If $G$ is a graph of order $n$ such that $\delta (G) ≥ (n-1)/2$ , then $\lambda(G) = \delta(G)$ 
Prove that if G is a graph of order n such that δ(G) ≥ (n-1)/2 , then λ(G) = δ(G).
where
δ(G)= minimum degree of the graph G
λ(G)= minimum edge cuts to disconnect graph G
κ(G)= minimum vertex cuts to disconnect graph G

I know by a theorem that $$κ(G)≤λ(G)≤δ(G)$$
I don't understand why λ(G) = δ(G) or how to get started.
Thank you very much for any assistance.
 A: Consider a minimal set of edges required to disconnect the graph.  This splits the graph into connected components; let $S$ be the smallest such component.  Defining $k = |S|$, note that we have $k \le \frac{n}{2}$.
Moreover, to have $S$ being a connected component, we must have removed all edges between $S$ and $V \setminus S$.  How many such edges must there be?
Well, each vertex in $S$ is incident to at least $\delta(G)$ edges.  At most $k-1$ of those edges can have their other endpoint in $S$, so at least $\delta(G) - (k-1)$ edges go to $V \setminus S$.  Summing over all $k$ vertices in $S$, we find
$e(S, V \setminus S) \ge k(\delta(G) - (k-1))$.
Hence $\lambda(G) \ge \min_{1 \le k \le \frac{n}{2}} k ( \delta - (k-1))$.
This is a quadratic in $k$, and so is minimised at the endpoints of the interval.  Since $k-1 \le \frac{n}{2} - 1 < \delta$, each factor is always a positive integer.  The minimum is thus attained when one of the factors (say $k$) is $1$, giving $\lambda(G) \ge \delta$.
As you mentioned, we also have $\lambda(G) \le \delta(G)$, since we can disconnect the graph by removing the edges incident to a minimum-degree vertex.
Thus $\lambda(G) = \delta(G)$.
