<s>[INST] Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
 [/INST] For $n = 2k$
$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k} + \frac{1}{2k+1} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k} \geq\\
\geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k+1 \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} = \frac{k+1}{3k} + \frac{k}{4k} \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12}
$$
For $n = 2k+1$
$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k+1} + \frac{1}{2k+2} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k+2} \geq\\
\geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} + \frac{1}{4k+1} + \frac{1}{4k+2} \geq \\ \geq \frac{k}{3k} + \frac{k}{4k} + \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12}
$$
</s>