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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.

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    $\begingroup$ Instead of adding the three equations, consider $\frac{a}{b+c}=\frac{a(c-b)}{c^2-b^2}=\frac{a(c-b)}{d}$. Similarly for the other two terms. $\endgroup$ – Lozenges Feb 14 '17 at 19:35
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$AP1$: $$a^2\\ b^2\\ c^2$$ $$\text{Common Difference, }\quad d=b^2-a^2=c^2-b^2\qquad \qquad $$

$XP2$: $$P=\frac a{b+c}\color{lightgrey}{\cdot\frac{(c-b)}{(c-b)}}=\frac {a(c-b)}{c^2-b^2}=\frac {a(c-b)}d\\ Q=\frac b{c+a}\color{lightgrey}{\cdot\frac{(c-a)}{(c-a)}}=\frac {b(c-a)}{c^2-a^2}=\frac {b(c-a)}{2d}\\ R=\frac c{a+b}\color{lightgrey}{\cdot\frac{(b-a)}{(b-a)}}=\frac {c(b-a)}{b^2-a^2}=\frac {c(b-a)}d\\$$ 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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$$\frac{2b}{a+c}=\frac{a}{b+c}+\frac{c}{a+b}\Leftrightarrow \\ 2b[b^2+b(a+c)+ac]=a[a^2+a(b+c)+bc]+c[c^2+c(a+b)+ab]\Leftrightarrow\\ 2b^3+2b^2(a+c)=a^3+a^2(b+c)+c^3+c^2(a+b)\Leftrightarrow\\ 2b^2(a+b+c)=a^2(a+b+c)+c^2(a+b+c)\Leftrightarrow \\ (a+b+c)(2b^2-a^2-c^2)=0$$

Can you finish?

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  • $\begingroup$ @Saksham: Is it clear? $\endgroup$ – Arnaldo Feb 15 '17 at 12:53
  • $\begingroup$ Not really. I don't get why you have used the equation which we have to prove as a part of your proof. $\endgroup$ – Saksham Feb 15 '17 at 14:30
  • $\begingroup$ @Saksham: I just did some manipulation using equivalence! You can go backward and get your proof! $\endgroup$ – Arnaldo Feb 15 '17 at 14:32
  • $\begingroup$ Ohh. So have you used what we had to prove to go back to the original question? $\endgroup$ – Saksham Feb 15 '17 at 14:37
  • $\begingroup$ @Saksham: yes, that is a common approach. $\endgroup$ – Arnaldo Feb 15 '17 at 14:53

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