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.

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).
