Let $G$ be a graph with $n$ vertices and $e$ edges. Let $m$ be the smallest positive integer such that $m \ge 2e/n$. Let $G$ be a graph with $n$ vertices and $e$ edges. Let $m$ be the smallest positive integer such that $m \ge 2e/n$.
Prove that $G$ has a vertex of degree at least $m$.
My approach is sum of all the degree of a graph is $2e$.
$d(v_1)+d(v_2)+......+d(v_n)=2e$.
then if we take the average, will get
$d(v)=2e/n$.
now this will be in between minimum degree and maximum degree in the graph.
Now $m$ also lies in between maximum and minimum degree of the graph.
From here how can we conclude the answer.
Please help me.

Suppose that there does not exist a vertex $v$ such that $\deg(v) \geq m$. Then the sum of the degrees of all vertices in $G$, $\sum_{i=1}^{n}v_i \leq (m-1)n$. Since $\sum_{i=1}^{n}v_i=2e$, we have $2e\leq(m-1)n$. Divide both sides by $n$ to get $\frac{2e}{n} \leq m-1$. So, we have $\frac{2e}{n}+1 \leq m$. This is a contradiction because if $\frac{2e}{n}$ is an integer, then $m=\frac{2e}{n}$ and if $\frac{2e}{n}$ is not an integer, $m=\frac{2e}{n}+c$ for some rational number $0<c<1$.
