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}
 A: $$\frac { \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 3n+1 }  }{ 2n+1 } >\frac { 2n+1 }{ n+1+n+2+...+3n+1 } \\ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 3n+1 } >\frac { { \left( 2n+1 \right)  }^{ 2 } }{ \frac { 3n+1+n+1 }{ 2 } \cdot \left( 2n+1 \right)  } =\frac { 4n+2 }{ 4n+2 } =1$$
A: 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
$$
1\le\frac1{n+1}+\cdots+\frac1{3n+1}
$$
A: You claim that $(n+1)+\cdots+(3n+1) = 2n((n+1)+(3n+1))$, which is not correct. 
First of all, there are $2n+1$ terms, not $2n$ (as user8734617 says in a comment).
Also, you need to multiply with the average of the first and the last term. 
So you actually get $$\frac{(2n+1)^2} { (2n+1) \left( \frac{(n+1)+(3n+1)}{2} \right)} = \frac{(2n+1)^2}{(2n+1)^2} = 1$$

For an alternative solution, try induction. 
A: We can use C-S by another way.
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}=\frac{1}{2n+1}+\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{3n+2-k}\right)\geq$$
$$\geq\frac{1}{2n+1}+\sum_{k=1}^n\frac{(1+1)^2}{n+k+3n+2-k}=\frac{1}{2n+1}+\frac{2n}{2n+1}=1.$$
A: 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
$$\begin{align}
a_1+\cdots+a_N&=(n+1)+\cdots+(n+N)\\
&=nN+(1+\cdots+N)\\
&=nN+{N(N+1)\over2}\\
&={N(2n+1+N)\over2}\\
&={N(N+N)\over2}\\
&=N^2
\end{align}$$
