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

score 1 · Answer 1 · edited Aug 11 '14 at 06:55

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.

edited Aug 11 '14 at 06:55

5xum

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answered Jan 22 '14 at 11:53

Ma Ming

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Hello,can you post your follow average methods? Thank you – math110 Jan 22 '14 at 11:56
@math110 Notice that $\int_{0}^{\pi} f(\theta)=\pi$, so there exists some $\theta$ such that $f(\theta)\ge 1$. – Ma Ming Jan 22 '14 at 11:59
why $\int_{0}^{\pi}f(\theta)d\theta=\pi$ – math110 Jan 22 '14 at 14:45
If you expand the formula as a sum of terms of the form $\exp(2k\theta+c)$, after integrating every term would be zero except the constant one. – Ma Ming Jan 23 '14 at 23:19

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

1 Answers1