# Proving that Hyperbolic tangent is less than some function which can be smaller than one

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 $x < 0$. – Robert Israel May 29 '12 at 19:02

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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Thanks Peter, i have found the derivative: $f'(x)=\frac\pi2(\frac1{(1+x^2)^{3/2}}-\sech^2\frac\pi2x)$. But still i am stack with showing that this is$>0$ for all $x\ge0$. Please help. – jm sol May 29 '12 at 17:44
Careful with the expression you're getting. Use the product rule, chain rule and power rule for the first function. The other is OK. – Pedro Tamaroff May 29 '12 at 17:56
Peter, even the first part is ok after simplification. Let me see what you are doing or can you work out the whole problem, i want too see. – jm sol May 29 '12 at 18:18
The statement is equivalent to $$(1+x^2)^{(3/2)}<\cosh^2(\pi x/2)$$ $$(1+x^2)^3<\cosh^4(\pi x/2)$$ which I guess you can easily prove with some Taylor expansion. – Pedro Tamaroff May 29 '12 at 18:51
Isn't there another way other than Taylor expansion? – jm sol May 29 '12 at 19:06