Answer with explanation:
It is given that, abc=1
[tex]\rightarrow \frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}\\\\\rightarrow \frac{b}{b+ab+1}+\frac{c}{c+bc+1}+\frac{a}{a+ac+1}\\\\abc=1\\\\\rightarrow \frac{b}{b+ab+abc}+\frac{c}{c+bc+abc}+\frac{a}{a+ac+abc}\\\\\rightarrow \frac{1}{1+a+ac}+\frac{1}{1+b+ab}+\frac{1}{1+c+bc}\\\\\rightarrow \frac{1}{abc+a+ac}+\frac{1}{1+b+ab}+\frac{1}{1+c+bc}\\\\\rightarrow \frac{1+a}{a(bc+1+c)}+\frac{c}{c+bc+1}\\\\\rightarrow\frac{1+a+ac}{a(bc+1+c)}\\\\\rightarrow\frac{1+a+ac}{abc+a+ac)}\\\\\rightarrow\frac{1+a+ac}{1+a+ac)}\\\\=1[/tex]
Hence proved.
