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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.

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Hint: Let $f(x)=\arctan x -\frac{1}{x^2+1}$. Calculate $f'(x)$, simplify. (One can also produce a geometric argument, if calculus is not allowed.) – André Nicolas Oct 27 '11 at 19:17
You can use the series at the origin as well $\frac{x}{1+x^2} = x - x^3 + x^5 - x^7 + \ldots \le x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots = \arctan(x)$. – Sasha Oct 27 '11 at 20:16
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. – JavaMan Oct 27 '11 at 20:37
up vote 10 down vote accepted

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 $0<c\leq x$ so

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

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Absolutely brilliant! Thank you so much! – IQlessThan70 Oct 27 '11 at 23:48

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.

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