# prove that if $G$ is a graph with $n$ vertices and $\delta(G) \geq (n - 1) / 2$ then $G$ is $\frac{n-1}{2}$-edge-connected

So I know $$G$$ is connected since $$\delta(G)\geq(n-1)/2$$, we can simply prove it by contradiction. My approach to this question is use contradiction. Suppose that $$G$$ is not $$\frac{n - 1}{2}$$- edge - connected, then let $$W$$ be a edge set s.t $$G\setminus W$$ is disconnected. Then there will be at least 2 component and the vertex $$v$$ in a smaller component will have a degree $$\leq n/2 - 1$$ if $$v$$ is not incident to any edge in $$W$$. This will contradict the assumption. However, I don't know how to continue the case when all vertex is incident to $$W$$.

Let $$V’$$ be the set of t the vertices in the smaller component and $$|V|=k\le n/2$$ vertices. Then each vertex $$v\in V’$$ is adjacent only to vertices of $$V’$$, than is degree of $$v$$ in the residual graph is at most $$k-1$$. Therefore at least $$\delta(G)-(k-1)\ge (n+1)/2-k$$ edges incident to $$v$$ belong to $$W$$, that is $$|W|\ge k((n+1)/2-k)\ge (n-1)/2.$$