If $(b + c) , (c + a) , (a + b)$ are in H.P. then find the relation between $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ . 
If $(b + c) , (c + a) , (a + b)$ are in H.P. then $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ are in $(i)$ A.P. $(ii)$ G.P. $(iii)$ H.P. $(iv)$ None of These.

What I Tried:- I have that $(b+c),(c+a),(a+b)$ are in H.P. .
$\implies \dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are in A.P. .
$\implies \dfrac{1}{c+a} - \dfrac{1}{b+c} = \dfrac{1}{a+b} - \dfrac{1}{c+a}.$
$\implies \dfrac{(b + c) - (c + a)}{(c + a)(b+c)} = \dfrac{(c+a) - (a+b)}{(a+b)(c+a)}.$
$\implies \dfrac{(b - a)}{(b + c)} = \dfrac{(c - b)}{(b + a)}.$
$\implies (b^2 - a^2) = (c^2 - b^2).$
$\implies 2b^2 = a^2 + c^2.$
So I was able to prove that $a^2 , b^2 , c^2$ are in A.P. , but I am not sure how to show the claim in the question.
Can anyone help me? Thank You.

Hint 1:
Add $1$ to each term as mentioned in the comments by @SathvikAcharya
Hint 2:
If $p,q,r$ is an arithmetic progression, $\lambda p,\lambda q,\lambda r$ is also an arithmetic progression where $\lambda$ is a real number
