Question

prove that the polynomials are nonnegative a^2-2ab+b^2+c^2

Answer 1

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.

Answer 2

To prove this equation non-negative, you will have to convert the equation in terms of perfect square form containing a,b and c.

Now,
a²+b²+c²-ab-bc-ca
= ½ • ( 2a²+2b²+2c²-2ab-2bc -2ca )
= ½ • ( a² -2ab +b² +b² -2bc +c² +c² -2ac +a² )
= ½ • { (a-b)² + (b-c)² + (c-a)² }

For any value of a,b,c
(a-b)² ≥ 0,
(b-c)² ≥ 0,
(c-a)² ≥ 0,

So,
a²+b²+c²-ab-bc-ca = ½ • { (a-b)² + (b-c)² + (c-a)² } ≥ 0 i.e. Non-negative [ proved ]