For $x>0$, Prove that $\dfrac{x}{1+x^2}<\tan^{-1}x < x$ Looking for an elegant way to do it.
I know one way to do it, will post soon
 A: 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 $y=0$.
A: 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 $\arctan(x) < x$.
A: 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 $\arctan x$.
A: 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) $
A: $$\arctan(x) = \int_0^x {1 \over 1 + y^2} \,dy$$
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,
$${x \over 1 + x^2} < \arctan(x) < x$$
A: Mean value theorem:
$$
\frac{\arctan x - \arctan 0}{x-0} = \arctan' c\text{ for some $c$ between $0$ and $x$.}
$$
$$
\text{Therefore }\arctan x = x\arctan' c = \frac{x}{1+c^2} > \frac{x}{1+x^2}
$$
$$
\text{and }\arctan x = x\arctan' c = \frac{x}{1+c^2}<x.
$$
A: 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 $x$.
