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Looking for an elegant way to do it.

I know one way to do it, will post soon

Holy cow
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7 Answers7

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Mean value theorem: $$ \frac{\arctan x - \arctan 0}{x-0} = \arctan' c\text{ for some $c$ between $0$ and $x$.} $$ $$ \text{Therefore }\arctan x = x\arctan' c = \frac{x}{1+c^2} > \frac{x}{1+x^2} $$ $$ \text{and }\arctan x = x\arctan' c = \frac{x}{1+c^2}<x. $$

Michael Hardy
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For the first inequality if you let $x=\tan(y)$ for $y\in(0\frac{\pi}{2})$ then not that:

$$\frac{\tan(y)}{1+\tan^{2}(y)}=\frac{\sin(y)}{\cos(y)}\cdot\cos^{2}(y)=\sin(y)\cos(y)=\frac{1}{2}\sin(2y)<y$$

Also the second inequality becomes:

$$\arctan(\tan(y))=y<\tan(y)$$

which follows by showing that the derivative of $\tan(y)-y$ is strictly positive in $y\in(0,\frac{\pi}{2})$ and noting that this function is $0$ at $y=0$.

Michael Hardy
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user71352
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You know that $\arctan (x)' = \frac{1}{1+x^2}$.

You can then check the sign of $\phi(x) = \arctan(x)-x$ and $\psi(x) = \arctan(x) - \frac{x}{1+x^2}$.

For example $\phi'(x) = \frac{-x^2}{1+x^2}<0$ and $\phi(0)=0$ so $\phi$ is always negative, ie $\arctan(x) < x$.

Matt B.
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Let $\displaystyle{\mathrm{f}(x)=\frac{x}{1+x^2}}$ and $\mathrm{g}(x)=\arctan x$. It is quite straightforward to show that $$\frac{\mathrm{df}}{\mathrm{d}x}-\frac{\mathrm{dg}}{\mathrm{d}x} < 0 \iff x\neq 0$$ Since $\mathrm{f}(0)=\mathrm{g}(0)=0$, and $\mathrm{f}(x)>0$ and $\mathrm{g}(x)>0$ for all $x>0$ we can conclude that $\mathrm{f}(x) < \mathrm{g}(x)$ for all $x>0$. You can do something similar by including $\mathrm{h}(x)=x$ and comparing it to $\arctan x$.

Fly by Night
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We need to show that $\dfrac{x}{1+x^2}<\tan^{-1}x< x$ for $x\in(0, \infty)$

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Since the expressions $\dfrac{x}{1+x^2}$, $\tan^{-1}x$ and $x$ are equal when $x=0$

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It is enough to show that their corresponding derivatives satisfy the Inequality.

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That is $$\dfrac{1-x^2}{(1+x^2)^2}<\dfrac{1}{1+x^2}<1$$

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Multiply throughout by $(1+x^2)^2$, To get

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$$1-x^2<1+x^2<(1+x^2)^2$$

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Note that this Inequality holds when $x>0$

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EDIT : Instead of Mean value theorem, I look at it this way,

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\color{blue}If $f(m) = g(m)$

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And $f'(x)>g'(x)$

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Then $f(x)>g(x)$ in $(m, n)$

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\color{red}Proof is simple.

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Consider $F(x) = f(x)-g(x)$

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Then $F'(x) = f'(x)-g(x)>0$, since $f'(x)>g'(x)$

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{\color{blue}\textbf{By definition}, $F(b)>F(a)$, if $b>a$ and $F(x)$ is Increasing.}

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Therefore, $F(x)>F(m)$

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$f(x)-g(x)>f(m)-g(x)$

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$f(x)>g(x) $

Holy cow
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  • This tacitly uses the mean value theorem: the standard way to show that if $f'>0$ on an interval then $f$ increases on that interval relies on the mean value theorem. For a problem like this I prefer to be explicit about that. – Michael Hardy Jul 13 '14 at 18:38
  • @Michael: this solution is very intuitive and obtained mechanically from the problem statement, by derivation. No need to reason about an extra variable $c$. –  Jul 14 '14 at 07:20
  • To be honest, I rarely think in terms of Mean value theorem (unless the the situation is in the standard form of mean value theorem). One can see that this Implication of $\textbf{Definition of an Increasing function}$ – Holy cow Jul 14 '14 at 13:56
  • colors not working, why? just why – Holy cow Jul 14 '14 at 14:05
  • The definition of an increasing function does not involve derivatives; it just says that whenever $a<b$ then $f(a)<f(b)$. Or if you like, the slopes of all of the secant lines are positive. – Michael Hardy Jul 14 '14 at 16:09
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$$\arctan(x) = \int_0^x {1 \over 1 + y^2} \,dy$$ Since the integrand is decreasing, it is minimized at $y = x$ and maximized at $y = 0$. Substituting these values into the integrand, we have $$\int_0^x {1 \over 1 + x^2} \,dy < \arctan(x) < \int_0^x 1\,dy$$ Equivalently, $${x \over 1 + x^2} < \arctan(x) < x$$

Zarrax
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For all $x>0$, $$\frac{1-x^2}{1+x^2}\frac{1}{1+x^2}<\frac{1}{1+x^2}<1.$$ Integrate from $0$ to $x$.