Can't understand the relation between expansion and path length I've been reading some graph theory lately and I came across the following:

In a graph with $n$ nodes, with expansion $\alpha$, for all pairs of nodes there is a path of length $O(\log n/\alpha)$, connecting them.

Could someone please explain how the above statement is derived, and possibly also provide an intuitive explanation for the same? Thank you!
P.S. Expansion $\alpha$ is defined as $\alpha  = \min_{S\subset V} \frac{\text{# of edges leaving S}}{\min(|S|,|V\backslash S|)}$
 A: The basic idea is to show by induction that $B_r(x)$, that is the set of vertices at distance at most $r$ from $x$, grows reasonably fast until it covers more than half of the vertices of $G$. Since this is also true of $y$, there is some suitably small $r$ for which $B_r(x)$ and $B_r(y)$ both cover more than half the vertices, so intersect at some point $z$, giving a path of length at most $2r$ from $x$ to $y$ via $z$.
This is easy for vertex expansion: you can show by induction that if $B_{r}(x)$ is at most half the vertices then it has at least $\alpha |B_r(x)|$ neighbours outside, so $|B_r(x)|\geq\min(n/2+1,(1+\alpha)^r)$.
However, what you have defined is edge expansion. The two notions are essentially equivalent for regular graphs, so perhaps there is an assumption in your quote that the graph is regular. I don't think the result holds in general for edge-expander graphs with large degree discrepancy.
See wikipedia for more details about the relationship between different types of expansion.
