Proof related to minimum and maximum degree of vertices of an undirected graph I don't know how to proceed for this problem. I would appreciate any help. Thanks!


  Let $\delta$ and $\Delta$ be the minimum and maximum degree of the
  vertices of an undirected graph G.
  
  Show that $\delta \leq \frac{2\left | E(G) \right |}{\left | V(G)
\right |}\leq \Delta $.
HINT: Use the basic fact relating degrees of vertices to number of edges: $$\sum_{v\in V(G)}\deg v=2|E(G)|\;.$$ What are the minimum and maximum possible values of the sum?
Let $e$ be the no. of edges.
Let $v$ be the no. of vertices.
Let $mn$ be min degree of vertex in graph.
Let $mx$ be the max degree of vertex in graph.
 every vertex's degree must lie between this min and max limit
$\min\leq \deg(v)\leq \max$
$\min\leq 2e\leq \max$  // using the above hint
thus for n vertices in a graph,
$n(\min)\leq 2e\leq n(\max)$ and so
$\min \leq 2\frac{e}{n}\leq \max$