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

Hint: We will first multiply and divide the given expression with 2 and then we will convert it in terms of square. Then we know that squares of numbers are always greater than or equal to zero. Thus using this information we will prove the given question.

Complete step-by-step answer:
The expression mentioned in the question is \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca.......(1)\].

Now dividing and multiplying equation (1) by 2 we get,

\[\Rightarrow \dfrac{2}{2}({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca).......(2)\]

Now multiplying 2 with all the terms in the numerator in equation (2) we get,

\[\Rightarrow \dfrac{(2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca)}{2}.......(3)\]

Now rearranging the terms in the form of squares in equation (3) we get,

\[\Rightarrow
\dfrac{({{a}^{2}}-2ab+{{b}^{2}})+({{b}^{2}}-2bc+{{c}^{2}})+({{c}^{2}}-2ac+{{a}^{2}})}{2}.......(4)\]

Now from equation (4) we can see that we have grouped all the terms in three form of

squares,

\[\Rightarrow \dfrac{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}}{2}.......(5)\]

Now from equation (5) we also know that the square of a number is always greater than or
equal to zero. Hence \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca\] is always non negative. Hence proved.

Note: Remembering the properties of the square of numbers is the key here. Also we need to remember \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] formula or else we might get confused how to proceed after equation (3). Also we in a hurry can make a mistake after equation (3) if we are unable to recognize the terms in the form of squares and hence we need to be very careful while doing this step.