<s>[INST] Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ . Problem. Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove:
$$a^{3}b+ b^{3}c+ c^{3}a\leqq 8$$
My solution in M&Y : (and I'm looking forward to seeing a nicer one(s), thanks for your interests !)
Because of the invariant of this circular permutation in $a, b, c$, suppose that $a\equiv \max\{\!a,\,b,\,c\!\}> 1$ :
$$\therefore\,ab+ bc+ ca= \frac{(a+ b+ c)^{2}- (a^{2}+ b^{2}+ c^{2})}{2}= \frac{3^{2}- 5}{2}= 2$$
Again, we have $5- a^{2}= b^{2}+ c^{2}\leqq (b+ c)^{2}= (3- a)^{2}$, therefore $a\geqq 2$ or $a\leqq 1\,(\!impossible\,\,!\!)$ .
$$\begin{align}
\therefore\,b,\,c\leqq 1\,\therefore\,a^{3}b+ b^{3}c+ c^{3}a & \leqq a^{3}b+ bc+ ca= ab(a^{2}- 1)+ 2\\ 
 & \leqq \frac{1}{4}(a^{2}+ 4\,b^{2})(a^{2}- 1)+ 2\\ 
 & \leqq \frac{1}{4}(5+ 3\,b^{2})(4- b^{2})+ 2\\ 
 & = 8- \frac{1}{4}(1- b^{2})(4- 3\,b^{2})\leqq 8
\end{align}$$
The equality condition $\{\!a= 2,\,b= 1,\,c= 0\!\}\bigcup\{\!a= 1,\,b= 0,\,c= 2\!\}\bigcup\{\!a= 0,\,b= 2,\,c= 1\!\}$/q.e.d
 [/INST] There is another solution by Rearrangement and AM-GM.
Indeed, let again $\{a,b,c\}=\{x,y,z\},$ where $x\geq y\geq z$.
Thus, $x+y+z=3,$ $xy+xz+yz=2$ and  $$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x^2+2xz+z^2)=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+2\cdot\frac{x+z}{2}}{3}\right)^3=4,$$
which gives $$a^2b+b^2c+c^2a\leq4-abc.$$
Id est,
$$\sum_{cyc}a^3b=(a+b+c)(a^2b+b^2c+c^2a)-\sum_{cyc}(a^2b^2+a^2bc)=$$
$$=3(a^2b+b^2c+c^2a)-\sum_{cyc}(a^2b^2+2a^2bc)+\sum_{cyc}a^2bc=$$
$$=3(a^2b+b^2c+c^2a)-4+3abc\leq3(4-abc)-4+3abc=8.$$
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