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.

Also, we can use a Tangent Line method:
$$\sum_{cyc}(a^3-a^2)=\sum_{cyc}(a^3-a^2-\ln{a})\geq0$$ because easy to see that
$$a^3-a^2-\ln{a}\geq0:$$
$$(a^3-a^2-\ln{a})'=3a^2-2a-\frac{1}{a}=$$
$$=\frac{3a^3-3a^2+a^2-a+a-1)}{a}=\frac{(a-1)(3a^2+a+1)}{a},$$
which gives $a_{min}=1$ and we are done! 
