Question

Prove that a2 + b2 + c2 - ab- bc- ca is always negative for all values of ab and c.​

Answer 1

Answer: Proof given below:

Step-by-step explanation:

a2 + b2 + c 2 - ab - bc - ca

= (1/2)(2a2 + 2b2 + 2c 2 - 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.

HOPE IT HELPS......