Let $x_1,\cdots,x_{4n}$ be positive integers such that for a fixed $m$
$$x_1+\cdots+x_{4n}=m$$
I want to prove that the maximum value the sums of the form
$$x_1x_2x_3x_4+x_5x_6x_7x_8+\cdots+x_{4n-3}x_{4n-2}x_{4n-1}x_{4n}$$ is reached when $n=1$ and $|x_i-x_j|\leq1$ for all $i,j\in\{1,2,3,4\}$. If one assumes the first condition ($n=1$) I already proved the second condition (the max is reached when the numbers are as close as they can be). But I'm having trouble proving the first part. For starters, I don't even know if it's true, but it sure seems like it
Edit: Just to be clear, what I'm asking is, over all possible $n$, which of those sums has the greatest maximum. In other words
$$\max_{n\geq1}\max_{x_1+\cdots+x_{4n}=m}(x_1x_2x_3x_4+\cdots+x_{4n-3}x_{4n-2}x_{4n-1}x_{4n})$$
Hint: Show that $$ \sum_{i=1}^n x_{4i-3}x_{4i-2}x_{4i-1}x_{4i} \leq (\sum_{i=1}^n x_{4i-3} )(\sum_{i=1}^n x_{4i-2} )(\sum_{i=1}^n x_{4i-1} )(\sum_{i=1}^n x_{4i} )$$
Hence, the value is maximized at $n =1$.