let $n=\mid V\mid$
base case: let $n=0$. Thus lemma becomes vacuously true since both bipartitions will contain the empty set thus not a bipartite graph.
Inductive step:
let $k\in\mathbb{N}$, assume $P(k)$ show P(k+1): $\forall G=(V,E), k+1 = \mid V \mid\Rightarrow \mid E\mid\leq\frac{(\mid k+1\mid)^{2}}{4}$, where G is bipartite
I'm having trouble the inductive step, here's my intuition of how I think I should approach. create a subproof to show max possible edges in bipartite graph is when both bipartitions of V has the same size. Then we can assume if a bipartite graph is complete then proposition is satisfied for case 1. I think for Case two, prove for when # of edges are less than the max?
There's no need to consider different cases to do the inductive step.
If for $k+1$ vertices $|E| \geq \frac{(k+1)^2}{4} +1$, remove a vertex with degree $ \leq \lfloor\frac{k + 1}{2}\rfloor$ (why does such a vertex exist?) to get a graph with $k$ vertices and more than $\frac{k^2}{4}$ edges, which contradicts $P(k)$.