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

[ 2\cdot\dfrac{c+a-b}{b}=\dfrac{b+c-a}{a}+\dfrac{a+b-c}{c} ]

Clearing and simplifying yields (2b(a+c)=ac(\cdots)) which reduces to (b) being the H.M. of (a) and (c), i.e. (a,b,c) in H.P.

Proved.