Let $f$ be a periodic function with period $2\pi$. Suppose $f$ is integrable. If $f$ is monotone decreasing on $(0,2\pi)$. Prove that $$ \forall n=1,2,\cdots,b_n:=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin nx \mathrm{d}x\geq0 $$
I tried to use mean value theorems for definite integrals. Write $m:=\inf_{[0,2\pi]}f$,then $$ \int_{-\pi}^\pi f(x)\sin nx \mathrm{d}x=[f(-\pi)-m]\int_{-\pi}^{\xi_1}\sin nx\mathrm{d}x+[f(0)-m]\int_{0}^{\xi_2}\sin nx\mathrm{d}x $$ The second term is always $\geq 0$, but the first term may be negative for even $n$, so I stuck here.
How should I go on? Also you can suggest your own way. Appreciate any help.