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.

As , $a^2,b^2,c^2$ & $\frac{1}{(b+c)}, \frac{1}{(c+a)}, \frac{1}{(a+b)}$ are in $AP$ then , $$\frac{2}{(c+a)}=\frac{1}{(b+c)}+\frac{1}{(a+b)}$$ $$\frac{2b}{(c+a)}=\frac{b}{(b+c)}+\frac{b}{(a+b)}$$ $$\frac{2b}{(c+a)}=\frac{ba+\color{green}{b^2+b^2}+bc}{(a+b)(b+c)}$$
Since we have $a^2,b^2,c^2$ are in $AP$ , we have : $$2b^2=a^2+c^2$$
So , finally $$\frac{2b}{(c+a)}=\frac{\color{blue}{a^2+ab}+\color{red}{c^2+cb}}{(a+b)(b+c)}$$ $$\frac{2b}{(c+a)}=\frac{\color{blue}{a(a+b)}+\color{red}{c(b+c)}}{\color{blue}{(a+b)}\color{red}{(b+c)}}$$ $$\color{green}{\frac{2b}{(c+a)}=\frac{a}{(b+c)}+\frac{c}{(a+b)}}$$
We're DONE . Hence its in $AP$ .
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