<s>[INST] Trig function bounded on interval (without calculus), prove that $x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$. 
If $0 \lt x \lt \dfrac{\pi}{2}$, prove that 
  $$x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$$

This question must be done without calculus. First, I tried splitting it into the intervals $(0,\pi/4)$ and $(\pi/4, \pi/2)$, hoping that, $\sin x$ was bound tightly enough on the interval that it'd be less than 3 even if $\cos x = 1$ (which doesn't work -- letting $\sin x = \dfrac{1}{\sqrt{2}}$ and $\cos x = 1$ produces a result greater than 3).
The other thing I noticed was that inside the square root sign, we have $\sqrt{9-x^3} = \sqrt{(3-x^{3/2})(3+x^{3/2})}$, and an $x^{3/2}$ appears in the first term, but I'm not sure how useful the similarity there is.
Advice on how to proceed?
 [/INST] By Cauchy Schwarz $$ 9 = \left(x^3+(9-x^3) \right) \left(\sin^2 x + \cos^2 x\right) \ge \left(x^{3/2}\sin x + \sqrt{9-x^3}\cos x\right)^2$$ and the result follows
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