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 $n=2k+1$

Hint By the monotonicity of $x \mapsto \frac{1}{x}$, we have
$$\frac{1}{n} + \dots + \frac{1}{2n} = \sum_{k=n}^{2n} \frac{1}{k} \geq \int_n^{2n} \frac{1}{x} \, dx.$$
A direct calculation of the latter integral yields an even sharper bound: 
$$\frac{1}{n} + \dots + \frac{1}{2n} \geq \ln 2 > \frac{7}{12}.$$
For the last inequality note that it follows from $e \leq 3$ that
$$e^{3/5} \leq 3^{3/5} <2;$$
hence $\ln 2>\frac{3}{5} > \frac{7}{12}$.
