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.