Prove that $H_n - H_m > \frac{n-m}{n}$ for $n, m \in \mathbb{N}$ and $n > m \geq 1$, by doing induction on $n$ and base case $n = m$ Problem:
Let $H_n = \sum_{i=1}^{n} \frac{1}{i} $ for all natural numbers $n \geq 1$.
Prove that $H_n - H_m \geq \frac{n-m}{n}$ for $n, m \in \mathbb{N}$ and $n > m \geq 1$, by doing induction on $n$, letting $m$ be any natural number, and starting with the base case $n = m$.
$--------------------------------------$
I have done the base case, but I am confused on how to start with the inductive step.
We want to show that
$H_{n+1} - H_m \geq \frac{n+1-m}{n+1}$
Proof:
Base Case:
Let $n = m$, then $H_m - H_m \geq \frac{m-m}{m} = 0 \geq 0$, this is true.
Inductive Step:
Let P(n) denote the statement $H_n - H_m \geq \frac{n-m}{n}$.
We need to show that $P(n) \implies P(n+1)$, so we need to prove  $H_{n+1} - H_m \geq \frac{n+1-m}{n+1}$
Then,
$H_{n+1} - H_m = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} + \frac{1}{n+1} - (\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{m})$
$= \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m} \geq \frac{1}{n+1} + \frac{n-m}{n}$, by the inductive hypothesis
$= \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m} \geq 1 + \frac{1}{n} + \frac{n-m}{n}$
$= \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m} \geq 1 + \frac{n+1-m}{n}$
which is $ > \frac{n+1-m}{n+1}$
I think this is a dud, since I proved $>$, and not $\geq$. Does anyone have any recommendations? Am I missing something?

The problem occurs when you are subtracting the sums
$H_{n+1} - H_m = 1 + \frac{1}{2} + ... + \frac{1}{n+1} - \Big(1 + \frac{1}{2} + ... + \frac{1}{m}) = $
$ = 1 + \frac{1}{2} + ... + \frac{1}{m} + \frac{1}{m+1} + ... + \frac{1}{n+1} - \Big(1 + \frac{1}{2} + ... + \frac{1}{m}) $
$ = \frac{1}{m+1} + ... + \frac{1}{n+1} $
since the first m terms of $H_{n+1}$ cancel out the terms of $H_m$. But this is not very helpful for induction. You must say the following,
$ H_{n+1} - H_m = \frac{1}{n+1} + H_n - H_m \geq \frac{1}{n+1} + \frac{n-m}{n} $
where the last inequality holds because of the induction hypothesis. Now, by rearranging terms it's easy to prove that
$ \frac{1}{n+1} + \frac{n-m}{n} \geq \frac{n+1-m}{n+1} $
Which implies that
$ H_{n+1} - H_m \geq \frac{n+1-m}{n+1} $
Now about using the equality or not is not a big problem. But, if u want to have only use ">", your first step in induction should be for $n=m+1$, and not $n=m$, since $m+1$ is the first natural number $>m$.
