Showing $\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$ for $0I was trying to solve the following exercise:

Show that for $0<x<1$ the following inequalities hold: $$\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$$

I think the Mean Value Theorem might be useful, but I don't know how to apply it.
 A: You can solve this with the mean value theorem. For some $y\in(x,1)~,~x>0$ with $f(x)=\arctan(x)$ it is true that:
$$f'(y)=\frac{f(1)-f(x)}{1-x}$$
However since the function $f'(x)=\frac{1}{1+x^2}$ is decreasing in $(0,1)$ it follows that
$$f'(1)<f'(y)=\frac{f(1)-f(x)}{1-x}<f'(x)$$
Can you finish from here doing some rearrangements?
A: You also can use integral to show the inequality. For $x\in(0,1)$, one has
$$ \frac{\pi}{4}-\arctan x=\int_x^1\frac{1}{1+t^2}dt \le\int_x^1\frac{1}{1+x^2}dt=\frac{1-x}{1+x^2} \tag1$$
and
$$ \frac{\pi}{4}-\arctan x=\int_x^1\frac{1}{1+t^2}dt\ge\int_x^1\frac{1}{2}dt=\frac12(1-x). \tag2$$
Putting (1) and (2) together, one will have the desired inequality.
A: $$
\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2} \\
-\frac{1-x}{1+x^2}<\arctan (x) - \frac{\pi}{4}<-\frac{1-x}{2} \\
\frac{1-x}{1+x^2}>\frac{\pi}{4} - \arctan (x)>\frac{1-x}{2} \\
\frac{1}{1+x^2}>\frac{\frac{\pi}{4} - \arctan (x)}{1-x}>\frac{1}{2} \\
$$
where the last step assumes $x<1$.
could you write down Taylor series for the arc-tangent around $1$?
