Prove the minimum directed cycle mean cost satisfies: $\lambda = \min_{i = 1,\ldots, n} \max_{0 \le k \le n-1} \left(\frac {p_i(n) - p_i(k)} {n-k}\right)$ using the Bellman-Ford algorithm.
Let $\mathcal G = (\mathcal V, \mathcal A)$ be directed graph with associated edge costs $c_{i,j}$ that has at least one directed cycle ($c_{i,j} = \infty$ for all $(i,j) \notin \mathcal A$).
Define the directed cycle mean cost to be $\frac {\{\text {sum of cost of arcs}\}} { \text {# arcs}}$.
Set $p_i(0) = 0$ and $p_i(t+1) = \min_{j \in O(i)} \{ c_{i,j} + p_j(t) \}$ for all $i\in \mathcal V$.
I've proven by induction that $p_i(t)$ denotes the length of the shortest walk start starts at $i$ and traverses $t$ arcs.
Now, I want to prove that the minimum directed cycle mean cost $\lambda$ satisfies:
$$\lambda = \min_{i = 1,\ldots, n} \max_{0 \le k \le n-1} \left(\frac {p_i(n) - p_i(k)} {n-k}\right)$$, where $n$ is the number of vertices in the graph $\mathcal G$.
I see that traversing $n$ arcs will always create a directed cycle. However, I don't know how to continue from here.