Question

14. Prove that a² + b2 + c2 - ab – bc – ca is always non-negative for all values of a, b
and c.​

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.

Step-by-step explanation:

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