how can I prove that $\frac{\arctan x}{x }< 1$? I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.
 A: We want to show that $\arctan(x) \leq x$ for all positive x (or vice-versa for negative x). Notice that at $x=0$, we can evaluate $\arctan(x) = 0$, so the functions are equal. Now, the derivative of $\arctan$ is $1/(1+x^2) < x' = 1$, and paired with our former observation, by a well-known theorem from calculus, this means that $\arctan(x) \leq x$ for positive x. (By the same theorem, this is in fact a strict inequality).
A: idm's solution written differently...  
For $x \ge 0$,
$$
\arctan x = \int_0^x \frac{dt}{1+t^2}
$$
(in some texts, this may even be the definition)
For $t>0$,
$$
\frac{1}{1+t^2} < 1
$$
so we conclude
$$
\arctan x \le \int_0^x 1\,dt = x
$$
with strict inequality except for $x=0$.  
Similar arguments when $x<0$.  I assume you do not intend to prove something for complex $x$...  For example, $|\arctan x| > |x|$ for $x$ purely imaginary (and nonzero).
A: Hint
Try the function $f(x)=\arctan{x} - x$. It's derivative is $\frac{-x^2}{1+x^2}$. Now use the monotonicity to get the required inequality.
A: $$\frac{\arctan x}{x}<1\iff \begin{cases}\arctan x<x&if\ x>0\\ \arctan x> x&if\ x<0\end{cases}$$
$$(\arctan x-x)'=\underbrace{\frac{1}{x^2+1}}_{< 1\ if\ x\neq 0}-1< 0$$ therefore $$f:x\longmapsto\arctan x -x$$ is strictly decrasing. Moreover $$\arctan 0-0=0,$$ therefore $(\arctan x-x)<0$ if $x>0$ and $(\arctan x-x)>0$ if $x<0$ because $f$ is continuous on $\mathbb R$. I let you conclude.
A: A basic proof without calculus:
We assume that $x>0$. Then we have to show that
$$\arctan(x)<x \tag{1}$$
But we  can concluded this from 
$$x \lt \tan(x), \;\; \forall x \in (0,\pi/2)  \tag{2}$$
If we substitute all occurrences of $x$ by $\arctan(x)$ in $(2)$ we get
$$\arctan(x) \lt \tan(\arctan(x))  \tag{3a}$$
The rhs of $(3a)$ can be simplified to $x$ because $\arctan$ is the inverse of $\tan$ and so we get $(1)$ from $(3)$.

Another possibility is to apply $\arctan$ to the equation $(2)$. Because $\arctan$ is a strictly increasing function we get
$$\arctan(x) \lt \arctan(\tan(x)) \tag{3b}$$
and from this we get $(1)$ again.

The validness of $(2)$ can be seen in the following picture:

The blue arc is the angle $x$ , the area of the yellow segment of the circle is therefore 
$$\pi \frac{x}{2\pi}=\frac{x}{2}$$
The green line is $\tan(x)$ and the (yellow + red) area of the triangle is
$$\frac{1\cdot \tan x}{2}=\frac{\tan x}{2}$$
So we have 
$$ \frac{x}{2}<\frac{\tan x}{2} \tag{4}$$
and therefore $(2)$.
The $\arctan$ of the length of the green line is the length of the blue arc.
We see that the $\arctan$ of a value greater $0$ is a value between $0$ and $\frac{\pi}{2}$
From the picture we see that
$$\arctan(-x)=-x \tag{5}$$
and so we have 
$$\frac{\arctan(x)}{x}=\frac{\arctan(-x)}{-x} \lt 1 \tag{6}$$
if $x<0$. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
By Mean Value Theorem: $\ds{\forall\ x\ \not=\ 0\,;\quad\exists\ \xi\ \mid\ 0\ <\ \xi\ <\ \verts{x}}$
and
\begin{align}
{\arctan\pars{x} \over x}={1 \over \xi^{2} + 1} < 1
\end{align}
A: For $0\lt x \lt \frac{\pi}{2}$ we have $x\leq\tan{x}$ This leads to $\frac{\arctan{x}}{x}\leq 1$
Similarly for $0\gt x \gt -\frac{\pi}{2}$ we have $x\geq\tan{x}$ and this leads to the same inequality. 
A: Note that $\tan 0 = 0$, $\tan' 0 = 1$ and $\tan'x = {1 \over \cos^2 x} > 1$ for $0<|x|< {\pi \over 2}$.
Then $\arctan 0 = 0$; the inverse function theorem gives
 $\arctan' 0 = 1$ and $0<\arctan' x < 1$ for $x \neq 0$.
Since $\arctan x = \arctan 0 + \arctan'( \xi ) x$ for some $ \xi \in (0,x)$ (or $(x,0)$, depending on sign of $x$),
we see that $\arctan x < x$ for $x >0$ and $\arctan x > x$ for $x <0$.
A: The inequality equivalent to 
$$\tan^{-1}(x)< x$$
take the derivative for both sides to get
$$\frac{1}{1+x^2}< 1$$ 
