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.

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).
