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I was trying to prove this inequality $\left|x\right| < \left|\tan(x)\right|$ in a neighborhood of $0$. I tried splitting into the four cases opening the modulus but still wasnt able to solve it. I dont know if it is possible using MVT or other calculus theorems. Sketching a graph I see it is true but how to prove it analitically?

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  • $\begingroup$ I assume you mean when $x\neq0$? I quick analytic proof is to note that by symmetry it is enough to look at when $x>0$. Then notice that $sec^2(x)>1$ for $\delta>x>0$. $\endgroup$ – Leon Sot Jan 29 '16 at 7:27
  • $\begingroup$ See this guide for how to mark up math nicely on this site. $\endgroup$ – Frentos Jan 29 '16 at 7:30
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This isn't a formal proof, but its a nice geometrical relation that shows the identity holds.


Edit: I'd like to thank Yves Daoust for pointing a flaw in my reasoning.

The proper way to view this image as evidence that the identity holds is to compare areas. Thea area of a sector of the unit circle subtended by an angle $\theta$ is $\frac{1}{2}\theta$. The are the right triangle with side lengths $(1,\tan(\theta),\sec(\theta))$ is $\frac{1}{2} \tan(\theta)$. Visually it is clear that the area of the triangle exceeds the area of the sector, which suggests that $\tan(\theta)>\theta$.


enter image description here

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  • $\begingroup$ I apologize for the sloppy image, I'll add a nicer one later when I have the tools available to make one at hand. $\endgroup$ – Spencer Jan 29 '16 at 8:08
  • $\begingroup$ Sorry but I don't see how this figure establishes a relation between $\theta$ and $\tan(\theta)$. $\endgroup$ – Yves Daoust Jan 29 '16 at 8:15
  • $\begingroup$ $\theta$ is the labeled arc length along the circle. Compare this length with the line segment which has a length equal to $\tan(\theta)$. I'll try to edit the image to make this more clear. $\endgroup$ – Spencer Jan 29 '16 at 8:16
  • $\begingroup$ How do you compare a curve and a straight line ? I can agree for the chord, but not for the arc. (Changing the figure won't help.) $\endgroup$ – Yves Daoust Jan 29 '16 at 8:17
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    $\begingroup$ +1 for the geometrical approach. FWIW here is a related proof of mine using lengths instead of areas. $\endgroup$ – dxiv Jan 29 '16 at 23:48
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We start from

$$\frac1{x^2+1}<1$$ for $x\in\left(0,\dfrac\pi2\right)$, and integrate

$$\int_0^x\frac{dt}{t^2+1}=\arctan(x)<\int_0^xdt=x.$$

As the tangent is a strictly increasing function, we have both

$$\arctan(x)<x$$ and $$x<\tan(x).$$

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Suppose $0<x<\pi/2$; by the mean value theorem, there exists $c\in(0,x)$ such that $$ \frac{\tan x}{x}=\frac{1}{\cos^2c} $$ so $$ \frac{x}{\tan x}=\cos^2c<1 $$ that yields $x<\tan x$.

The case $-\pi/2<x<0$ is now easy.

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  • $\begingroup$ That's probably the answer expected by the OP. $\endgroup$ – Yves Daoust Jan 29 '16 at 10:42
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$\tan x=\sin x/\cos x$ and so it's odd. Suppose you prove $x<\tan x$ for $0<x<a$; it follows than $|x|<|\tan x|$ for $-a<x<0$. We have $d(x-\tan x)/dx=-\tan^2 x<0$, so $x-\tan x< 0-\tan 0=0$ and $\tan x$ is $> x$ for small positive $x$.

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Preliminary observation: The inequality you say is not strict for $x=0$. We shall therefore prove that it is strict for $x\in(-\varepsilon,0)\cup(0,\varepsilon)$.

Since both functions are odd, you just need to consider the case $x>0$.

$$\lim_{x\to 0}\frac{\tan x-x}{x^3}\stackrel{\text{l'H}\hat{\text{o}}\text{pital}}=\lim_{x\to 0}\frac{1+\tan^2x-1}{3x^2}=\lim_{x\to 0}\frac13\left(\frac{\tan x}{x}\right)^2=\frac13$$

This implies that, for sufficiently small $\varepsilon>0$ and for every $x\in(0,\varepsilon)$, it holds $$\frac{\tan x-x}{x^3}>\frac16$$

Hence, for $x\in (0,\varepsilon)$, it holds $$\tan x>x+\frac{x^3}{6}>x$$.

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