(a,b,c,d) in H.P. ⇒ (\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}) in A.P.

Let reciprocals be (A,B,C,D) in A.P.

Need (3(a-b)(c-d)=(b-c)(a-d)).

Express in terms of (A,B,C,D) and use equal common differences — the identity follows.

Proved.