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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.

Orion Pax
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    How can it be proven without using its derivative if it says you can use Taylor series? – battletwink69 Jul 10 '19 at 08:11
  • Taylor Series involve derivatives. I've posted an answer that avoids use of derivatives and relies on only elementary (i.e., pre-calculus) analysis. Using that analysis shows that $$\frac1{\arctan(x)}-\frac1x<\frac x2$$Interestingly, using derivatives, we can show $$\frac1{\arctan(x)}-\frac1x<\frac x3$$ – Mark Viola Jul 10 '19 at 20:22
  • @battletwink69 Sorry for the misleading question. I have updated the problem. – Orion Pax Jul 13 '19 at 06:02

3 Answers3

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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.

Adam Latosiński
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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

$$\frac1{\arctan(x)}-\frac1x\le \frac x3$$

Mark Viola
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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 $$f(0)=0$$

Bernard
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Dr. Sonnhard Graubner
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