Disjoint paths of length at most L and the number of nodes to remove to vanish this property - Inequality Given an graph $G$ and two, randomly chosen, non-adjacent nodes, let's call them $s$ and $t$,
we make the following two notations:

$P_L(s,t;G)$ = the maximum number of internally disjoint paths (paths that don't have any node in common, except s and t) with a length at most $L$ (L ranges from 1 to |G|).

$K_L(s,t;G)$ = the minimum size of a set of nodes (s and t not included in the sets) with this property - if we remove  the set of nodes, the result is that there are no more paths of length at most $L$, from $s$ to $t$.

I have to prove the inequality
$$ P_L(s,t;G) \le K_L(s,t;G) $$
and specify the values of $L$ (L ranging from 2 to |G|) for which the relation satisfies the equality too, for any chosen graph G and nodes s,t.

The first idea that comes to my mind is to start a discussion based on the values of $L$ and try to prove the relation by induction. Any ideas how?
Hint: Suppose that $P_1,\ldots,P_t$ are internally disjoint $(s,t)$-paths of length at most $L$. Show that you have to remove at least one vertex from each path to make the distance from $s$ to $t$ larger than $L$.

When $L = |G|$, the classical Menger's theorem states that equality holds.