We have, $a\left(\frac{1}{b}+\frac{1}{c}\right)$ , $b\left(\frac{1}{c}+\frac{1}{a}\right)$ , $c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P. ⇒ $\frac{a(b+c)}{bc}$ , $\frac{b(a+c)}{ca}$ , $\frac{c(a+b)}{ab}$ are in A.P. ⇒ $\frac{a(b+c)}{bc}$ + 1 , $\frac{b(a+c)}{ca}$ + 1 , $\frac{c(a+b)}{ab}$ + 1are in A.P. [adding 1 to each term] ⇒ $\frac{ab+ca+bc}{bc}$ , $\frac{ab+bc+ca}{ca}$ , $\frac{ca+bc+ab}{ab}$ are in A.P. ⇒ $\frac{1}{bc} , \frac{1}{ca} , \frac{1}{ab}$ are in A.P. [Dividing by ab + bc + ca] ⇒ a, b, c are in A.P. [multiplying by abc] Hence proved.