<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] one approach is to pair off the integers in the harmonic sum as $n+k$ and $2n-k$. so if $n$ is even we have:
$$ 
S_n = \sum_{k=0}^{\frac{n}2-1}\left(\frac1{n+k} + \frac1{2n-k} \right) + \frac2{3n}
$$
now 
$$\frac1{n+k} + \frac1{2n-k} = \frac{3n}{(n+k)(2n-k)} \ge \frac{4n}{3n^2} = \frac4{3n}
$$
and there are $\frac{n}2$ such terms, so we have:
$$
S_n \ge \frac23 + \frac2{3n}
$$
when $n$ is odd we arrive (i think) at the numerically similar estimate
$$
S_n \ge \frac{6 \left(1+\frac1{n}\right)}{9-\frac1{n^2}}
$$
</s>