Question

Prove that a2 + b2 + c2 - ab - ac - bc is always non-negative for all the values of a, b and c.

Answer 1

Answer:

a² + b²  + c² - ab - ac - bc is non-negative

Step-by-step explanation:

Prove that a2 + b2 + c2 - ab - ac - bc is always non-negative for all the values of a, b and c.

a² + b²  + c² - ab - ac - bc

Multiply & divide by 2

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

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

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

=(1/2) ( (a -b)² + (a -c)² + (b-c)² )

Square of any number number is non - negative

=> (a -b)² ,  (a -c)² & (b-c)² are non - negatives

sum of non - negatives is non - negative

=> (a -b)² + (a -c)² + (b-c)² is non negative

non - negative divided by any non-negative number is also non- negative

=> (1/2) ( (a -b)² + (a -c)² + (b-c)² ) is non - negative

=> a² + b²  + c² - ab - ac - bc is non-negative