prove $\left(\frac{1}{\arctan x}-\frac1x\right)\lt x$ for $x\gt 0$ How to prove $\left(\frac{1}{\arctan x}-\frac1x\right)\lt x$ for $x\gt 0$ without the method of using its derivative? 
It appeared as a part of a math paper. The author said it could be easily proven by its Taylor's series. But I couldn't figure it out. 
Revision: I am sorry for the confusion of using Taylor's series but not its derivative. 
The original paper said that 

Prove $$\lim \limits_{x \to 0} \left( \cfrac{1}{arctanx}-\frac{1}{x} \right) = 0$$
Substituting arctan x by its Taylor's series yields: 
$$\frac{1}{\arctan x}-\frac 1x\ = \frac{x-arctan x}{x\cdot arctan x} = \cfrac{x-\left(x-\cfrac{x^3}{3}+\cfrac{x^5}{5}\cdots\right)}{x\cdot \left(x-\cfrac{x^3}{3}+\cfrac{x^5}{5}\cdots\right)}=x\cdot\cfrac{\cfrac 13-\cfrac{x^2}{5}+\cdots}{1-\cfrac{x^2}{3}+\cdots}\longrightarrow 0$$
  as $x \longrightarrow 0$, since the limit of the fraction is $1/3$. 
the inequality can be easily extracted from the proof above. Here is another, independent, proof. Consider function $f(x)=\left(1+x^2\right)arctanx-x$. Since $f(x)=2x arctanx \gt0$ for $x\gt0$ and $f(0)=0$, we have $f(x)\gt0$ for all $x\gt0$

So, I was trying to ask how the inequality could be proved by another method rather than the "independent" method using derivative given by the author. 
Sincere apology for the incomplete question. 

You can rewrite it as $$ \frac{1}{\arctan x} < \frac{1}{x}+x$$
$$ \arctan x > \frac{x}{1+x^2}$$
Let $x=\tan y$, $y\in(0,\frac\pi 2)$. We need to show that
$$ y > \frac{\tan y}{1+ \tan^2 y}= \frac{\sin y\cos y}{\cos^2y+\sin^2 y} = \frac12\sin 2y$$
That is $$ 2y > \sin 2y$$
which is a well known fact.
