Let $M$ be an $n\times n$ matrix such that all of its entries are non-negative and sum to $A$.
Can we prove $\det(M)\leq \frac{A^n}{n^n}$?
I am looking for cool solutions, please go for it.
My boring solution:
We proceed via induction, Let $A_j$ be the matrix obtained by removing the $n$th row and $j$th column.Let $b_1,b_2,\dots b_n$ be the determinants of these matrices, then the determinant of $M$ is $\sum\limits_{j=1}^n(-1)^{n+j}a_ib_i$, if $a_1+a_2+\dots+a_n=s$ and the maximum of the absolute values of $b_1,b_2,\dots b_n$ is $m$ then clearly this sum is less than or equal to $sm$. Notice that the sum of the terms inside the matrix corresponding to $m$ is at most $A-s$, so by the induction hypothesis this product is at most $\frac{(A-s)^{n-1}}{(n-1)^{n-1}}s$. It is easy to show with calculus that this is maximized when $s=\frac{A}{n}$.
Here is an alternative solution:
We have Hadamard's inequality, $| \det M | \le \prod_k \|M e_k\|$.
We see that $\|Me_k\|= \sqrt{\sum_i M_{ik}^2 } $ and we can maximise $f(M)=\prod_k \sum_i M_{ik}^2$ subject to $\sum_i \sum_j M_{ij} = A$ and $M_{ij} \ge 0$. Let $M$ be a maximiser (which exists since the feasible set is compact).
It is clear that by taking diagonal entries with the value ${A \over n}$ that the $\max$ is $>0$. In particular, all columns of the maximising $M$ are non zero.
Note that each column contains exactly one non zero entry. To see this, suppose there are two non zero entries $M_{i_1j}, M_{i_2j}$, then notice that if we let $\phi(t) = (M_{i_1j}+t)^2 + (M_{i_2j}-t)^2 = \phi(0)+ 2t(M_{i_1j} - M_{i_2j}) + 2 t^2$, then there is always some $t$ such that $M_{i_1j}+t \ge 0, M_{i_2j}-t \ge 0$ and $\phi(t) > \phi(0)$, which contradicts the maximality of $M$.
Hence the problem reduces to maximising $g(x) = x_1^2 \cdots x_n^2$ subject to $\sum_k x_k = 0$ and $x_k \ge 0$. Then at a solution $x$ we have $x_1=\cdots =x_n$. To see this, suppose $x_1 >x_2$ and let $\eta(t) = (x_1-t)^2(x_2+t)^2 = (x_1 x_2+ t(x_1-x_2) + t^2)^2$. Then $\eta'(0) = 2 x_1 x_2 (x_1-x_2) >0$ and so $g$ can increased, which contradicts maximality. Hence we obtain $x_k ={A \over n}$.
Hence we have $|\det M| \le ({A \over n})^n$, and by choosing a diagonal matrix with entries ${A \over n}$, we see that we have equality.