Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I proceed to the results? Please help me to proceed. Thanks in advance.

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.
