I have to prove that $ x - \frac{x^3}{3} < \arctan(x) < x - \frac{x^3}{6} $ is true if $ 0 < x \leq 1 $
I tried to convert the second inequality into the one I'm trying to prove. Since $ \arctan(x) $ is a monotonically increasing function, applying it to each part of the second inequality should not change the relationships: $$ 0 < x \leq 1 \Rightarrow \arctan(0) < \arctan(x) \leq \arctan(1) $$
Then I calculated the 3rd degree Taylor polynomial of $ \arctan(x) $ centered at $ x = 0 $ and at $ x = 1 $ to approximate the left and right sides of the inequality:
$$ \arctan(0) \approx x - \frac{x^3}{3} $$
$$ \arctan(1) \approx \frac{\pi}{4} + \frac{1}{2} (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3$$
So it's evident that $ x - \frac{x^3}{3} < \arctan(x) $ is correct. But I'm stuck trying to prove that $ \arctan(x) < x - \frac{x^3}{6} $
We have that
$$f(x)=\arctan x- x+\frac16 x^3 \implies f'(x)=\frac{x^2(x^2-1)}{2(x^2+1)}\le 0$$
with $f(0)=0$.