I'm having a hard time solving the following problem:
Prove by induction that $\forall\ n\in\mathbb{N} $:
$$\sum_{i=1}^{n} \frac{n+i}{i+1}\ge n$$
I checked the base case, and then made it up to here in the inductive step:
$\sum_{i=1}^{n+1} \frac{n+1+i}{i+1}\gt \frac{\sum_{i=1}^{n+1} n+1+i}{\sum_{i=1}^{n+1} i+1}\ge \frac{(\sum_{i=1}^{n} n+i)+(2n+1)+(2n+2)-(n+1)}{(\sum_{i=1}^{n} i+1)+(n+2)}$
How do I use the inductive hypothesis to get to: $$\sum_{i=1}^{n+1} \frac{n+1+i}{i+1}\gt n+1$$
As Michael Rozenberg states in a question comment, it's fairly obvious without induction as each term is $\ge 1$ (since $n + i \ge i + 1$) and there are $n$ terms. However, if you wish to use induction, then here is one way to do it to prove that
$$\sum_{i=1}^n \cfrac{n + i}{i + 1} \ge n \tag{1}\label{eq1}$$
First, check the base case of $n = 1$, with both sides giving $1$, so it's true. Next, assume \eqref{eq1} is true for all $n \le k$ for some $k \ge 1$, in particular that
$$\sum_{i=1}^k \cfrac{k + i}{i + 1} \ge k \tag{2}\label{eq2}$$
For $n = k + 1$, note that $\frac{k + 1 + i}{i + 1} \gt \frac{k + i}{i + 1}$ for all $1 \le i \le k$. Thus,
$$\sum_{i=1}^{k + 1} \cfrac{k + 1 + i}{i + 1} \gt \sum_{i=1}^k \cfrac{k + i}{i + 1} + \cfrac{\left(k + 1\right) + k + 1}{k + 1 + 1} \tag{3}\label{eq3}$$
Using \eqref{eq2} and that $2k + 2 \gt k + 2$, the right side of \eqref{eq3} gives that
$$\sum_{i=1}^k \cfrac{k + i}{i + 1} + \cfrac{2k + 2}{k + 2} \gt k + 1 \tag{4}\label{eq4}$$
Putting \eqref{eq3} and \eqref{eq4} together gives that
$$\sum_{i=1}^{k + 1} \cfrac{k + 1 + i}{i + 1} \gt k + 1 \tag{5}\label{eq5}$$
This shows it's true for $n = k + 1$, thus completing the inductive steps.