Find this maximum of this $\frac{\int_{0}^{\pi}f(x) \, dx}{\int_{0}^{\pi} f(x)\sin x\,dx}$ 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.

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.
