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.

Proof:$$a^2+b^2\geq 2ab$$so:$$3(a^2+b^2+c^2)=a^2+b^2+c^2+2(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)=(a+b+c)^2=36$$Done!
