Is a graph with minimum degree $\delta$, $\delta$-edge-connected? Let G be a simple $\lambda$-edge-connected graph with $n$ vertices and minimum degree $\delta$. Prove that if $\delta \ge n/2$ then $\delta=\lambda$.
What i thought was to use the Whitney theorem: if for every pair of (distinct) vertices (u,v) there are at least k edge independent paths between u and v then the graph is k-edge connected.  
Let's say in our graph we chose 2 distinct vertices $u,v$. Then it must be $N(u) \cap N(v) \neq \emptyset$
(because if $N(u) \cap N(v) = \emptyset$ then $|N(U) \cup N(v)|=|N(u)|+|N(v)|=n$, but u,v were not accounted for). 
Then there must be exactly 2 nodes that are both connected to u and v.
The only case where this is not true is if $v\in N(u)$ or vice versa. 
From here i was thinking of seeing if i can find the n/2 paths between pairs of vertices based on the rough description of what the graph must look like, but it looks difficult to do so. Am i on the right track or do i need to use another method?
 A: This would be my approach: 
Let $G=(V,E)$. It is obvious that $\lambda \leq \delta$, because removing the edges incident to some vertex $v$ of minimal degree will disconnect $G$. So the question is why not strictly less? The start is pretty straight forward:
Assume $\lambda < \delta$. Edge-connectivity $\lambda$ simply means we have a partition $V=A\,\dot\cup\, B$ with exactly $\lambda$ edges between $A$ and $B$. Lets say $A$ is the smaller (not bigger) partition class, hence $|A|\leq|B|\Rightarrow |A|\leq n/2$. Every $v\in A$ can have at most $|A|-1$ neighbors in $A$, hence must have at least $\delta-|A|+1\geq 1$ neighbors in $|B|$. This makes at least $|A|\cdot(\delta-|A|+1)\geq|A|$ edges from $A$ to $B$ (therefore, note $\lambda\geq |A|$). We will see that these are too many. First, conclude
$$(*)\qquad\lambda \geq |A|\cdot(\delta-|A|+1)>|A|\cdot(\lambda-|A|+1). $$
If $|A|=1$, we have $\delta=\lambda$. Therefore assume $|A|> 1$. Some playing with $(*)$ will give you $\lambda <|A|$. This contradicts $\lambda\geq |A|$. $\square$
