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 $n=2k+1$

Let $k$ be a positive integer and let $x$ be such that $x \in [k-1,k]$. You have
$$
n+k-1 \leq x +n
$$ and 
$$
\int_{k-1}^k \frac{1}{x+n} dx \leq \int_{k-1}^k \frac{1}{n+k-1} dx =\frac{1}{n+k-1}
$$ thus, summing from $k=1$ to $n+1$, $n\geq1$, we get
$$
\int_{0}^{n+1} \frac{1}{x+n} dx \leq \sum_{k=1}^{n+1}\frac{1}{n+k-1}=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}
$$ or
$$
\ln \left(2+\frac{1}{n}\right) \leq \frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}
$$
But $$ e^{7/12} \leq 2 < 2+\frac{1}{n}.$$
