How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?

Try $(a-b)^2+(b-c)^2+(c-a)^2 \ge0$

Compute lhs, divide by two and rearrange.

This is a specific form of Cauchy-Schwarz inequality.

Let $x = (a, b, c)$ and $y = (b, c, a)$ as vectors.

The inequality is $ | \left< x,y \right>| \le \|x\|\|y\|. $ with standard inner product definition. One neat trick to prove this is using an auxilary parameter $t,$ and expanding $$ \| x+ty \|^2 = \left< x+ty,x+ty \right> = \|x\|^2 + 2 \left< x,y \right>t +\|y\|^2t^2.$$ We know, this being a square, is non-negative. Therefore, the discriminant of the polynomial in $t$ is less or equal to zero. Which is $\left< x,y \right>^2 - (\|x\|\|y\|)^2 \le 0.$ Substituting the values for $x$ and $y$ will do the job.

This is also a consequence of the Rearrangement inequality.