# Prove the following inequalities: $\tan ^{-1}x < x$ for all $x>0$

Note: This is part 3rd of a 4 part question, split as 4 separate questions here.

Toolbox:
• (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
• (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$
• A function $f(x)$ is said to be increasing on an interval I $f(x_1) \leq f(x_2)$ whenever $x_1 < x_2$
• It is strictly increasing if $x_1 < x_2 =>f(x_1) < f(x_2)$
• A function $f$ is said to be decreasing on an interval I if $f(x_1) \geq f(x_2)$ whenever $x_1 < x_2$
• It is strictly decreasing if $x_1 < x_2 =>f(x_1) > f(x_2)$
$\tan ^{-1} x < x$ for all $x>0$
Let $f(x) =\tan ^{-1} x -x$
Step 1:
$f'(x)=\large\frac{1}{1+x^2}$$-1 \qquad x > 0 for x >0 \large\frac{1}{1+x^2} <1 \therefore f '(x) =\large\frac{1}{1+x^2}$$ -1 <0$
Step 2:
$f'(x) <0=>f(x)$ is strictly decreasing
Step 3:
$\therefore f(x) < f(0)$ for $x > 0$
=>$\tan ^{-1} x -x < 0$ for $x > 0$
=>$\tan ^{-1} x < x$ for $x > 0$