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?
 A: You can take the derivative to show that it is always negative except at $0$ where it's zero.  Though that's not obvious, so you take the derivative again to show the same thing.  Until you get something obvious, then work your way back up.
Ok, I think you got it.  So I'll add the answer now.
The first three derivatives are:
$$f'(x)=3 x^2 - 2\sin^2(x) - \tan^2(x)$$
$$f''(x)=6 x - 4\cos(x)\sin(x) - 2\sec^2(x)\tan(x)$$
$$f'''(x)=-2(5+2\cos(2x))\tan^4(x)$$
Just plug and play to see that $f(0)=0$, $f'(0)=0$, and $f''(0)=0$.
You can see that $5+2\cos(2 x)\ge 3$ and $\tan^4(x)\ge 0$.  Therefore $f'''(x)\le 0$.
As long as $f'''(x)$ is continuous, which is true up to $x&lt{\pi\over 2}$, then we have $f''(x)&lt0$ for $0&ltx&lt{\pi\over 2}$.
Rinse and repeat.  $f'(x)$ is zero at zero and then decreasing, since $f''(x)$ is negative after zero.  So $f'(x)&lt0$ for $0&ltx&lt{\pi\over 2}$.  Now $f(x)$ is zero at zero and decreasing, since $f'(x)$ is negative after zero.
Therefore $f(x)&lt0$ for $0&ltx&lt{\pi\over 2}$.
Update:
@Aryabhata asked how I got that expression for the third derivative.  It was by factoring a polynomial.  Without any messing around, you get for the third derivative:
$$6 - 4\cos^2(x) + 4\sin^2(x) - 2\sec^4(x) - 4\sec^2(x)\tan^2(x)$$
You then convert everything to sines, where here we replace $\sin(x)$ with $s$:
$$6 - 4(1-s^2) + 4 s^2 - {2\over (1-s^2)^2} - {4\over (1-s^2)}{{s^2}\over(1-s^2)}$$
Now you pull out a common numeric factor and denominator, and then expand to get:
$$-{2\over(1-s^2)^2}(7 s^4 - 4 s^6)$$
and factoring:
$$-{2 s^4\over(1-s^2)^2}(7 - 4 s^2)$$
There are some obvious trig identities we can apply to simplify that, but for the purposes of this problem, we'll stop right there.  We can already see by inspection that $f'''(x)\le 0$, which is all that is needed for the answer above.
(Note: when I did this problem originally, I replaced everything with cosines instead of sines.  That made factoring the polynomial much harder.  When I saw the answer, I realized I should have used sines instead.  So when I wrote up this answer, I used sines to make myself look smarter than I really am.  We all do that when we write up proofs, right?  We look for the shortest path, probably taking $17$ different dumb detours before writing the final elegant-looking solution.  Or maybe that's just me.)
A: 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
$$x^3 - \frac{x^5}{2}  + \frac{13x^7}{120}  - \frac{41x^9}{3024} \gt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$
