I'm reading Bondy and Murthy's Graph Theory, and I'm doing the proposed exercise in the title. I've tried to do the following:
$m$: Edges
$n$: Vertices
A simple graph with $n$ vertices has a maximum of $m=(n-1)+(n-2)+\dots+(n-n)$ and hence
$$m=(n-1)n-\frac{(n-1)(n)}{2}=\frac{n^2-n}{2}$$
Which is $(n-1)$ times the number of $n$'s and the sum of the first $(n-1)$ natural numbers. Knowing that the maximum number of edges in a simple graph is ${n \choose 2}$, we can write:
$$\begin{eqnarray*}
{m}&\leq&{{n \choose 2}}\\
{}&&{}\\
{\frac{n^2-n}{2}}&\leq&{\frac{n!}{2(n-2)}}\\
{}&&{}\\
{n^2-n}&\leq&{\frac{n!}{(n-2)!}}\\
{}&&{}\\
{n^2-n}&\leq&{n \cdot (n-1)}\\
{}&&{}\\
{n^2-n}&\leq&{n^2-n}\\
{}&&{}\\
{n^2-n}&\leq&{n^2-n}
\end{eqnarray*}$$
Best Answer
Assume that $m\gt{n\choose 2}$ where the order and size of $G$ is $n$ and $m$, respectively. We know that $K_n$ has exactly ${n\choose 2}$ edges. Since $m\gt{n\choose 2}$ this implies that $G$ contains at least $1$ multiple edge which contradicts the fact that $G$ is simple. Thus for all simple graphs $G$ we know that $m\leq {n\choose 2}$.