There is an inequality between the product of all entries of the matrix and the permanent of a matrix with non-negative entries. The geometric-arithmetic mean inequality states that
$(x_1\dots x_n)^{1/n}\leq\frac{1}{n}(x_1+\dots+x_n)$ whenever $x_1,\dots,x_n$ are non-negative. Therefore, $n(x_1\dots x_n)^{1/n}\leq x_1+\dots+x_n$ whenever $x_1,\dots,x_n$ are non-negative.


Suppose $A$ is a matrix with non-negative entries. Then

$$\text{per}(A)=\sum_{\sigma\in S_{n}}\prod_{k=1}^{n}a_{k,\sigma(k)}\geq n!\cdot(\prod_{\sigma\in S_{n}}\prod_{k=1}^{n}a_{k,\sigma(k)})^{1/n!}=n!\cdot\big((\prod_{i,j}a_{i,j})^{(n-1)!}\big)^{1/n!}$$
$$=n!\cdot(\prod_{i,j}a_{i,j})^{1/n}.$$

Here, equality is reached if and only if every entry in $A$ is the same. While this inequality is easy to prove, the Van der Waerden's conjecture is a result that was proven in 1980 that strengthens this inequality.

If $A$ is doubly stochastic, then by again applying the geometric-arithmetric mean inequality, we obtain
$\prod_{i,j}a_{i,j}\leq n^{-n^2}.$

Van der Waerden's conjecture states that
$$n!\cdot(\prod_{i,j}a_{i,j})^{1/n}\leq\frac{n!}{n^n}\leq \text{per}(A).$$

For stochastic matrices, the product of all entries can be interpreted in terms of Markov chains.

Observation: Suppose that $(X_r)_r$ is an irreducible aperiodic Markov chain with underlying set $\{1,\dots,n\}$ and with transition matrix $A$. Furthermore, suppose that every entry in $B$ is $1/n$.

For almost all tuples $(y_r)_r\in\{1,\dots,r\}^{\omega}$, we have
$$\lim_{N\rightarrow\infty}P(X_0=y_0,\dots,X_N=y_N)^{1/N}=\prod_{i,j}a_{i,j}^{n^{-2}}\leq 1/n.$$