$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$ Let $m,n\in\mathbb N-\{1,2\}$ such that $m\ne n.$ How to show that 
$$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}?$$
Please help me. I am clueless.
 A: I am going to prove this for cases. First if $m < n$:
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{m}-\frac{1}{n}
$$
For all $n\in \mathbb{N}$
$$
\frac{2}{n+2} > 0
$$
Then
\begin{eqnarray}
(n-1) + \frac{2}{n+2}  &>& (n-1)  \\
\frac{n^2+n-2 +2}{n+2}  &>& n-1 \\
\frac{n^2+n}{n+2} &>& n-1  \\
\end{eqnarray}
For the axioms of Peano (I think), $n-1 \geq m$, then:
\begin{eqnarray}
\frac{n^2+n}{n+2} &\geq & m \\
\frac{1}{m} &\geq & \frac{n+2}{n^2+n} \\
\frac{1}{m} &\geq & \frac{2n+2-n}{n(n+1)} \\
\frac{1}{m} &\geq & \frac{2}{n} - \frac{1}{n+1}  \\
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{m} - \frac{1}{n} &\geq & \frac{1}{n} - \frac{1}{n+1} \\
\end{eqnarray}
And if $m >n$:
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{n}-\frac{1}{m}
$$
For the axioms of Peano:
\begin{eqnarray}
m &\geq & n+1  \\
\frac{1}{n+1} &\geq & \frac{1}{m}  \\
-\frac{1}{m} &\geq & -\frac{1}{n+1}  \\
\end{eqnarray}
Then 
$$
\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}} = \frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-\frac{1}{n+1}$$
Thus, $\forall m,n\in\mathbb{N}$ with $m\neq n$
$$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$$
