<s>[INST] Show $f(x) = x^3 - \sin^2{x} \tan{x} < 0$ on $(0, \frac{\pi}{2})$ This is the last of a homework problem set from Principles of Mathematical Analysis (Ch. 8 #18(a)) that I've been working/stuck on for a few days:

Define $f(x) = x^3 - \sin^2{x}\tan{x}.$
  Find out whether it is positive or negative for all $x \in (0, \frac{\pi}{2})$, or whether it changes sign. Prove your answer.

I've thought of a couple possible ways to solve it but have gotten stuck each time.


*

*Power series: This would be super easy, since the polynomial is gone and all of the other terms are negative. The problem: I'd have to calculate the $\tan$ power series and show various properties of the Bernoulli numbers. Big hassle.

*Straight up approximation: I've spent most of the time on this method, but I always overestimate, causing me to lose the strict inequality. This was using various trig identities and just basic stuff like $\sin{x} < x < \tan{x}$ on $(0, \frac{\pi}{2})$.

*Integrating and more approximating. Running into the same difficulty as above; overestimation.


I'm kind of just running around in circles at this point, probably missing something simple. Can I get a hint?
 [/INST] Power series approach works, though it requires some numerical computation. Since you seemed to be interested in asymptotics and approximations, you might find this interesting.
It is enough to show that
$$ \sin ^3 x \gt x^3 \cos x,\quad 0 \lt x \lt \frac{\pi}{2} $$
Now the power series for $\displaystyle \sin^3 x$ is
$$ \sin^3 x = x^3 - \frac{x^5}{2}  + \frac{13x^7}{120}  - \frac{41x^9}{3024}  + \frac{671x^{11}}{604800}- \frac{73x^{13}}{1140480} + \dots   $$
Note that $\displaystyle \sin^3 x = -\frac{1}{4} (\sin 3x - 3 \sin x)$ and so the $\displaystyle (2k+1)^{th}$ derivative at $\displaystyle 0$ is
$$ c_{2k+1} = \frac{(-1)^{k+1}}{4}(3^{2k+1} - 3)$$
Now for $\displaystyle k \ge 4$, it we can show that
$$ \left|\frac{c_{2k+1}}{(2k+1)!}\right| \gt \left|\frac{\pi^2}{4} \frac{c_{2k+3}}{(2k+3)!}\right|$$
(It is enough to verify for $\displaystyle k=4$, as $\displaystyle 4|c_{2k+1}| (2k+2)(2k+3) - \pi^2|c_{2k+3}|$ in an increasing function of $\displaystyle k$).
Notice that $\displaystyle \frac{c_{2k+1}}{(2k+1)!}$ is nothing but the coefficient in the Taylor series.
Thus for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$, we have that (alternating series with decreasing absolute value of the terms)
$$ \sin ^3 x \gt x^3 - \frac{x^5}{2}  + \frac{13x^7}{120}  - \frac{41x^9}{3024} $$
Similarly for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$ we have that
$$ x^3 \cos x \lt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$
Now it is easy to see that for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$ we have that
$$x^3 - \frac{x^5}{2}  + \frac{13x^7}{120}  - \frac{41x^9}{3024} \gt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$
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