a, A₁, A₂, b are in A.P. ⇒ A₁ – a = b – A₂ ⇒ A₁ + A₂ = a + b           ...(i) 

a, G₁, G₂, b are in G.P. ⇒ \(\frac{G_1}{a}\) = \(\frac{b}{G_2}\) ⇒ G₁ G₂ = ab                       ....(ii)

a, H₁, H₂, b are in H.P.

⇒ \(\frac{1}{a}\), \(\frac{1}{H_1}\), \(\frac{1}{H_2}\),\(\frac{1}{b}\) are in A.P.

⇒ \(\frac{1}{H_1}\)- \(\frac{1}{a}\) = \(\frac{1}{b}\) - \(\frac{1}{H_2}\)

⇒ \(\frac{1}{H_1}\) + \(\frac{1}{H_2}\) = \(\frac{1}{a}\) + \(\frac{1}{b}\)

⇒ \(\frac{H_1+H_2}{H_1H_2}\) = \(\frac{a+b}{ab}\) = \(\frac{A_1+A_2}{G_1G_2}\)              (From (i) and (ii))

⇒ \(\frac{G_1G_2}{H_1H_2}\) = \(\frac{A_1+A_2}{H_1+H_2}\). Hence proved.

Now, \(\frac{1}{a}\), \(\frac{1}{H_1}\), \(\frac{1}{H_2}\) are in A.P        (∵ a, H₁, H₂, are in H.P.)

⇒ \(\frac{1}{H_1}\) - \(\frac{1}{a}\) = \(\frac{1}{H_2}\) - \(\frac{1}{H_1}\)

⇒ \(\frac{2}{H_1}\) - \(\frac{1}{H_2}\) = \(\frac{1}{a}\)                             ......(iii)

Also, \(\frac{1}{H_1}\), \(\frac{1}{H_2}\),\(\frac{1}{b}\) are in A.P       (∵ H₁, H₂, b are in H.P.)

⇒ \(\frac{1}{H_2}\) - \(\frac{1}{H_1}\) = \(\frac{1}{b}\) - \(\frac{1}{H_2}\) ⇒ \(\frac{2}{H_2}\) - \(\frac{1}{H_1}\) = \(\frac{1}{b}\)       .....(iv)

Eq. (iii) + 2 x Eqn. (iv) ⇒\(\bigg(\)\(\frac{2}{H_2}\) - \(\frac{1}{H_1}\)\(\bigg)\) +\(\bigg(\)\(\frac{4}{H_2}\) - \(\frac{2}{H_1}\)\(\bigg)\)= \(\frac{1}{a}\) + \(\frac{2}{b}\)

⇒ \(\frac{3}{H_2}\) = \(\frac{b + 2a}{ab}\) ⇒ H₂ = \(\frac{3ab}{2a+b}\)

Eq. (iv) + 2 x Eq. (iii) ⇒ \(\bigg(\)\(\frac{2}{H_2}\) - \(\frac{1}{H_1}\)\(\bigg)\) +\(\bigg(\)\(\frac{4}{H_2}\) - \(\frac{2}{H_1}\)\(\bigg)\)= \(\frac{1}{b}\) + \(\frac{2}{a}\)

⇒ \(\frac{3}{H_1}\) = \(\frac{a + 2b}{ab}\) ⇒ H₁ = \(\frac{3ab}{a+2b}\)

∴ \(\frac{G_1G_2}{H_1H_2}\) = \(\frac{ab}{\frac{3ab}{(a+2b)}\times\frac{3ab}{(2a+b)}}\) = \(\frac{(2a+b)(a+2b)}{9\,ab}\) Hence proved.