# Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$

The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $$x = 0$$). This attempt failed because firstly it really is true but after that we get opposing inequality. Also, I have noticed that $$\lim \frac{x}{1 + x \cdot \frac{2}{\pi}} = \lim \arctan(x) = \frac{\pi}{2}$$ as $$x \rightarrow + \infty$$ but it didn't lead me to solution. I also tried to apply Taylor's formula but it didn't help much either.

So, what are available ways to solve this problem?

A possible way is to set $$x = \tan t$$ with $$t \in \left(0,\frac{\pi}{2}\right)$$.

So, the inequality is equivalent to

$$\frac{\tan t}{1+\frac 2{\pi} \tan t}< t \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$

or after rearranging

$$\tan t < \frac t{1-\frac 2{\pi}t} \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$

But this is true because for $$t \in \left(0,\frac{\pi}{2}\right)$$ we have $$\sin t < t$$ and because of the concavity of $$\cos$$ on this interval we have $$\cos t > 1-\frac 2{\pi} t$$. Hence,

$$\tan t = \frac{\sin t}{\cos t} < \frac t{1-\frac 2{\pi} t} \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$

With the idea of @trancelocation: the map $$x\mapsto x/(1 + a x)= \frac{1}{a} ( 1 - \frac{1}{1+a x})$$ is increasing on $$(0, \infty)$$ ( for $$a>0$$), so the inequality is equivalent to

$$x < \frac{ \arctan x}{ 1 - \frac{\arctan x}{\pi/2}} \textrm{ or } \tan t < \frac{t}{1- \frac{t}{\pi/2}}$$

We have the product expansions

$$\cos t = \prod_{n=1}^{\infty} \left(1- \frac{t^2}{(n-1/2)^2\pi^2} \right), \ \ \sin t = \prod_{n=1}^{\infty} \left(1- \frac{t^2}{n^2 \pi^2}\right)$$ so

$$\frac{\tan t }{t} = \frac{1}{1- (\frac{t}{\pi/2})^2} \cdot\prod_{n=1}^{\infty} \frac{\left(1- \frac{t^2}{n^2 \pi^2}\right)}{ \left(1- \frac{t^2}{(n+1/2)^2 \pi^2}\right)}< \frac{1}{1- (\frac{t}{\pi/2})^2}$$ for $$0< t <\pi/2$$. So we have in fact

$$$$\boxed{\color{indigo}{ \frac{1-(\frac{t}{\pi})^2 }{1- (\frac{t}{\pi/2})^2}<\frac{\tan t}{t} < \frac{1}{1- (\frac{t}{\pi/2})^2} < \sec t }}$$$$ for real $$t$$, $$0<|t|< \frac{\pi}{2}$$.

In the above inequalities, we have also the corresponding inequalities for the corresponding coefficients of the Taylor series, clear if we examine the product formulas.

From the second inequality we get $$\begin{eqnarray} \boxed{\color{indigo}{ \frac{\frac{\pi}{2} x}{ \sqrt{x^2 + (\frac{\pi}{4 })^2} + \frac{\pi}{4 }} < \arctan(x) }} \end{eqnarray}$$ for $$x > 0$$, an improvement over the inequality $$\frac{\frac{ \pi }{2} x}{ x+\frac{\pi}{4}+\frac{\pi}{4}} < \arctan x$$.

Hint: let $$f(x):=\arctan x-\tfrac{x}{1+2x/\pi}$$, whose unique maximum on $$x>0$$ you're welcome to locate. For $$0, $$f(x)\sim(\tfrac13+\tfrac{2}{\pi})x^3$$; for $$x\gg1$$,$$f(x)=\tfrac{\pi}{2}-\arctan\tfrac1x-\tfrac{\pi/2}{1+\pi/(2x)}\sim\tfrac{\pi^2-4}{4x}.$$Now sketch $$f$$.

Consider the function $$f$$ defined by $$f(x): = \arctan x - \frac{x}{{1 + \frac{2}{\pi }x}}$$ for all $$x>0$$. It is easy to check that $$f'(x) = \frac{{x((4 - \pi ^2 )x + 4\pi )}}{{(x^2 + 1)(2x + \pi )^2 }}.$$ If $$0 < x \le \frac{{4\pi }}{{\pi ^2 - 4}} = 2.14 \ldots$$, then $$f'(x) \geq 0$$, i.e., $$f$$ is increasing on this range. Thus, $$f(x) > \mathop {\lim }\limits_{x \to 0 + } f(x) = 0 \Leftrightarrow \arctan x > \frac{x}{{1 + \frac{2}{\pi }x}}$$ for $$0 < x \le \frac{{4\pi }}{{\pi ^2 - 4}}$$. If $$x > \frac{{4\pi }}{{\pi ^2 - 4}}$$, then $$f'(x) < 0$$, i.e., $$f$$ is decreasing on this range. Hence, $$f(x) > \mathop {\lim }\limits_{x \to + \infty } f(x) = 0 \Leftrightarrow \arctan x > \frac{x}{{1 + \frac{2}{\pi }x}}$$ for $$x > \frac{{4\pi }}{{\pi ^2 - 4}}$$.