Key idea. Add the squares of all three pairwise differences.

1. Expand and add:

   \[\begin{aligned}(a-b)^2+(b-c)^2+(c-a)^2\\=2a^2+2b^2+2c^2-2ab-2bc-2ca.\end{aligned}\]

2. Divide by 2 and rearrange:

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

3. The right side is non-negative, proving the inequality.

4. Equality requires every square to be zero. Thus \(a=b\), \(b=c\), and hence \(a=b=c\). Conversely, equal numbers do give equality.

Remember. Several differences often suggest a sum of several squares.