The sum-of-squares method

Computation.
$$a^2+b^2+c^2-ab-bc-ca=\frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right].$$

Verification. Expand the RHS:
$$\frac12[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)]=a^2+b^2+c^2-ab-bc-ca.$$

Conclusion. The difference is $\ge0$, so the inequality holds. Equality iff $a=b=c$.

Remark. This is the fundamental "SOS" (sum of squares) identity. Many symmetric inequalities can be reduced to such decompositions.