Claim: Let $G=A\cup B$ be a balanced bipartite graph with $e(A,B) \geq n/2 $ then $G$ has a path of length $n/2$.
I know about the erdos-gallai theorem that would net a path of length $n/4$. By noting that $d(G)=2E(G)/V(G)\geq n^{2}/4n$.
I suspect that the condition of being bipartite forces the ecistence of a longer path, and I am yet unaware of such a result or a counterexample.
Part of my intuition is from the fact that considering a disjoint union of 2 copies of $K_{n/4-1,n/4}$ which are edge-maximal biparite graphs not containing such a path, we then have these two subgraphs and two yet to be connected vertices on $A$, also:
$$ 2 e(K_{n/4-1,n/4}) = \frac{n^{2}}{8} - \frac{n}{2} < n^2/8 $$
And adding any edge would form a path of the desired length. Any help on how to go about proving this, or a reference for such a resukt would be greatly appreciated.
I think i have solved it, by noting the main result from this paper, which states:
Theorem: Let $c>1$, let $G=(A,B,E)$ be a bipartite graph with parts $|A|=a$, $|B|=b$, which does not contain an even path of length $2l$, with $l>c$, then $$ e(G)\leq \begin{cases} ab, \text{ if }c\geq a \\ bc,\text{ if }2c>a>c\\ (a+b-2c) , \text{ if }a \geq 2c \end{cases} $$
By replacing $a=b=n/2$, $c=n/4 -1$, we get that if $G$ doesn't contain a path of length $n/2$, then $e(G)\leq \frac{n}{2}(\frac{n}{4}-1) = \frac{n^{2}}{8} - \frac{n}{2}$ < $\frac{1}{2}\cdot \frac{n}{2}\cdot\frac{n}{2}$