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

a^{2} + b^{2} + c^{2} - ab - bc - ca

= (1/2)(2a^{2} + 2b^{2} + 2c^{2} - 2ab - 2bc - 2ca)

= (1/2)(a^{2} - 2ab + b^{2} + b^{2}- 2bc + c^{2} + c^{2} - 2ca + a^{2})

= (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.