Question: If $a(\frac1b+\frac1c), b(\frac1a+\frac1c),$ and $c(\frac1b+\frac1a)$ are in Arithmetic Progression, then prove that $a, b, c$ are in Arithmetic Progression.
My attempt:-
Let $a, b, c$ be in Arithmetic Progression.
$$2b = a + c$$
$$(a + c)/b = 2$$
$$(2b - a)/c = 1$$
$$(2b - c)/a = 1$$
$$(a + c)/b = (2b - a)/c + (2b - c)/a$$ $$(a/b) + (c/b) = (2b/c) - (a/c) + (2b/a) - (c/a)$$
$$a((1/b)+(1/c)) + c((1/b)+(1/a)) = 2b((1/a)+(1/c))$$ Therefore, $a(\frac1b+\frac1c), b(\frac1a+\frac1c),$ and $c(\frac1b+\frac1a)$ are in Arithmetic Progression. Which was given, so we can say that $a, b, c$ are in Arithmetic Progression.
I just want to know if my solution is correct or not. If not then please provide correct solution.
Hint:
As $b\left(\dfrac1c+\dfrac1a\right)-a\left(\dfrac1b+\dfrac1c\right)=c\left(\dfrac1a+\dfrac1b\right)-b\left(\dfrac1c+\dfrac1a\right)$
$\iff b\left(\dfrac1c+\dfrac1a\right)+1-\left(a\left(\dfrac1b+\dfrac1c\right)+1\right)=c\left(\dfrac1a+\dfrac1b\right)+1-\left(b\left(\dfrac1c+\dfrac1a\right)+1\right)$
But $b\left(\dfrac1c+\dfrac1a\right)+1=b\underbrace{\left(\dfrac1c+\dfrac1a+\dfrac1b\right)}$ assuming $b,\dfrac1b\ne0$
If $\dfrac1c+\dfrac1a+\dfrac1b\ne0,$ cancelling it from both sides to find $$b-a=c-b$$