Trig function bounded on interval (without calculus), prove that $x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$.

Question

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?

Answer 1

For all $0\le x\le 3^{2/3}$ (which is $\ge\frac\pi2$) \begin{align}&x^{3/2}\sin x+(9-x^3)^{1/2}\cos x=\\&=\sqrt{(x^{3/2})^2+9-x^3}\left(\frac{x^{3/2}}{\sqrt{(x^{3/2})^2+9-x^3}}\sin x+\frac{(9-x^3)^{1/2}}{\sqrt{(x^{3/2})^2+9-x^3}}\cos x\right)=\\&=3\left(\frac{x^{3/2}}{3}\sin x+\frac{(9-x^3)^{1/2}}{3}\cos x\right)=3\sin\left(x+\arccos\frac{x^{3/2}}3\right)\le 3\end{align}

Answer 2

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

Answer 3

Note that $$(9-x^3)/9 +x^3/9 =1$$

Let $$(9-x^3)/9 =\sin ^2 t$$ and $$x^3/9 =\cos ^2 t$$

Thus $$ \cos t \sin x +\sin t \cos x =\sin (x+t) \le 1$$

$$ x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$$

Answer 4

It is just trig: Any expression $A\sin(\theta)+B\cos(\theta)=C\sin(\theta+\omega)$ for a phase shift $\omega$ with $\tan(\omega)=\frac{B}{A}$ and $C^2=A^2+B^2.$ Here $C=3.$

Answer 5

Let $$\alpha=\sqrt{x^3},~~~~\beta=\sqrt{9-x^3};~~~~u=\sin x,~~~~ v=\cos x; ~~~~~~~~\.$$

You may see that $\alpha,\beta,u,v>0$ when $x \in (0,\pi/2)$, and$$\alpha^2+\beta^2=9;~~~~~u^2+v^2=1.$$

Thus, by Cauchy's inequality,$$x^{3/2}\sin x + \sqrt{9-x^3}\cos x=\alpha u+\beta v \leq \sqrt{(\alpha^2+\beta^2)(u^2+v^2)}=3.$$

Till here, you have to show that the equality could hold. Notice that Cauchy's inequality require that the equality holds if and only if $$\frac{\alpha}{u}=\frac{\beta}{v}.$$

How to show this would hold?