<s>[INST] Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$ $$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$
I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ where $a_1,a_2,a_3\in \{a,b,c,d,e\}$.
 [/INST] Notice that for all $x,y$ we have $$x^2+xy+y^2 = {1\over 2}(x^2+(x+y)^2+y^2)\geq 0$$
Now we have:
$$3(a-b) = (b-e)\color{red}{\Big(} \underbrace{b+c+d}_x )^2+\underbrace{(b+c+d)(c+d+e)}_{xy}+( 
\underbrace{ c+d+e}_y)^2\color{red}{\Big)} $$
so $${\rm sign} (a-b) = {\rm sign} (b-e) $$ and similary for:
\begin{align} {\rm sign} (b-c) &= {\rm sign} (c-a)\;\;\;\;(2) \\
{\rm sign} (c-d) &= {\rm sign} (d-b)\;\;\;\;(3) \\
{\rm sign} (d-e) &= {\rm sign} (e-c) \;\;\;\;(4)\\
{\rm sign} (e-a) &= {\rm sign} (a-d) \;\;\;\;(5)\\
\end{align}

*

*If $a>b$ then $b>e$ so $a>e$ and from $(5)$ we have $d>a$ so $d>e$ and now from $(4)$ we have $e>c$ and now $d>c$ so from $(2)$ we have $b>d$ so $b>c$ and now from $(2)$ we have $c>a$. So we have $$a>b>d>e>c>a$$ A contradiction.

*If $b>a$ we again get a contradiction, so $a=b=c=d=e=:x$ and all we need is to solve $x=9x^3$...

</s>