Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$ I have this inequality that I don't know how to prove. Possibly the inequality between means might be useful. For $n \in \mathbb{N}$: 
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$$
 A: Write this as
$$ \sum_{j=1}^{2n+1} \dfrac{1}{n+j} \ge \int_1^{2n+2} \dfrac{dx}{n+x} = \ln \left(\frac{3n+2}{n+1}\right)$$
Now  $\dfrac{3n+2}{n+1} \ge e$ for $n > \dfrac{e-2}{3-e} \approx 2.549$.  Do the cases $n=1$ and $2$ separately.  
A: We induct on $n$. For $n=1$, it's trivial.
Now assume it holds for $n=k$, then
$$
\frac{1}{(k+1)+1}+\cdots+\frac{1}{3(k+1)}+\frac{1}{3(k+1)+1}=\frac{1}{k+2}+\cdots+\frac{1}{3k+4}
$$
$$
=(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})-\frac{1}{k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}
$$
Since the first expression is at least $1$ by the inductive hypothesis, it suffices to show
$$
\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\ge \frac{1}{k+1}
$$
But by Cauchy-Schwarz we obtain
$$
(3k+2+3k+3+3k+4)(\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4})\ge9
$$
Which is equivalent to what we want after dividing by $9(k+1)$. So the inductive hypothesis always holds.
We can also use user8268's hint and apply Cauchy or AM-HM directly:
$$
(k+1+\cdots+3k+1)(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})\ge (2k+1)^2
$$
So it suffices to show
$$
(2k+1)^2\ge (k+1+\cdots+3k+1)=(1+\cdots+3k+1)-(1+\cdots+k)
$$
or
$$
4k^2+4k+1\ge \frac{(3k+1)(3k+2)}{2}-\frac{k(k+1)}{2}=\frac{8k^2+8k+2}{2}=4k^2+4k+1
$$
but this is clearly true.
A: In general, if $f$ is a convex function and $p_j$ are non-negative coefficients such that $\sum p_j = 1$ then
$$
\sum_j p_j \, f(x_j) \geq f(\sum_j p_j \, x_j).
$$
(This is Jensen's inequality.)  Take $f(x) = 1/x$, $p_j = 1/(2n + 1)$ and $x_j = n + j$ for $j \in \{1, \dotsc,  2n+1 \}$ and your inequality follows.
A: It follows immediately from arithmetic mean harmonic mean inequality.
Recall that given a set of positive numbers, $\{a_k\}_{k=1}^{k=m}$, we have that
$$\dfrac{\displaystyle\sum_{k=1}^{m} a_k}m \geq \dfrac{m}{\displaystyle \sum_{k=1}^m \dfrac1{a_k}}$$
$$\dfrac{\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k}}{2n+1} \geq \dfrac{2n+1}{\displaystyle \sum_{k=1}^{2n+1} (n+k)}$$
$$\displaystyle \sum_{k=1}^{2n+1} (n+k) = n(2n+1) + \dfrac{(2n+1)(2n+2)}2 = (2n+1)^2$$
Hence,
$$\dfrac{\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k}}{2n+1} \geq \dfrac{2n+1}{(2n+1)^2}$$
Hence, $$\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k} \geq 1$$
A: By C-S we obtain:
 $$\sum_{k=1}^{2n+1}\frac{1}{n+k}=\sum_{k=0}^{n-1}\left(\frac{1}{n+1+k}+\frac{1}{3n+1-k}\right)+\frac{1}{2n+1}\geq$$
$$\geq\sum_{k=0}^{n-1}\frac{(1+1)^2}{n+1+k+3n+1-k}+\frac{1}{2n+1}=\sum_{k=0}^{n-1}\frac{2}{2n+1}+\frac{1}{2n+1}=1.$$
Done!
