How to prove a result of expander graphs? The following is a definition of expander graphs. Let $X = X(V, E)$ be a regular graph of degree $k$, where $V$ is the set of vertices, $|V|=n$, $E$ is the set of edges. $X$ is called an $(n, k, c)$-expander if for every subset $A$ of $V$, $$|\partial A| \geq c(1 - \frac{|A|}{n})|A|. \quad (1)$$ Here $\partial A = \\{y \in V \mid d(y, A)=1\\}$ is the boundary of $A$ and $d$ is the distance function on $X$. In the book of Alexander Lubotzky, Discrete Groups, Expanding Graphs and Invariant Measures, Page 2, Proposition 1.1.4, it is said that (i) If $X$ is an $(n, k, c)$-expander, then $h(X) \geq c/2$. Here $$h(X) = \operatorname{inf}_{A, B \subseteq V, A \cup B=V, A \cap B = \emptyset} \frac{|E(A,B)|}{min(|A|, |B|)}.$$ I am trying to prove this. (1) implies that $$ \frac{|\partial A|}{(n-|A|)|A|} \geq c/n. $$ But how could we prove that $h(X) \geq c/2$? Thank you very much.

Let $A \subseteq V$ such that $h(X) = \frac{|E(A, B)|}{min(|A|, |B|)}$, $B=V - A$. We have $ | \partial A | \leq |E(A, B)|$. Since $A \cup B = V, A \cap B = \emptyset$, $max(|A|, |B|) \geq n/2$. Therefore $$ c/2 \leq \frac{n|\partial A|}{2|B| |A|} \leq \frac{n|E(A, B)|}{2|B| |A|} \\\ = \frac{n|E(A, B)|}{2 max(|B|, |A|)min(|B|, |A|)} \leq \frac{|E(A, B)|}{min(|B|, |A|)} = h(X). $$