Let $G$ be a connected bigraph with $p$ vertices for $p>1$. Assume that a minimal edge cover $E$ of $G$ contains $n$ edges and a maximal matching $M$ of $G$ contains $m$ edges. Show that $m+n=p$.
I found the first half of this proof on my own: Prove that the sum of minimum edge cover and maximum matching is the vertex count
Couldn't manage the $\geq$ part though. The accepted answer is probably the expected solution.
Let A be the part of the bipartition with more vertices, and let B be the other part, so that there are only edges between A and B (note that possibly |A|=|B|. Since the graph is connected, there must be an edge adjacent to every vertex in B, so n=|B|. Now we want to show that |A|=m (notice that m cannot be bigger than the size of A). Suppose there is some vertex in A that is not covered by a maximal matching. Then there must also be some vertex in B which isn't covered since B is larger than A. Since the graph is connected, there must be a path between these two vertices that alternates between vertices in A and vertices in B. Swapping the edges in the matching with those that are not in the matching creates a bigger matching. So we can eventually get a matching with size m. Thus p=|A|+|B|=m+n.