I give a proof of a question in AoPSonline https://artofproblemsolving.com/community/c7h492604_sum_k0nfracsinkxk
$\textbf{What I want to know is: is there some mistake in my proof, any help will welcome!}$
The LHS is just Fejer-Jackson inequality. Let $$f_n(x)=\sum_{k=1}^n\frac{\sin kx}{k},\quad x\in[0,\pi]$$ For the RHS, we can deal with as follows: consider $$f'_n(x)=\sum_{k=1}^{n}\cos kx=\frac{\cos\frac{n+1}{2}x\sin\frac{n}{2}x}{\sin \frac{x}{2}},x\in(0,\pi),$$ then $$f'_n(x)=0\iff \cos\frac{n+1}{2}x\sin\frac{n}{2}x=0,\quad x\in(0,\pi),$$ the critical points contained in $(0,\pi)$ of $f_n$ are listed below: $$\text{critical points}:\frac{\pi}{n+1}<\frac{2\pi}{n}<\frac{3\pi}{n+1}<\cdots.$$ Remark：The number of critical points of $f_n$ is $n$ if $n$ is odd, and is $n-1$ if $n$ is even. It can be shown that $$f_n\left(\frac{\pi}{n+1}\right)=\sum_{k=1}^n\frac{\sin\left(\frac{k\pi}{n+1}\right)}{k} =\max_{x\in[0,\pi]}f_n(x).$$ Note that: $\frac{\pi}{n+1}$ is the smallest critical point of $f_n$. We claim that:
(1) sequence $\left\{f_n\left(\frac{\pi}{n+1}\right)\right\}$ is strictly increasing（can refer$\sum_{k=1}^{n}\frac1{k}\sin(\frac{k\pi}{n+1})$ is increasing）;
(2) $$\lim_{n\to\infty}\sum_{k=1}^n\frac{\sin\left(\frac{k\pi}{n+1}\right)}{k} =\int_{0}^{\pi}\frac{\sin x}{x}dx;$$ (3) $$f_n(x)<\int_{0}^{\pi}\frac{\sin x}{x}dx\quad x\in(0,\pi).$$
Due to $f_n$ is strictly increasing on $[0,\frac{\pi}{n+1}]$ , we know $$f_{n+1}\left(\frac{\pi}{n+2}\right)>f_{n+1}\left(\frac{\pi}{n+1}\right) =f_{n}\left(\frac{\pi}{n+1}\right).$$ $$\sum_{k=1}^n\frac{\sin\left(\frac{k\pi}{n+1}\right)}{k} =\frac{1}{n+1}\sum_{k=1}^n\frac{\sin\left(\frac{k\pi}{n+1}\right)}{\frac{k}{n+1}}\to \int_{0}^{\pi}\frac{\sin x}{x}dx \quad (\text{Riemann\ sum})$$ So we conclude: $$f_n(x)\leq f_n\left(\frac{\pi}{n+1}\right)<\sup_{n}f_n\left(\frac{\pi}{n+1}\right)=\lim_{n\to\infty}\sum_{k=1}^n\frac{\sin\left(\frac{k\pi}{n+1}\right)}{k} =\int_{0}^{\pi}\frac{\sin x}{x}dx.$$
