If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$ Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$.
I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.

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!
