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.
We need to prove that $$\sum_{cyc}\left(a^3-a^{\frac{7}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}\right)\geq0$$ and since $$(3,0,0)\succ\left(\frac{7}{3},\frac{1}{3},\frac{1}{3}\right),$$ it's true by Muirhead.
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!