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: 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,
$$\tan t = \frac{\sin t}{\cos t} < \frac t{1-\frac 2{\pi} t} \text{  for } t \in \left(0,\frac{\pi}{2}\right)$$
A: 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
\begin{equation}
\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 
}} 
\end{equation}
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 $\frac{\frac{ \pi }{2} x}{  x+\frac{\pi}{4}+\frac{\pi}{4}} < \arctan x$.
A: Hint: let $f(x):=\arctan x-\tfrac{x}{1+2x/\pi}$, whose unique maximum on $x>0$ you're welcome to locate. For $0<x\ll1$, $f(x)\sim(\tfrac13+\tfrac{2}{\pi})x^3$; for $x\gg1$,$$f(x)=\tfrac{\pi}{2}-\arctan\tfrac1x-\tfrac{\pi/2}{1+\pi/(2x)}\sim\tfrac{\pi^2-4}{4x}.$$Now sketch $f$.
A: 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 $x > \frac{{4\pi }}{{\pi ^2  - 4}}$.
