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. 
 A: 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.
A: Using elementary (i.e., pre-calculus) analysis only in THIS ANSWER, I developed the bounds 
$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{\sqrt{1+x^2}} \le \arctan(x)\le x} \tag1$$
for $x\ge 0$.
Rearranging $(1)$, we see that for $x>0$
$$\begin{align}
\frac{1}{\arctan(x)}-\frac1x&\le \frac{\sqrt{x^2+1}-1}{x}\\\\
&= \frac{x}{\sqrt{x^2+1}+1}\\\\
&< \frac x2\\\\
&<x
\end{align}$$
as was to be shown!  

So, from $(1)$ we actually found the much tighter bound 
$$\frac1{\arctan(x)}-\frac1x < \frac x2$$

And interestingly, using derivatives, we can show that 
$$\frac1{\arctan(x)}-\frac1x\le \frac x3$$
A: Your inequalitiy simplifies to $$x^2\arctan(x)-x+\arctan(x)>0$$ let
$$f(x)=x^2\arctan(x)-x+\arctan(x)$$ then $$f'(x)=2x\arctan(x)+\frac{x^2-x^2-1+1}{x^2+1}>0$$ and $$f(0)=0$$
