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

Given as (a – b), (b – c), (c – a) are in G.P



(b – c)2 = (a – b)(c – a)


As we have to prove :(a + b + c)2 = 3(ab + bc + ca) so we proceed as follows:


b2 + c2 – 2bc = ac – a2 – bc + ab


a2 + b2 + c2 = ac + ab + bc


Add 2(ac + ab + bc) to both sides:


a2 + b2 + c2 + 2(ac + ab + bc) = ac + ab + bc + 2(ac + ab + bc)


(a + b + c)2 = 3(ab + bc + ca)


Hence Proved.


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