Question

14. Prove that a^2+b^2+c^2-ab-bc-ca is always non-negative for all values of a, b
and c
ANSWERS​

Answer 1

To prove:

a²+b²+c²-ab-bc-ca.

Solution:

>> a²+b²+c²-ab-bc-ca

Take ½ as common.

>> ½(2a²+2b²+2c²-2ab-2bc-2ca)

Split 2a²=a²+a², 2b²=b²+b² and 2c²=c²+c²

>> ½(a²+a²+b²+b²+c²+c²-2ab-2bc-2ca)

>> ½[(a²+b²-2ab)+(a²+c²-2ca)+(b²+c²-2bc)]

From identity a²+b²-2ab=(a-b)²

>> ½[(a-b)²+(a-c)²+(b-c)²]

Since, squares of integers are always positive. Their sum will be also in positive and dividing then by 2 will give a answer as positive number.

Hence proved,

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