}{1+b+\frac{1}{c}}=\frac{c}{1+bc+c}$

Third term $=\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+c+bc}[\because abc=1\Rightarrow bc=\frac{1}{a}]$

On adding all terms,

$\frac{bc}{1+c+bc}+\frac{c}{1+bc+c}+\frac{1}{1+c+bc}=\frac{1+c+bc}{1+c+bc}=1$