Question

Prove that:

${\displaystyle \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1}$

Answer 1

Consider the left hand side:

${\displaystyle \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}}$

${\displaystyle =\frac{1}{1+\frac{x^b}{x^a}+\frac{x^c}{x^a}}+\frac{1}{1+\frac{x^a}{x^b}+\frac{x^c}{x^b}}+\frac{1}{1+\frac{x^b}{x^c}+\frac{x^a}{x^c}}}$

${\displaystyle =\frac{1}{\frac{x^a+x^b+x^c}{x^a}}+\frac{1}{\frac{x^b+x^a+x^c}{x^b}}+\frac{1}{\frac{x^c+x^b+x^a}{x^c}}}$

${\displaystyle =\frac{x^a}{x^a+x^b+x^c}+\frac{x^b}{x^b+x^a+x^c}+\frac{x^c}{x^c+x^b+x^a}}$

${\displaystyle =\frac{x^a+x^b+x^c}{x^a+x^b+x^c}}$

= 1

Therefore left hand side is equal to the right hand side. Hence proved.