Question:
Assmue that $\int_0^\pi f(x)\,dx$ and $\int_0^\pi f(x)\sin x\,dx$ is convergence,and $f(x)>0,\forall x\in(0,\pi)$ Find this maximum as possible for all function $f$ $$I=\dfrac{\int_0^\pi f(x)\,dx}{\int_0^\pi f(x)\sin{x}\,dx}$$
show that： $$I\le\dfrac{4}{\pi}?$$
I think this problem is interesting,But I can't.
Hint: let $$f_n(x):= n\chi_{(0,1/n)} + n^{-1}\chi_{[1/n,\pi]}, $$ with $\chi$ denoting the indicator (i.e. for any set $E$, $\chi_E=1$ on $E$, $\chi_E=0$ outside $E$). You get $$\int_0^\pi f_n(x)dx = 1+ (\pi-1/n)/n ,$$ while (for large $n$) $$\int_0^\pi |f_n(x)\sin x| dx \leq n\int_0^{1/n} x dx + (\pi-1/n)/n ... $$
Similar to the already posted counterexample: for every positive $t$, consider $$f_t(x)=\min\{t,1/\sin x\},$$ then, when $t\to+\infty$, $$\int_0^\pi f_t(x)\mathrm dx\to\int_0^\pi\frac{\mathrm dx}{\sin x}=+\infty,$$ while $$\int_0^\pi f_t(x)\sin(x)\mathrm dx\to\int_0^\pi\mathrm dx=\pi,$$ hence the ratios considered in the question are unbounded.
