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How can I prove this inequality?

$\frac{x}{x^2+1}\leq \arctan(x) \, , x\in [0,1].$

Thank you so much for tips! Sorry if this is just stupid.

  • 3
    By the way, this is far from a silly question. Even if it was, I spent too many years being worried about asking silly questions in front of others, and I worrying about these things can only hinder growth! Good luck with what you are working on.2011-10-27

5 Answers 5

10

Let $x\in [0,1]$. By Mean Value Theorem there is a point $c$ between $0$ and $x$ such that

$ \frac{d}{dx}\arctan(x)|_{x=c}=\frac{\arctan(x)-\arctan(0)}{x-0}. $

But, $\arctan(0)=0$ and $\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$. Thus

$ \frac{1}{1+c^2}=\frac{\arctan(x)}{x}. $

Multiply both sides by $x$ to obtain

$ \frac{x}{1+c^2}=\arctan(x). $

But $0 so

$ \frac{x}{1+x^2}\leq\arctan(x). $

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    Absolutely brilliant! Thank you so much!2011-10-27
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Although this is more or less a variant of considering the derivative, I think it's a neater way of putting it:

Using the inequality $(1+x^2)^{-1} \le (1+s^2)^{-1}$ for all $0\le s \le x$, we obtain

$\frac{x}{1+x^2} = \int_0^x \frac{ds}{1+x^2} \le \int_0^x \frac{ds}{1+s^2} = \arctan(x)$

for all $x\ge0$.

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Put $f(x):=(1+x^2)\arctan x-x$. Then f'(x)=2x\arctan x+\frac{1+x^2}{1+x^2}-1=2x\arctan x, hence for all $x\geq 0$ we have f'(x)\geq 0 and $f$ is increasing on $\left[0,+\infty\right($. Therefore, we get for all $x\geq 0: \, f(x)\geq f(0)=0$ and $x\leq (1+x^2)\arctan x$, so $\frac x{1+x^2}\leq \arctan x$ for all $x\geq 0$.

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Let $u=\arctan(x)$. Then $u\in[0,\frac \pi 4]$. We have $x = \tan u$ and $1+x^2 = 1+\tan^2 u = sec^2 u$, so $\frac{x}{1+x^2} = \frac{\tan u}{\sec^2 u} = \sin u \cos u = \frac{\sin 2u}2$. Letting $v=2u$, you need to show that:

$\sin v \leq v$

for $v\in[0,\frac{\pi}2]$

But $v-\sin v = \int_0^v ({1-\cos t}) dt$

And $1-\cos t$ is non-negative on $[0,\frac{\pi}2]$, so $v-\sin v\geq 0$.

You could go directly from $\sin v \leq v$ to your inequality by substitution, using $v=2\arctan x$ by using that $\sin(\arctan x) = \frac{x}{\sqrt{1+x^2}}$ and $\cos( \arctan x) = \frac {1}{\sqrt{1+x^2}}$

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In general, if $f(0)=g(0)$, then you can prove that $f(x) \le g(x)$ for $0\le x\le 1$ by proving that f'(x) \le g'(x) for $0\le x\le 1$. (The validity of this fact is a consequence of the mean value theorem.) Here, the derivatives in question are $(1-x^2)/(1+x^2)^2$ and $1/(1+x^2)$, and the required inequality is easy to prove.