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 $A.P.$

(A) 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 $A.P.$
$\therefore b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)$
$\Rightarrow \frac{b(a+c)}{ac}-\frac{a(b+c)}{bc}=\frac{c(a+b)}{ab}-\frac{b(a+c)}{ac}$
$\Rightarrow \frac{b^2a+b^2c-a^2b-a^2c}{abc}=\frac{c^2a+c^2b-b^2a-b^2c}{abc}$
$\Rightarrow b^2a-a^2b+b^2c-a^2c=c^2a-b^2a+c^2b-b^2c$
$\Rightarrow ab(b-a)+c(b^2-a^2)=a(c^2-b^2)+bc(c-b)$
$\Rightarrow ab(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+bc(c-b)$
$\Rightarrow (b-a)(ab+cb+ca)=(c-b)(ac+ab+bc)$
Assuming $ab+bc+ca \neq 0$,we get $b-a=c-b$.
Thus,$2b=a+c$,which implies $a, b, c$ are in $A.P.$