Prove or disprove the inequality $\tan x \geq x + x^3$ Given the inequality $\tan x \geq x + x^3$. Prove or disprove it.
$x \in (0, \pi/2)$.
Hints would be appreciated.
 A: Let $x=\frac{\pi}{4}$.
$$\tan(x)=1$$
$$x+x^3 > 1$$
A: The MacLaurin series of $\tan$ is given by
$$\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$$
Hence taking the difference $x^3 + x - \tan(x)$ becomes
$$\frac{2x^3}{3} - \frac{2x^5}{15} - \cdots$$
You could then suspect that for small $x$ (especially when $x^3 \gg x^5$), we will have that $x^3 +x > \tan(x)$.
For example $x = \pi/6$ gives you that:


*

*$x^3 + x = \pi/6 + \pi^3/216 \approx 0.66715$

*$\tan(x) = \tan (\pi/6) = 1/\sqrt{3} \approx 0.55735$

A: Consider the functions $f(x)=\frac{\sin x}{x}$ and $g(x)=(1+x^2)\cos x$. 
They both are continuous functions on $(0,\pi/2]$ and $\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}g(x)$. 
$f(x)$ is decreasing in a right neighbourhood of zero, while $g(x)$ is increasing.
On the other hand, $f(\pi/2)>g(\pi/2)$, hence there is some $\xi\in(0,\pi/2)$ such that $f(\xi)=g(\xi)$.
By rearranging, you get that your inequality cannot hold on $(0,\pi/2)$, since $\tan(x)-(x+x^3)$ has a sign change in a neighbourhood of $\xi$.
