Prove that $\tan x \geq x + \frac{x^3}{3}$ for $x \in [0, \frac{\pi}{2})$ I am teaching someone basic real analysis. Working through the past exams, they've found the following problem:
Prove that $\tan x \geq x + \dfrac{x^3}{3}$ for $x \in \left[0, \dfrac{\pi}{2}\right)$.
What's the simplest way to prove it? I've found two possible solutions, but both seem too complex for the course level:


*

*Use Taylor expansion of the $\tan$ function. While Taylor series were covered in the course, this specific one was not.

*Plug $x = \arctan y$, and prove that
$f(y) = y - \arctan y - \dfrac{\arctan^3 y}{3}$
is monotonically increasing through derivation.
This works, but it's also fairly tricky.
Is there a simpler solution that I am missing?
 A: The two members are equal when $x=0$. We can differentiate and get
$$\tan^2x+1\ge1+x^2$$ or
$$\tan x\ge x.$$
Again, the two members are equal when $x=0$ and
$$\tan^2x+1\ge1.$$
Below, the functions and their derivatives.


Caution: the proof works backwards, starting from $(\tan x)'\ge (x)'\land\tan(0)\ge0$.
A: A simple, if somewhat tedious, solution, using just basic notions of calculus, is to let $f(x)=\tan x-x-{x^3\over3}$ and compute the first three derivatives:
$$\begin{align}
f'(x)&=\sec^2x-1-x^2\\
f''(x)&=2\sec^2x\tan x-2x\\
f'''(x)&=4\sec^2x\tan^2x+2\sec^4x-2
\end{align}$$
We note that $f(0)=f'(0)=f''(0)=0$ and $f'''(x)\ge0$ for all $x\in[0,\pi/2)$ (since $|\sec x|\ge1$). From $f''(0)=0$ and $f'''(x)\ge0$ it follows that $f''(x)\ge0$ for all $x\in[0,\pi/2)$; from $f'(0)=0$ and $f''(x)\ge0$ it follows that $f'(x)\ge0$ for all $x\in[0,\pi/2)$; finally, from $f(0)=0$ and $f'(x)\ge0$ it follows that $f(x)\ge0$ for all $x\in[0,\pi/2)$.
A: In my view we can see that $\tan x \geq x$ geometrically in the interval $[0,\pi/2)$
From this we get of course that $\tan^2(x) \geq x^2$ and then $\tan^2(x)+1 \geq x^2+1$. After, by integration,  we have
$$
\int\limits_0^x {\left( {\tan ^2 t + 1} \right)} dt \geqslant \int\limits_0^x {\left( {1 + t^2 } \right)} dt
$$
and from this we have the required inequality

NB: we have the initial required inequality after comparing the areas of the right triangle with the "blue" leg with the area of the circular sector with "red" arc.
A: $f(x)=\tan(x)$ is an odd function fulfilling the differential equation $f'(x)=1+f(x)^2$, hence
$$ f''(x) = 2f'(x)f(x) = 2f(x)(1+f(x)^2), $$
$$ f'''(x) = 2(1+f(x)^2)^2+4 f(x)^2(1+f(x)^2), $$
$$ \ldots $$
and by induction we have that $f^{(2m)}(0)=0$ while $f^{(2m+1)}(0)\in\mathbb{N}^+$. The radius of convergence of the Maclaurin series is $\frac{\pi}{2}$ since $\tan(x)$ is a meromorphic function and $\pm\frac{\pi}{2}$ are the closest singularities (simple poles) to the origin. In particular, for any $x\in\left[0,\frac{\pi}{2}\right)$ and any $N\in\mathbb{N}$ we have
$$ \tan(x) > \sum_{k=0}^{N}\frac{f^{(2k+1)}(0)}{(2k+1)!}\,x^{2k+1} $$
with your inequality being the $N=1$ instance.
A: let $f(x)=\tan(x)-x-\frac{x^3}{3}$
$f'(x)=\sec^2(x)-1-x^2=\tan^2(x)-x^2\geq 0$, because  $\tan(x)\geq x$ for $x\in[0,\frac{\pi}{2})$ 
means $f$ is increasing, and so, see that, $f(x)\geq f(0)=0$
