Prove that $\frac{\pi}{2}-x<\tan^{-1}(x)<\frac{\pi}{2}-x+\frac{x^3}{3}$

Question

Prove that for every $x>0$, it is true: $$\frac{\pi}{2}-x<\tan^{-1}(x)<\frac{\pi}{2}-x+\frac{x^3}{3}$$ We can split it into two statements:

1. $\frac{\pi}{2}-x<\tan^{-1}(x)$

2. $\tan^{-1}(x)<\frac{\pi}{2}-x+\frac{x^3}{3}$

After moving expressions from both sides

1. $x + \tan^{-1}(x) > \frac{\pi}{2}$

2. $x + \tan^{-1}(x) - \frac{x^3}{3}<\frac{\pi}{2}$

What should be done next? Do I have to calculate the limit as $x \to 0^{+}$? I suppose, that I also need to know derivatives of left side inequality?

Also, could it be possible, that inequality is not necessarily true fro every $x>0$? I think it may not be actually true (it's example from my textbook)

Answer 1

Here is a plot of $\arctan x$, $\frac{\pi}2 - x$ and $\frac{\pi}2-x+\frac{x^3}{3}$. From the drawing, it's clear the inequality is true for $x>x_0$, with $x_0 \approx 1$.

But is it really the inequality you want to prove? If you replace $\arctan x$ with $\mathrm{arccot}\;x$, it looks much more interesting (see 2nd plot below).