Let be $G$ a graph of order $n$. Show that if $\delta(G) = \frac{n}{2}$, then $\lambda(G) = \delta(G)$ I was reading the book graph theory by harary, and he prove the upper bound for the edge connectivity, and mentions that the equality holds when $\delta(G) = \frac{n}{2}$, Any ideas how to prove it.

Let the order of the graph be $2m$.
Clearly $\lambda(G)\leq \delta(G)=m$.
We need only prove that removing $m-1$ edges does not disconnect the graph.
To do this it clearly suffices to show that every connected component of the new graph has a vertex of degree at least $m$.
So pick $v$ a vertex in $G$. we will prove that one of the neighbours of $v$ (in the new graph) has degree at least $m$. Let $k$ be the number of edges removed from $v$. Then $v$ has at least $m-k$ neighbours in the new graph. And at most $m-k-1$ more edges can be removed, so one of the neighbours of $v$ in the new graph has degree at least $m$.
