For every nontrivial connected graph, $\kappa (G)\le \lambda (G)$

Choose an edge cut $F$ of size $\lambda (G)$. Let $A$ be the vertex set of one component of $G-F$ and put $B=V\setminus A$. Then $A$ and $B$ are nonempty, and every edge of $G$ between $A$ and $B$ belongs to $F$.

If $\mathrm{\mid }F\mathrm{\mid }\ge n-1$, deleting all but one vertex gives a vertex cut of size $n-1\le \mathrm{\mid }F\mathrm{\mid }$, so $\kappa (G)\le \lambda (G)$.

Suppose $\mathrm{\mid }F\mathrm{\mid }<n-1$. The bipartite set of all pairs with one endpoint in $A$ and one in $B$ has $\mathrm{\mid }A\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }\ge \mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }-1=n-1$ members. Hence not every such pair is an edge of $G$, for all cross edges lie in $F$. Choose nonadjacent $a\in A$ and $b\in B$.

For each edge $e\in F$, choose one endpoint $s(e)$ different from both $a$ and $b$; this is possible because $e\mathrm{\ne }\{a,b\}$. Put $S=\{s(e):e\in F\}$. Then $a,b\mathrm{\notin }S$, every edge of $F$ meets $S$, and $\mathrm{\mid }S\mathrm{\mid }\le \mathrm{\mid }F\mathrm{\mid }$.

Any path from $a\in A$ to $b\in B$ must use an edge between $A$ and $B$, hence an edge of $F$ by step 1.1, and therefore a vertex of $S$ by step 3.1. Thus $G-S$ has no $a$ to $b$ path and is disconnected, so $S$ is a vertex cut.

In the case $\mathrm{\mid }F\mathrm{\mid }<n-1$, step 4.1 gives $\kappa (G)\le \mathrm{\mid }S\mathrm{\mid }\le \mathrm{\mid }F\mathrm{\mid }=\lambda (G)$; step 2.1 gives the same inequality in the remaining case.