Prove that a2+b2+c2-ab-bc-ca is always non-negative for all valuesof a, b and c.
a2 + b2 + c2 - ab - bc - ca
= (1/2)(2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca)
= (1/2)(a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2)
= (1/2)[(a-b)2 + (b-c)2 + (c-a)2]
Now we know that, the square of a number is always greater than or equal to zero.
Hence, (1/2)[(a-b)2 + (b-c)2 + (c-a)2] ≥ 0 => it is always non-negative.