2025/003) If $\frac{1}{a + b}$, $\frac{1}{b + c}$ and $\frac{1}{c + a}$ are in A.P., prove that $b^2, a^2$ and $c^2$ are in A.P.

We have from the properties of AM

$\frac{1}{b + c} - \frac{1}{a + b} =   \frac{1}{c + a} - \frac{1}{b + c}$

Or $\frac{a-c}{(b+c)(a+b)} = \frac{b-a}{((a+c)(b+c)}$

Or $\frac{a-c}{a+b }= \frac{b-a}{a+c}$

Or $(a-c)(a+c) = (a+b)(b-a)$

Or $a^2-c^2 = b^2-a^2$ 

Or $b^2+c^2= 2a^2$   hence $b^2, a^2$ and $c^2$ are in A.P.