How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$ How do I prove the following inequality
$$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$
without the help of induction?
Thanks for any help!!

You don't need calculus here (although it would make things easier). Let $$S=\frac{1}{n+1}+\frac {1}{2n}+\frac{1}{n+3}+...+\frac{1}{2n}$$
and write it both ways, à la Gauss:
$$\begin{align}
2S &= \frac{1}{n+1}+\frac {1}{n+2}\,\,\,+\,\,\frac{1}{n+3}\,+...+\,\,\,\frac{1}{2n} \\
&\,\,\,\,+\frac{1}{2n}\,\,\;+\frac {1}{2n-1}+\frac{1}{2n-2}+...+\frac{1}{n+1}\\
&=\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-(k-1)}\right)\\
&<\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-k}\right)\\
&=\sum_{k=1}^n\frac{3n}{(n+k)(2n-k)}\\
&\le\sum_{k=1}^n\frac{3n}{2n^2} \,\,\,\,\text{because}\, (n+k)(2n-k) \ge 2n^2 \text{ if } 1 \le k \le n\\
&=\frac{3}{2n}\sum_{k=1}^n1\\
&=\frac32
\end{align}$$
Hence $S < \dfrac{3}{4}$.
