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let $x_{k}>0(k=1,2,3,\cdots,n)$ show that $$\left(\sum_{k=1}^{n}x_{k}\tan{k}\right)\left(\sum_{k=1}^{n}x_{k}\cot{k}\right)\le n^3\sum_{k=1}^{n}x^2_{k}$$

My try use AM-GM and Cauchy-Schwarz inequality we have $$LHS\le \dfrac{(\sum_{k=1}^{n}(\tan{k}+\cot{k})x_{k})^2}{4}\le\dfrac{\sum_{k=1}^{n}x^2_{k}\sum_{k=1}^{n}(\tan{k}+\cot{k})^2}{4}=\dfrac{2n+\sum_{k=1}^{n}(\cot^2{k}+\tan^2{k})}{4}\sum_{k=1}^{n}x^2_{k}$$

math110
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    CS is enough, applied to the vectors in $\mathbb{R}^n$ whose $k$-th components are $\sqrt{x_k|\tan k|}$ and $\sqrt{x_k|\cot k|}$. – Jack D'Aurizio Dec 22 '19 at 17:08
  • @JackD'Aurizio With your method I get $$\left(\sum_{k=0}^n x_k\right)^2\le\left(\sum_{k=0}^n x_k |\tan(k)|\right)\left(\sum_{k=0}^n x_k |\cot(k)|\right)$$ which is not exactly what we want Am I doing something wrong? – Maximilian Janisch Dec 22 '19 at 22:25
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    Please check: $n=12$, $x_1 = x_2 = \cdots = x_{10} \to 0$ and $x_{11} = x_{12} = 1$. – River Li Dec 23 '19 at 02:25
  • @RiverLi,maybe the $n^2$ take place of $n^3$ is right? – math110 Dec 26 '19 at 04:33
  • @inequality One might be able to use bounds of the form $tan(k)<Ck$ and $cot(k)<Ck$ for some$C$ to complete your proof. See for example https://arxiv.org/pdf/1902.08817.pdf The constant $C$ might be chosen arbitrarily close to 1 as $k$ grows I think. – Marco Dec 30 '19 at 16:02
  • For those interested, I have tried the following. For a given $n$, let the max of $\tan(k)$ and the max of $\cot(k)$ for $k$ on the interval $[1, n]$ be $a$ and $b$ respectively. Then of course the left sum is less than $$ab\left(\sum_{k=1}^n x_k\right)^2$$ which is always less than or equal to $$n ab\sum_{k=1}^n x_k^2$$ and thus if $ab\leq n^2$ the proof of complete. Unfortunately, as early as $n=22$ for $\tan(14)$ and $\cot(22)$ this fails. Hopefully this saves someone some time. – Kraigolas Dec 31 '19 at 18:10
  • @Marco: "See for example"?! The claims in that paper should be treated with skepticism. – Dap Dec 31 '19 at 22:09
  • @Dap: Any result should be treated with skepticism unless it is verified by oneself. The bounds for $1/|\sin k|$ in the paper do check out and they are relevant to this problem. Any solution of this problem will have to do with bounding $\tan k$ and $\cot k$. – Marco Jan 01 '20 at 18:11
  • It is likely that there is no known answer to this problem. Although $\tan(n)$ is unbounded (see https://math.stackexchange.com/questions/1056119/), that doesn't mean that we know anything about how close we come to some $(k+1/2)\pi$ with growing $n$. The bound constant will depend on the irrationality measure of π, which is unknown. See discussions in here: https://math.stackexchange.com/questions/2977461/ – Andreas Jan 05 '20 at 22:31

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