For $0 < x < \pi/2 $, prove that $x + \frac{x^{3}}{3} <\tan x $

Question

I proved as follows:

Let $f(x) = \tan x - x - \frac{x^{3}}{3}.$

Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x > \tan x - x,$ since $\\tan x > x,$ $f''(x) > 0$ so $f'(x)$ is increasing.

Since $f '(0) = 0,$ $f '(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$

So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$

But my proof seems complex. Is there any simpler way to prove this?

Answer 1

Differentiate the given inequation three times

$$1+x^2\stackrel?<\tan^2x+1,$$

$$x\stackrel?<\tan x(\tan^2x+1),$$

$$1\stackrel?<(\tan^2x+1)^2+2\tan^2x(\tan^2x+1).$$

The final one is obviously true and equality holds at $x=0$, so that the second holds, by integration. That one holds at $x=0$, so that the first holds. And as the first also holds at $x=0$, so does the given one.

You can generalize the reasoning with more terms of the Taylor expansion.

Answer 2

Perhaps this:$$f'(x)=\tan^2(x)-x^2.$$Now, use the fact that$$\left(\forall x\in\left(0,\frac\pi2\right)\right):\tan(x)>x>0.$$

Answer 3

For $0 < x < \pi/2$ we have $\tan'(x) = 1 + \tan^2(x)$ and therefore $$ \tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x 1 \, du = x \, . $$ It follows that $$ \tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x (1 +u^2) \, du = x + \frac 13 x^3 \, . $$

This process can be repeated to get better lower bounds: $$ \tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x (1 + (u + \frac {u^3}3)^2) \, du = x + \frac 13 x^3 + \frac{2}{15} x^5 + \frac{1}{54} x^7 \, . $$