Question

Prove that a + b ² + c²ab-bc-ca is
always non- negative for all values of a,band c​

Answer 1

Answer:

a2 + b2 +c2 - ab - bc - ca

2/2(a2 + b2 +c2 - ab - bc - ca)

1/2(2a2 + 2b2 +2c2 - 2ab - 2bc - 2ca)

1/2(a2 + b2 - 2ab +b2 +c2 - 2bc + c2 + a2 - ca)

1/2[(a - b)square + (b - c)square + (c - a)square]

As squares of any real numbers can't be negative, therefore is always non negative for all values of a, b, c.

Answer 2

Step-by-step explanation:

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.