If (a − b), (b − c), (c − a) are in G.P., then prove that (a + b + c)² = 3 (ab + bc + ca)

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Solution

$(a - b), (b - c) and (c - a) are in G . P . ∴ {(b - c)}^{2} = (a - b) (c - a) \Rightarrow b^{2} - 2 b c + c^{2} = a c - b c + a b - a^{2} \Rightarrow a^{2} + b^{2} + c^{2} = a b + b c + c a . . . . . . . (i) Now, LHS = {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = a b + b c + c a + 2 a b + 2 b c + 2 c a [Using (i)] = 3 a b + 3 b c + 3 c a = 3 (a b + b c + c a) = RHS$

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