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}

We can use C-S by another way.
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}=\frac{1}{2n+1}+\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{3n+2-k}\right)\geq$$
$$\geq\frac{1}{2n+1}+\sum_{k=1}^n\frac{(1+1)^2}{n+k+3n+2-k}=\frac{1}{2n+1}+\frac{2n}{2n+1}=1.$$
