<s>[INST] 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. 
 [/INST] $AP1$: $$a^2\\ b^2\\ c^2$$
 $$\text{Common Difference, }\quad d=b^2-a^2=c^2-b^2\qquad \qquad $$
$XP2$:
$$P=\frac a{b+c}\color{lightgrey}{\cdot\frac{(c-b)}{(c-b)}}=\frac {a(c-b)}{c^2-b^2}=\frac {a(c-b)}d\\
Q=\frac b{c+a}\color{lightgrey}{\cdot\frac{(c-a)}{(c-a)}}=\frac {b(c-a)}{c^2-a^2}=\frac {b(c-a)}{2d}\\
R=\frac c{a+b}\color{lightgrey}{\cdot\frac{(b-a)}{(b-a)}}=\frac {c(b-a)}{b^2-a^2}=\frac {c(b-a)}d\\$$
If $XP2$ is an AP, then $P+R=2Q$.   
Testing both sides:
$$\text{LHS}=P+R=\frac {a(c-b)+c(b-a)}d=\frac {bc-ba}d=2\cdot\frac {b(c-a)}{2d}=2Q=\text{RHS}$$
Hence $XP2$ is also an $AP$, if $AP1$ is an $AP$. 

NB - I had worked this out independently but credit to the hint given by @Lozenges in the comments. 
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