Prove that a2+b2+c2-ab-bc-ca is always non-negative for all valuesof a, b and c - Maths - Polynomials

Question

Prove that
a2+b2+c2-ab-bc-ca
is always non-negative for all valuesof a, b and c

Suresh Anand · Accepted Answer

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

Prove that a2+b2+c2-ab-bc-ca is always non-negative for all valuesof a, b and c.

Prove that a²+b²+c²-ab-bc-ca is always non-negative for all valuesof a, b and c.