Prove that they form an arithmetic progression

Question

If $\frac{1}{c+b},\frac{1}{c+a},\frac{1}{a+b}$ this is an arithmetic progression, prove that $a^2,b^2,c^2 $ is an arithmetic progresson as well.

Answer 1

Subtract the fractions in pair and express an arithmetic progression:

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

or

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

Answer 2

Hint: $m \geq n \geq p $ are in an arithmetic progression iff $m+p=2n$

Your problem is symmetric with respect to $a,b,c$, hence you can assume that $a\geq b \geq c$

If you apply the hint, what do you get?

Answer 3

Hint: Use that $$\frac{\frac{1}{c+b}+\frac{1}{a+b}}{2}=\frac{1}{c+a}$$ this is equivalent to

$$\frac{a^2+c^2}{2}=b^2$$