<s>[INST] 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 $n \in \mathbb{N}$: 
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$$
 [/INST] 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.
</s>