Question

prove that a^2+b^2+c^2-ab-bc-ca is always a non negative term, moreover it is =0 if a=b=c​

Answer 1

Answer:

Given,

a2+b2+c2−ab−bc−ca

multiply and divide by 2

=22×(a2+b2+c2−ab−bc−ca)

=2a2−2ab+b2+b2−2bc+c2+c2−2ac+a2

=2(a−b)2+(b−c)2+(c−a)2

square of a number is always greater than or equal to zero.

∴(a−b)2+(b−c)2+(c−a)2≥0

and 

(a−b)2+(b−c)2+(c−a)2=0 when a=b=c

Hence, a2+b2+c2−ab−bc−ca is always non negative