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 $1<k<n$ How do I find the maximum value of
$$\sum_{cyc} \prod_{i=1}^{k} x_{i}$$ for $1&lt; k&lt;n$
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 $k&lt;n$.

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 $1&lt;k$.

$n=3, k=2,3$ is just AM-GM.
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 $$x_1=x_2=\dfrac{a}{2}\text{ and }x_3=x_4=\ldots=x_n=0\,.$$
Whenever you have an expression that is symmetrical over all variables, if you have only one point with extreme value it is achieved when all variables are equal.
Therefore max or min the sum of $k$-cyclic products of $n$ elements is $n(\frac{a}{n})^k$ and if you take $n&gt;1, x_1=x_2=x_{n-1}=0, x_n=a$ you have the sum equal to $0 &lt; n(\frac{a}{n})^k$, that means that $n(\frac{a}{n})^k$ is the maximum value.