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After I've read the Problem 4136 from [1], and I think (I've computed it in the same way that in the solutions) that the same statement is right for the following expression where $a,b$ and $c$ are real numbers $>0$ $$b\operatorname{gd}(a)+c\operatorname{gd}(b)+a\operatorname{gd}(c)<\frac{\pi\sqrt{3}}{2}\sqrt{a^2+b^2+c^2},$$ where $\operatorname{gd}(x)$ denotes the Gudermannian function, see the Wikipedia Gudermannian function, I wodered what should be interesting or feasibles cyclic inequalities invovling particular values of the Gudermannian function and/or its inverse.

Question. Can you create a cyclic inequality, with good mathematical content, involving an expression of particular values of the Gudermannian function and/or its inverse? Many thanks.

You can take the expression involving this/these special function/s in the way that you think that the resulting inequality has good mathematical content.

References:

[1] Daniel Sitaru and Mihaly Bencze, Problem 4136, Crux Mathematicorum, Volume 43, Number 4, April 2017.

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  • $\begingroup$ The web page of Crux Mathematicorum (I believe that is a journal of the Canadian Mathematical Society) is cms.math.ca/crux from which you can see its Digital Archive. $\endgroup$
    – user142929
    Sep 4, 2019 at 17:05
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    $\begingroup$ it would be nice if you retype here the problem 4136 and specify what exactly holds for your expression $\endgroup$ Sep 4, 2019 at 17:15
  • $\begingroup$ I'm sorry @FedorPetrov but I do it with all respect to the mentioned journal and authors, that is a free online journal: search for Crux Mathematicorum and you can find the problem in the corresponding section of solved problems. On the other hand I did the calculations for my expression and I can deduce the same inequality of the statement of the Problem 4136. $\endgroup$
    – user142929
    Sep 4, 2019 at 17:46
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    $\begingroup$ @user142929 if you can't be bothered to link to it, why would anyone here be bothered to answer your question?! $\endgroup$
    – ssx
    Sep 4, 2019 at 17:52
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    $\begingroup$ @user142929 ok and what is disrespectful in citing the statement of the problem here?! $\endgroup$ Sep 4, 2019 at 17:54

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By Wikipedia, $\text{gd}(x)=2\arctan e^x-\pi/2<\pi/2$ for real $x$. So, $$b\,\text{gd}(a)+c\,\text{gd}(b)+a\,\text{gd}(c) <\frac\pi2(b+c+a)\le\frac{\pi\sqrt3}2\,\sqrt{a^2+b^2+c^2},$$ by the arithmetic mean--quadratic mean inequality. Thus, we have the inequality in question.

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  • $\begingroup$ Many thanks for your answer. $\endgroup$
    – user142929
    Sep 6, 2019 at 7:53

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