Question

Prove that ${\tan }^{-1}x<x$, for all $x>0$

Answer 1

Let $f\left(x\right)={\tan }^{-1}\left(x\right)-x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x^2}-1=\frac{1-1-x^2}{1+x^2}$
$=\frac{-x^2}{1+x^2}=<0$ for all $x>0$.
$\therefore f\left(x\right)$ is strictly decreasing for all $x>0$.
$\therefore f\left(x\right)<f\left(0\right)$ for all $x>0$.
$\Rightarrow {\tan }^{-1}x-x<{\tan }^{-1}\left(0\right)-0$ for all $x>0$
$\Rightarrow {\tan }^{-1}\left(x\right)-x<0-0$ for all $x>0$
$\Rightarrow {\tan }^{-1}\left(x\right)-x<0$ for all $x>0$
$\Rightarrow {\tan }^{-1}\left(x\right)<x$ for all $x>0$
Hence, proved.