Let $ G = (V, E) $ be a simple directed graph and let $ s $ and $ t $ be vertices. Prove that if the length of the shortest path between $ s $ and $ t $ is equal to $ L $, then there are no more than $ \left(\frac{|V|}{L}\right)^2 $ edge disjoint paths between $ s $ and $ t $.
UPD: I made significant progress a while ago.
The solution is indeed connected with network flow.
Let's give every edge capacity of $ 1 $. In such network maximum flow between two vertices equals to amount of edge disjoint paths. Let maximum flow from $ s $ to $ t $ be $ \alpha $.
Let $ V_i $ be a set of vertices such that the shortest path from $ s $ to any vertex in $ V_i $ is equal to $ i $.
Let maximum flow between $ V_i $ and $ V_{i+1} $ be $ \alpha_i $.
The amount of non-empty paths between $ V_i $ and $ V_{i+1} $ is no more than $ |V_i| \cdot |V_{i+1}| $. On the other hand $ \alpha_i $ is greater than or equal to maximum flow between $ s $ and $ t $. The capacity of every edge is $ 1 $ so $ \alpha_i $ is no more than the amount of paths between $ V_i $ and $ V_{i+1} $. Therefore $$ \alpha \leq \alpha_i \leq |V_i| \cdot |V_{i+1}| $$
This means that for every $ i \in \mathbb{Z},\ 0 \leq i < L $: $$ |V_i| \geq \sqrt{\alpha} \qquad\text{or}\qquad |V_{i+1}| \geq \sqrt{\alpha} $$ So at least $ L / 2 $ sets have cardinality of no less than $ \sqrt{\alpha} $.
For all $ i,j $ such that $ i \neq j \quad V_i \cap V_j = \emptyset $ therefore $ \sum_{i=0}^{L-1} |V_i| \leq |V| $.
$$ \begin{gather*} \frac{L}{2}\sqrt{\alpha} \leq \sum_{i=0}^{L-1} |V_i| \leq |V| \\ \alpha \leq 4 \left(\frac{|V|}{L}\right)^2 \end{gather*} $$
This comes pretty close but I have no idea how to get rid of the constant.