Background: I am trying to use simple induction to prove that any bipartite graph with n vertices has no more than $\frac{n^2}{4}$ edges, and I find that the following question can be helpful here to help me finish the induction and thus complete the proof, but I get stuck.
How to prove that in a bipartite graph, you can always choose a vertex with $\le\lfloor\frac{|V|}{2} \rfloor$ edges incident with it?
In a bipartite graph, the vertices are partitioned into two "parts" such that there are no edges between two vertices in the same part. Thus, each vertex is adjacent to at most all the vertices in the other part. The two parts cannot both contain more than $\lfloor \frac{|V|}{2} \rfloor$ vertices since then the bipartite graph would have $> 2\frac{|V|}{2}=|V|=n$ vertices, which is false. So there must be at least one vertex with at most $\lfloor \frac{|V|}{2} \rfloor$ edges adjacent to it.