How prove this $\left|\prod_{k=1}^{n}\sin{(\theta-a_{k})}\right|\ge\dfrac{1}{2^n}$

Question

let $a_{1},a_{2},a_{3},\cdots,a_{n}\in(0,\pi)$, show that there exsit $\theta\in(0,\pi)$ such $$\left|\prod_{k=1}^{n}\sin{(\theta-a_{k})}\right|\ge\dfrac{1}{2^n}$$

My try: $$\Longleftrightarrow |\prod_{k=1}^{n}2\sin{(\theta-a_{k})}|\ge 1$$

Then I can't prove it,Thank you very much

Answer 1

Hint: Expand

$$\prod \sin(\theta-a_k)=\prod \frac{1}{2i}\left[e^{i(\theta-a_k)}-e^{i(\theta-a_k)}\right]$$

Let $f(\theta)=\prod\left[e^{2i(\theta-a_k)}-1\right]$. It suffices to show that there exists some $\theta$, such that $|f(\theta)|\ge 1$. Prove it by average(integral) method.