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}

The HM-AM (or Jensen's Inequality applied to $\frac1x$ and a discrete measure) says
$$
\begin{align}
\frac1{2n+1}\left(\frac1{n+1}+\cdots+\frac1{3n+1}\right)
&\ge\frac1{\frac1{2n+1}\left((n+1)+\cdots+(3n+1)\right)}\\
&=\frac1{2n+1}
\end{align}
$$
Therefore,
$$
\frac1{n+1}+\cdots+\frac1{3n+1}\ge1
$$

We can also prove this by the Cauchy-Schwarz Inequality
$$
\begin{align}
(\overbrace{1+\cdots+1}^{2n+1\text{ terms}})^2&
\le\left(\frac1{n+1}+\cdots+\frac1{3n+1}\right)\left((n+1)+\cdots+(3n+1)\right)\\
&=\left(\frac1{n+1}+\cdots+\frac1{3n+1}\right)(2n+1)^2
\end{align}
$$
which also implies
$$
1\le\frac1{n+1}+\cdots+\frac1{3n+1}
$$
