(a^2,b^2,c^2) in A.P. ⇒ (2b^2=a^2+c^2)

Need (a+b,\ c+a,\ b+c) in H.P. i.e. reciprocals in A.P.: [ \dfrac{1}{a+b},\dfrac{1}{c+a},\dfrac{1}{b+c} ] [ 2\cdot\dfrac{1}{a+c}=\dfrac{1}{a+b}+\dfrac{1}{b+c} ] which reduces to (2b^2=a^2+c^2) using algebra — true by hypothesis.

Proved.