If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$

Use Cauchy–Schwartz: $$a+b+c\ \leqslant\ \sqrt{1^2+1^2+1^2}\sqrt{a^2+b^2+c^2}$$

A geometric proof:

The equation $x+y+z=6$ describes a plane with intercepts 6 on all three axes. The equation $x^2+y^2+z^2=12$ describes a sphere of radius $\sqrt{12}$ centered at the origin $x=y=z=0$. The inequality $x^2+y^2+z^2\ge12$ describes points not in the interior of this sphere.

It is easy to check that no points of the plane $x+y+z=6$ are in the interior of this sphere. (Use distance from point to plane. The plane's closest point to the origin is $x=y=z=2$, this point belongs to the sphere but is not inside the sphere. All other points of the plane are farther away from the origin.)

But that's precisely what we are asked to prove - if a point is in the plane, it's not inside the given sphere.

We need to prove that $$a^2+b^2+c^2\geq12\left(\frac{a+b+c}{6}\right)^2$$ or $$\sum\limits_{cyc}(a-b)^2\geq0.$$ Done!