Maximum of $\sum_{cyc}x_1x_2x_3\cdots x_k$ given $x_1+x_2+\ldots+x_n=a$ for non-negative $x_1,\ldots,x_n$ for a positive integer $n$ and $1
How do I find the maximum value of
$$\sum_{cyc} \prod_{i=1}^{k} x_{i}$$ for $1< k<n$
given that $x_1,x_2,\ldots,x_n$ are non-negative reals satisfying $$x_1+x_2+\ldots+x_n = a$$ for an integer $n\ge1$?


*

*The case where $k=n$ can be solved by straightforward $AM-GM$ hence why I put $k<n$.


*If $n=3, k=2$. It's well-known that $x^2+y^2+z^2 \ge xy+xz+yz$ so $3(xy+xz+yx)\leq (x+y+z)^2$.


*$k=1$ is trivial hence why I am interested in $1<k$.


*$n=3, k=2,3$ is just AM-GM.
 A: Partial Result (when $k$ divides $n$).  The maximum value in this case is $\left(\dfrac{a}{k}\right)^k$.  However, I think the answer is still $\left(\dfrac{a}{k}\right)^k$ even when $k$ does not divide $n$.
Here we assume that $x_1,x_2,\ldots,x_n\geq 0$ (if the variables are strictly positive, then there are cases where the maximum does not exist).  Suppose that $k$ is a divisor of $n$.  Write $n=mk$ for some positive integer $m$.  Let
$$S:=\sum_{i=1}^n\,x_ix_{i+1}x_{i+2}\cdots x_{i+k-1}\,,$$
where the indices are considered modulo $n$.  Observe that
$$S\leq \prod_{r=1}^{k}\,\left(x_r+x_{r+k}+x_{r+2k}+\ldots+x_{r+(m-1)k}\right)\,.$$
By the AM-GM Inequality,
$$S\leq \left(\frac{1}{k}\,\sum_{r=1}^k\,(x_r+x_{r+k}+x_{r+2k}+\ldots+x_{r+(m-1)k})\right)^k\,.$$
Consequently,
$$S\leq \left(\frac{1}{k}\,\sum_{i=1}^n\,x_i\right)^k=\left(\frac{a}{k}\right)^k\,.$$
The maximum value is attained, for example, when $$x_1=x_2=\ldots=x_k=\dfrac{a}{k}\,,$$ and $$x_{k+1}=x_{k+2}=\ldots=x_n=0\,.$$

Another Partial Result (when $k=2$ and $n\geq 5$ is an odd integer).  The maximum value is also $\left(\dfrac{a}{k}\right)^k=\left(\dfrac{a}{2}\right)^2$.
To show this, suppose without loss of generality that $x_1$ is the smallest value among $x_1,x_2,\ldots,x_n$.  Observe that
$$S\leq S_1S_2\,,$$
where
$$S:=x_1x_2+x_2x_3+\ldots+x_{n-1}x_n+x_nx_1\,,$$
$$S_1:=x_1+x_3+x_5+\ldots+x_{n}\,,$$
and
$$S_2:=x_2+x_4+x_6+\ldots+x_{n-1}\,.$$
This is because all terms $x_1x_2,x_2x_3,\ldots,x_{n-1}x_n$ in $S$ appear in $S_1S_2$, and the term $x_nx_1$ in $S$ is less than or equal to the term $x_nx_2$ in $S_1S_2$.  Therefore,
$$S\leq \left(\frac{S_1+S_2}{2}\right)^2=\left(\frac{a}{2}\right)^2\,.$$
The maximum value is attained, for example, when $$x_1=x_2=\dfrac{a}{2}\text{ and }x_3=x_4=\ldots=x_n=0\,.$$
A: It's the W.Janous's open problem in the general.
The case $n=5$ and $k=3$.
We'll prove that $$abc+bcd+cde+dea+eab\leq\frac{1}{25}(a+b+c+d+e)^3$$ for non-negatives $a$, $b$, $c$, $d$ and $e$.
Indeed, let $e=\min\{a,b,c,d,e\}$.
Thus, by AM-GM we obtain:
$$abc+bcd+cde+dea+eab=e(a+c)(b+d)+bc(a+d-e)\leq$$
$$\leq e\left(\frac{a+c+b+d}{2}\right)^2+\left(\frac{b+c+a+d-e}{3}\right)^3\leq\frac{1}{25}(a+b+c+d+e)^3$$ because for a proof of the last inequality it's enough to assume $a+b+c+d+e=5,$
which gives $$(e-1)^2(e+8)\geq0.$$
A solution for $n=6$ and $k=4$ see here: How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$? and here: How prove this inequality $(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$
Also, you can see here: https://artofproblemsolving.com/community/c6h112980p663097
