TITLE: Verifying $\int_0^\pi  \sin(x) /2(\sin(x/(2n+1)) \,dx \leq \pi$
QUESTION [6 upvotes]: I'm having trouble verifying this inequality.  It goes like this (appears in Giaquinta, Mathematical analysis, linear and metric structures, page 445):
$$
\int_{0}^{\pi} \cfrac{\sin(x)}{\sin\left(\frac{x}{2n+1}\right)} dx \leq\frac{ 2(n+1)\pi}{2n+1} \leq 2\pi
$$
Of course, the last inequality is obvious.  The first one, however, I can't show.  I've tried bounding $\sin(x)$ by $1$, and then calculating the integral with mathematica, but it comes out unbounded.  When I put $n=1,2,3...$ or any finite number in mathematica, the result is numerically true, but I want to show this for any "$n$", and mathematica gives me a very complicated function (depending on $n$) with imaginary units and hypergeometric functions.  I guess I'm missing out a very simple argument here.  Any ideas?
Edit: I have edited so that the formula is identical to the one of the book.

REPLY [3 votes]: Giaquinta and Modica are right, although they are wrong. Let me explain: the book suggests to compute $G_n \left( \frac{2 \pi}{2n+1} \right)$, where
$$
G_n(x) = \int_0^x D_n(t)\, dt
$$
and $D_n$ is the standard Dirichlet kernel. In their computation there is a very stupid mistake: they change variable and they forget to change the differential! Here is a corrected version of their formula on page 445:
$$
\begin{align*}
\|G_n\|_\infty &= G_n \left( \frac{2 \pi}{2n+1} \right) = \int_0^{2\pi/(2n+1)} \frac{\sin (n+1/2)s)}{\sin (s/2)}ds\\
&= \frac{2}{2n+1} \int_0^\pi \frac{\sin t}{\sin \frac{t}{2n+1}}dt \leq \frac{2}{2n+1} \cdot (2n+1)\pi = 2\pi.
\end{align*}
$$
We have used the fact that $t \mapsto \frac{\sin t}{\sin \frac{t}{2n+1}}$ is decreasing, so that it is always less that its limit as $t \to 0^+$.
I hope my answer will be useful.