Question

3x + 1)
14. Prove that a² + b2 + c2 - ab – bc – ca is always non-negative for all values of a, b
and c.
​

Answer 1

Step-by-step explanation:

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ANSWER

Given,

a

2

+b

2

+c

2

−ab−bc−ca

multiply and divide by 2

=

2

2

×(a

2

+b

2

+c

2

−ab−bc−ca)

=

2

a

2

−2ab+b

2

+b

2

−2bc+c

2

+c

2

−2ac+a

2

=

2

(a−b)

2

+(b−c)

2

+(c−a)

2

square of a number is always greater than or equal to zero.

∴(a−b)

2

+(b−c)

2

+(c−a)

2

≥0

and

(a−b)

2

+(b−c)

2

+(c−a)

2

=0 when a=b=c

Hence, a

2

+b

2

+c

2

−ab−bc−ca is always non negative