(\dfrac{1}{a+b},\ \dfrac{1}{c+a},\ \dfrac{1}{b+c}) in A.P.

[ 2\cdot\dfrac{1}{c+a}=\dfrac{1}{a+b}+\dfrac{1}{b+c} ] [ \dfrac{2}{a+c}=\dfrac{(b+c)+(a+b)}{(a+b)(b+c)}=\dfrac{a+2b+c}{(a+b)(b+c)} ] [ 2(a+b)(b+c)=(a+c)(a+2b+c) ]

Expand and simplify to obtain (2b^2=a^2+c^2), i.e. (a^2,b^2,c^2) in A.P.

Proved.