Proving $\Delta(G)-\delta(G)\geq2$ for a non-regular graph $G$

Question

I'm trying to prove that $\Delta(G)-\delta(G)\geq2$ holds for a non-regular graph $G$ of order $n$ and size $m=rn/2$ such that $r \in Z$ and $1 \leq r \leq n-2$.

Here's my attempt:

$$ n/2 \leq m \leq n(n-2)/2 $$ so the average degree of a vertex $2m/n$ must be bounded between $$ 1 \leq 2m/n \leq n-2 $$

Now what i can conclude is that $\delta(G) \geq 1$ and $\Delta(G) \leq n-2$.

From here I'm not sure how to show that $\Delta(G)-\delta(G)\geq2$.

Answer 1

Let $G=(V,E)$ be a graph which is not regular, with $n$ vertices, and $m=\frac{nr}{2}$ edges, where $r\in\mathbb N, 1\leq z\leq n-2$.

Prop 1: the average degree $\bar d(G) = r$

By the Handshaking lemma, the average degree $\bar d(G) = 2\frac{nr}{2n}=r$.

Prop 2: there is a vertex $v\in V$ such that $deg(v)\neq r$.

This is true because the graph is not regular.

Case 1: $deg(v)> r$
there must be vertex $u\in V$ s.t $deg(u)<r$ otherwise, the average degree is bigger than $r$.
We get $\Delta(G)-\delta(G)\geq deg(v)-deg(u)\geq 2$

Case 2: $deg (v) < r$
we get exactly the same thing by symmetry.