How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
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 $$ \| x+ty \| = \left< x+ty,x+ty \right> = \|x\| + 2 \left< x,y \right>t +\|y\|t^2.$$ We know, this being a square, is greater or equal to zero. Therefore, the discriminant of the polynomial in $t$ is less or equal to zero. Which is $\left< x,y \right>^2 - \|x\|\|y\| \le 0.$ Substituting the values for $x$ and $y$ will do the job.