Question

abc -3abc =(1/2abc)+(2a+2b +2c+2ab +2bc +2ca )
Prove that LHS=RHS

Answer 1

$\huge\red{A}\huge\green{N}\huge\pink{S}\huge\blue{W}\huge\purple{E}\huge\orange{R}$

Consider, a^2 + b^2 + c^2 - ab - bc - ca

= 0

Multiply both sides with 2, we get 2(a^2 + b^2 + c^2 - ab - bc - ca) = 0

⇒ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca

= 0

⇒ (a^2 - 2ab + b^2) + (b^2 − 2bc + c^2)+ (c^2 - 2ca + a^2) = 0

⇒ (a−b)^2 + (b − c)^2 + (c − a)^2 =0 -

Here the sum of the terms is a non-negative term which means all the individuals are also positive.