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This seems quite simple and is certainly numerically true (I have tested it), but I cannot find a proof for the following inequality. Let $\{a_j\}_{j=1}^n$ be a sequence of real numbers then $$\sum_{j=1}^n \tanh(a_j)a_j \geq \frac{1}{n}\sum_{j=1}^n \sum_{k=1}^n \tanh(a_k)a_j.$$

I don't think it has much to do with the $\tanh$ function as any sign-preserving function should do. Equality clearly occurs when the sequence is constant. All I can think to do is bound the average by its maximum but that's weaker than what I want. AM-GM doesn't help, maybe Jensen? Or maybe it's just something simple and I don't see it (it has to be right? it looks so simple). Any help is appreciated. Thanks.

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  • $\begingroup$ This is interesting. Do we also have the constraint $a_j\geq 0$ or not? $\endgroup$ – Jack D'Aurizio Jun 7 '17 at 0:22
  • $\begingroup$ No we don't, but it seems to me that that's a good thing because the summands of the lhs are always positive while the ones on the rhs might not be. $\endgroup$ – John Doe Jun 7 '17 at 0:27
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The RHS can only increase by replacing any $a_j$ with $\left|a_j\right|$, hence we may assume $a_j\geq 0$ WLOG.
Then we have to prove

$$n \sum_{j=1}^{n}a_j\tanh(a_j) \geq \left(\sum_{j=1}^{n}a_j\right)\left(\sum_{k=1}^{n}\tanh(a_k)\right)\tag{1} $$ Since $f(x)=\tanh(x)$ is a positive and increasing function on $\mathbb{R}^+$, the sequences $a_1,a_2,\ldots,a_n$ and $\tanh(a_1),\tanh(a_2),\ldots,\tanh(a_n)$ are ordered in the same way, and we may further assume $a_1\geq a_2 \geq \ldots \geq a_n$ without loss of generality. In such a case, $(1)$ follows from Chebyshev's sum inequality.

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  • $\begingroup$ Really nice, thank you! $\endgroup$ – John Doe Jun 7 '17 at 1:18

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