<s>[INST] AM>HM Problem $\frac{1}{n+1}+...+\frac{1}{3n+1}>1$ I am having difficulty solving one of the problems from "Problems in Mathematical Analysis I" -  W. J. Kaczor;M. T. Nowak .
It's a problem 1.2.5 b), and it goes like this:

1.2.5. For $n \in \mathbb{N}$, verify the following claims: 
$$\tag{b} \qquad \dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3} + \ldots + \dfrac{1}{3n + 1} \, > \, 1$$

In solutions it says: "Use the arithmetic-harmonic mean inequality!"
I tried to apply it on whole inequality but got:
\begin{align}
& \dfrac{\frac{1}{n+1}+\ldots+\frac{1}{3n+1}}{2n}\, >\, \dfrac{2n}{n+1+\ldots+3n+1} \\ 
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{8n^2}{2n(n+1+3n+1)} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{2n}{2n+1}
\end{align}
 [/INST] This is just a different presentation of the answers already given.  I find it helpful to assign a variable to the number of terms in the sum, which makes some of the algebra easier to keep track of.
The AH inequality can be written as
$${1\over a_1}+\cdots+{1\over a_N}\ge{N^2\over a_1+\cdots+a_N}$$
In our case, we have
$${1\over n+1}+\cdots+{1\over3n+1}={1\over n+1}+\cdots{1\over n+(2n+1)}$$
suggesting we let $N=2n+1$, in which case
$$\begin{align}
a_1+\cdots+a_N&=(n+1)+\cdots+(n+N)\\
&=nN+(1+\cdots+N)\\
&=nN+{N(N+1)\over2}\\
&={N(2n+1+N)\over2}\\
&={N(N+N)\over2}\\
&=N^2
\end{align}$$
</s>