I would like to know the solution of the following 1992 AMM problem:
Let $f$ be a continuous non-negative function defined on the square $[0,1]^2$. Show that $$ \int_0^1\int_0^1\int_0^1\int_0^1f(x_1, y_1)f(x_2, y_1)f(x_1, y_2)\,dx_1\,dx_2\,dy_1\,dy_2 \ge \left(\int_0^1\int_0^1f(x,y)\,dx\,dy\right)^3. $$
There is also the following note:
The integral on the right may be thought of as the integral of $f(x_1, y_1)\cdot f(x_2, y_2)\cdot f(x_3, y_3)$ as all variables range over the interval $[0,1]^3$. The integral on the left would not be changed if one were to integrate if further over $[0,1]$ with respect to $x_3$ and $y_3$ not occuring in the expression. The problem deals with comparing to expressions, each of which is the integral over the same space of a product of three variants of $f(x,y)$. The general question underlying this problem is to determine when the relation of the size of the integrals depends on the variants being integrated and not on the function $f$.
I had an idea of a different approach to the problem. Consideration of the Darboux sums of $f$ provides a discrete form of this inequality. Let $N$ be a positive integer and let $a_{k,m}$, $1\le k, m \le N$ be a two-dimensional sequence of positive numbers. The question is to show that $$ N^2\sum_{k,m,n,l}a_{k,m}a_{n,m}a_{k,l}\ge \left(\sum_{k,m}a_{k,m}\right)^3, $$ where all variables range over the segment $1,\ldots, N$. However, the discrete inequality does not seem to be much easier.
UPD. The discrete inequality can be rewritten in more elegant way. Let $c_j = \sum_i a_{i,j}$ and $r_i = \sum_j a_{i,j}$ then we need to prove $$ N^2\sum_{i,j}a_{i,j}r_i c_j\ge \left(\sum_{i,j}a_{i,j}\right)^3. $$
UPD2. My colleague Leonid Gorbunov suggested the following idea. Let us construct a bipartite graph with columns and rows as vertices. Let us draw an edge between column $i$ and row $j$ if $a_{i,j} > 0$. The claim is that we can get rid of all cycles in this graph decreasing the considered sum. Indeed, if there is a cycle $r_1c_1\ldots r_nc_n$ then we can reduce all elements $a_{r_i,c_i}$ by some $\delta$ and increase all $a_{c_i,r_{i + 1}}$ by the same $\delta$. The change of our sum will be linear in $\delta$ hence we can take $\delta$ so that the sum does not increase and one of $a_{i,j}$ becomes zero. Therefore we can assume that the graph is a tree and there are at most $2n - 1$ nonzero elements in the table. Surprisingly, it is still not clear how to finish the proof.
The complete reference: Juan Bosco Romero Marquez, Ivan Vidav, Edgar A. Ramos, Douglas B. West, Richard Sinkhorn, David M. Bloom, Eric Freden, et al. “Problems: 10202-10210.” The American Mathematical Monthly 99, no. 3 (1992): 265–66. https://doi.org/10.2307/2325065.