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We know that the hyperbolic tangent function, $\tanh x$, is less than one. I want to show that it is also less than a function which can be smaller than one. In particular, I want to prove that \begin{align}\tanh \frac\pi 2x\le\frac\pi2\frac x{\sqrt{1+x^2}}.\end{align} If $\frac x{\sqrt{1+x^2}}\ge\frac2\pi$, then $\frac\pi2\frac x{\sqrt{1+x^2}}\ge\frac\pi2(\frac2\pi)=1\ge\tanh \frac\pi 2x$ and the inequality holds. My question is: How can i show that this inequality is also true for the case $\frac x{\sqrt{1+x^2}}<\frac2\pi$ ?.

I am working on something, and if i show this last case then I will be done. Thanking you in advance.

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    Presumably you want this for $x \ge 0$. It's false for x < 0.2012-05-29

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A simple solution would be using theory of differential calculus. Note that the case for $x<0$, the conclusion is not true. You can argue as follows:

Define $f$ as $f(x) = \frac {\pi} 2 \frac{x}{\sqrt{1+x^2}}-\tanh\frac \pi 2x$

then show using basic calculus theory that $f$ is increasing for all $x \in \Bbb R$ and $f(0)=0$. This means that

$\frac {\pi} 2 \frac{x}{\sqrt{1+x^2}} > \tanh\frac \pi 2x \text{ ; when } x>0$ $\frac {\pi} 2 \frac{x}{\sqrt{1+x^2}} < \tanh\frac \pi 2x \text{ ; when } x<0$

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    It doesn't come to mind other now. Since $\cosh$ is simply a sum of exponentials, the expansion shouldn't be hard.2012-05-29