Proof in Arithmetic Progression My maths teacher at school asked a question which I am finding difficult to crack down. 
We are given that $a^2 , b^2$ and $c^2$ are in AP. We need to prove that $\frac{a}{b+c} , \frac{b}{a+c}$ and  $\frac{c}{a+b}$ are in AP. 
This is what I tried. 
Let the common difference of the AP be d .  So, 
$b^2 - a^2 = d   \implies b-a =  \frac{d}{a+b} ........(1)$
Similarly  
$c^2 - b^2 = d   \implies c-b = \frac{d}{b+c}. ........(2)$
Also, 
$a^2 - c^2 = -2d     \implies  a-c =  \frac{-2d}{a+c} .........(3) $
Now adding the three equations,
$$0 = \frac{d}{a+b} + \frac{d}{b+c} - \frac{2d}{a+c}
\implies \frac{2d}{a+c} = \frac{d}{a+b} + \frac{d}{b+c}$$ 
$$\implies \frac{2}{a+c} = \frac{1}{a+b} + \frac{1}{b+c}$$ 
So, $\frac{1}{a+b}, \frac{1}{a+c}$ and $\frac{1}{b+c}$ are in AP. How should I go further? Or if I am going wrong anywhere, please tell. 

$$\frac{2b}{a+c}=\frac{a}{b+c}+\frac{c}{a+b}\Leftrightarrow \\
2b[b^2+b(a+c)+ac]=a[a^2+a(b+c)+bc]+c[c^2+c(a+b)+ab]\Leftrightarrow\\
2b^3+2b^2(a+c)=a^3+a^2(b+c)+c^3+c^2(a+b)\Leftrightarrow\\
2b^2(a+b+c)=a^2(a+b+c)+c^2(a+b+c)\Leftrightarrow \\
(a+b+c)(2b^2-a^2-c^2)=0$$
Can you finish?
