(a,b,c) in G.P. ⇒ (b^2=ac)

Need (\log_a x,\log_b x,\log_c x) in H.P.

i.e. (\dfrac{1}{\log_a x},\dfrac{1}{\log_b x},\dfrac{1}{\log_c x}) in A.P.

But (\dfrac{1}{\log_a x}=\log_x a), so (\log_x a,\log_x b,\log_x c) in A.P.

Since (a,b,c) in G.P., (\log a,\log b,\log c) in A.P. (any base), hence (\log_x a,\log_x b,\log_x c) in A.P.

Proved.