$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$

Question

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.

Answer 1

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 $m\neq n$

$$\frac{1}{\min\{m,n\}}-\frac{1}{\max\{m,n\}}\ge\frac{1}{n}-\frac{1}{n+1}$$