Let $f$ : $(-\displaystyle \frac{\pi}{2}, \frac{\pi}{2})\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f(0)=0$ and
$f'(x)\geq 1+(f(x))^{2}$ holds for all $x\displaystyle \in(-\frac{\pi}{2}, \frac{\pi}{2})$ .
Show that $$ |f(x)|\geq|\tan x| $$ holds for all $x\displaystyle \in(-\frac{\pi}{2}, \frac{\pi}{2})$
Let $g(x)=\arctan f(x)-x$ for $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Clearly we have $$ \forall\, x \in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right),\qquad g'(x)=\frac{f'(x)}{1+f^2(x)}-1\geq0 $$ So, $g$ is strictly increasing on $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, with $g(0)=0$. this proves that $$ \eqalign{0\leq x<\frac{\pi}{2}\Longrightarrow 0\leq x\leq \arctan f(x) \Longrightarrow 0\leq \tan x\leq f(x)\cr -\frac{\pi}{2}< x\leq0\Longrightarrow \arctan f(x)\leq x\leq 0 \Longrightarrow f(x)\leq \tan x\leq 0 } $$ That is $|f(x)|\geq |\tan x|$ for $x \in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.
