Here is a useful theorem on bipartite graphs.
Theorem. Let $G = (X\cup Y,E)$ be a bipartite graph with $n$ vertices in each side and positive weights on the edges. If for each vertex $v \in X\cup Y$, the sum of weights of edges adjacent to $v$ is exactly $1/n$, then the graph admits a perfect matching.
Proof. For every integer $k \geq 1$, and for every subset of $k$ vertices in $X$, the sum of weights near them is $k/n$, so they must be adjacent to at least $k$ vertices in $Y$. By Hall's marriage theorem, the graph admits a perfect matching.
I am looking for an analogue of this theorem on tripartite hypergraphs. My first attempt was:
Let $H = (X\cup Y \cup Z,E)$ be a tripartite hypergraph with $n$ vertices in each side and positive weights on the hyperedges. If for each vertex $v \in X\cup Y\cup Z$, the sum of weights of edges adjacent to $v$ is exactly $1/n$, then the graph admits a perfect matching.
But this is not correct even for $n=2$. For example, consider the hypergraph with hyperedges $(x_1,y_1,z_1),(x_2,y_2,z_1),(x_1,y_2,z_2),(x_2,y_1,z_2)$ where each hyperedge has a weight of $1/4$. The sum of weights near each edge is $1/2$, but it does not admit a perfect matching.
I conjecture that the analogous theorem requires to make one of the sets larger. For example, suppose $|X|=|Y|=n$ and $|Z|=2n$; the sum of weights is $1/n$ near each vertex in $X$ or $Y$ and $1/2n$ near each vertex in $Z$. In this case, is a perfect matching (a matching of size $n$) guaranteed to exist?
