<s>[INST] 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.
 [/INST] $$
\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2} \\
-\frac{1-x}{1+x^2}<\arctan (x) - \frac{\pi}{4}<-\frac{1-x}{2} \\
\frac{1-x}{1+x^2}>\frac{\pi}{4} - \arctan (x)>\frac{1-x}{2} \\
\frac{1}{1+x^2}>\frac{\frac{\pi}{4} - \arctan (x)}{1-x}>\frac{1}{2} \\
$$
where the last step assumes $x<1$.
could you write down Taylor series for the arc-tangent around $1$?
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