<s>[INST] For $a^3+b^3+c^3=3$ prove that $a^4b+b^4c+c^4a\leq3$ 
Let $a$, $b$ and $c$ be non-negative numbers such that $a^3+b^3+c^3=3$. Prove that:
  $$a^4b+b^4c+c^4a\leq3$$

This inequality similar to the following.

Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
  $$a^3b+b^3c+c^3a\leq3,$$
  which follows from the following identity.
  $$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\frac{1}{2}\sum_{cyc}(a^2-b^2-ab-ac+2bc)^2.$$

I tried Rearrangement.
Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, $$a^4b+b^4c+c^4a=a^3\cdot ab+b^3\cdot bc+c^3\cdot ca\leq x^3\cdot xy+y^3\cdot xz+z^3\cdot yz=$$
$$=y(x^4+y^2xz+z^4)$$
and I don't see what is the rest.
Thank you!
 [/INST] The Buffalo Way works. However it is not nice.
Let me describe Michael Rozenberg's solution. We have
\begin{align}
\sum_{\mathrm{cyc}} a^4b &\le \frac{1}{3}\sum_{\mathrm{cyc}} (a^{9/2}b^{3/2} + a^{9/2}b^{3/2} + a^3)\\
&= \frac{2}{3} \sum_{\mathrm{cyc}} a^{9/2}b^{3/2} + \frac{1}{3} (a^3+b^3+c^3)\\
&\le \frac{2}{9} (a^3+b^3+c^3)^2 + \frac{1}{3} (a^3+b^3+c^3)\\
&= 3
\end{align}
where we have used Vasc's inequality $(x^2+y^2+z^2)^2\ge 3(x^3y+y^3z+z^3x)$
to obtain $\sum_{\mathrm{cyc}} a^{9/2}b^{3/2} \le \frac{1}{3}(a^3+b^3+c^3)^2$.
This inequality can be used to prove the following inequality:
Let $x,y,z>0$ and $x^7+y^7+z^7=3$. Prove that
$\frac{x^4}{y^3}+\frac{y^4}{z^3}+\frac{z^4}{x^3}\ge 3$.
Proof: Using AM-GM, we have $7\frac{x^4}{y^3} + 9 x^{28/3} y^{7/3} \ge 16 x^7$
which results in $$7 \sum_{\mathrm{cyc}} \frac{x^4}{y^3} + 9 \sum_{\mathrm{cyc}} x^{28/3} y^{7/3} \ge 16 (x^7+y^7+z^7). $$
It suffices to prove that $\sum_{\mathrm{cyc}} x^{28/3} y^{7/3} \le 3$.
Let $a = x^{7/3}, \ b = y^{7/3}, \ c = z^{7/3}$. The condition becomes $a^3+b^3+c^3 = 3$.
We need to prove that $a^4b+b^4c+c^4a \le 3$. We are done.
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