Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $

This is the same as proving that $$(a^2+b^2+c^2)(abc)^{1/3}\le a^3+b^3+c^3$$ for all positive $a$, $b$, $c$. This is AM/GM. We get $$a^{7/3}b^{1/3}c^{1/3}\le\frac{7a^3+b^3+c^3}9$$ applying AM/GM to seven copies of $a^{3}$ and one each of $b^{3}$ and $c^{3}$. Cyclicly permute and add.