If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.,$ prove that $a, b, c$ are in $A.P.$

It is given that $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.$

$\therefore b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)$

$\Rightarrow \frac{b(a+c)}{a c}-\frac{a(b+c)}{b c}=\frac{c(a+b)}{a b}-\frac{b(a+c)}{a c}$

$\Rightarrow \frac{b^{2} a+b^{2} c-a^{2} b-a^{2} c}{a b c}=\frac{c^{2} a+c^{2} b-b^{2} a-b^{2} c}{a b c}$

$\Rightarrow b^{2} a-a^{2} b+b^{2} c-a^{2} c=c^{2} a-b^{2} a+c^{2} b-b^{2} c$

$\Rightarrow a b(b-a)+c\left(b^{2}-a^{2}\right)=a\left(c^{2}-b^{2}\right)+b c(c-b)$

$\Rightarrow a b(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+b c(c-b)$

$\Rightarrow(b-a)(a b+c b+c a)=(c-b)(a c+a b+b c)$

$\Rightarrow b-a=c-b$

Thus, $a, b$ and $c$ are in $A.P.$