How prove $\left(\sum_{i=0}^{n-1}a_{i}\cos{\frac{2k\pi i}{n}}\right)^2\le (n^2-1)\left(\sum_{i=0}^{n-1}a_{i}\sin{\frac{2k\pi i}{n}}\right)^2$

Asked Mar 06 '14 at 09:49

Active Mar 06 '14 at 10:02

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let $n$ is odd number, and $a_{i},(i=0,1,\cdots,n-1)$ is $\{0,1,2,\cdots,n-1\}$ arrangement,and $k=0,1,2,\cdots,n-1$

prove or disprove $$\left(\sum_{i=0}^{n-1}a_{i}\cos{\dfrac{2k\pi i}{n}}\right)^2\le (n^2-1)\left(\sum_{i=0}^{n-1}a_{i}\sin{\dfrac{2k\pi i}{n}}\right)^2$$

My idea: $$\Longleftrightarrow \left(\sum_{i=0}^{n-1}a_{i}\cos{\dfrac{2k\pi i}{n}}\right)^2+\left(\sum_{i=0}^{n-1}a_{i}\sin{\dfrac{2k\pi i}{n}}\right)^2\le n^2\left(\sum_{i=0}^{n-1}a_{i}\sin{\dfrac{2k\pi i}{n}}\right)^2$$ This inequality is from my frend ask me,maybe this is true,But I can't prove it.Thank you

edited Mar 06 '14 at 10:02

asked Mar 06 '14 at 09:49

math110

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Obviously false for $n>1$: Set $a_i=1, k=0$ then the LHS is $n^2$ but the RHS is zero. – gammatester Mar 06 '14 at 10:00
Oh,I have edit,Thank you – math110 Mar 06 '14 at 10:05
1

The updated assumption that $(a_{i})$ is permutation of $(0..n-1)$ is still too weak for $k=0$. Set $a_i=i,,$ the LHS is $(n(n-1)/2)^2,,$ but the RHS is still zero. I guess you must exclude $k=0$ for $n>1$. – gammatester Mar 06 '14 at 10:31

How prove $\left(\sum_{i=0}^{n-1}a_{i}\cos{\frac{2k\pi i}{n}}\right)^2\le (n^2-1)\left(\sum_{i=0}^{n-1}a_{i}\sin{\frac{2k\pi i}{n}}\right)^2$

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