Prove that $\vert\sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx)\vert < \left(\frac{\sqrt{3}}{2}\right)^n$

Question

For $x$ in $\mathbb R^*$ and $n$ in $\mathbb N$, define

$$U_n = \sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx) = \prod_{k=0}^n\sin(2^k x)$$

Prove that $$\vert U_n\vert \leq \left(\frac{\sqrt{3}}{2}\right)^n$$

Can anyone help please.

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(Edited by @River Li) I found it was B6(c) in the 34th Putnam (1973):

B-6: On the domain $0 \le \theta \le 2\pi$:

(a) Prove that $\sin^2\theta \sin 2\theta$ takes its maximum at $\pi/3$ and $4\pi/3$ (and hence its maximum at $2\pi/3$ and $5\pi/3$).

(b) Show that $$\left|\sin^2\theta \, \Big[\sin^3(2\theta) \cdot \sin^3(4\theta) \cdots \sin^3(2^{n - 1}\theta)\Big]\, \sin (2^n\theta)\right|$$ takes its maximum at $\theta = \pi/3$. (The maximum may also be attained at other points.)

(c) Derive the inequality: $$\sin^2\theta \cdot \sin^2(2\theta) \cdot \sin^2(4\theta) \cdots \sin^2(2^n\theta)\le (3/4)^n.$$

See: The American Mathematical Monthly, Vol. 81, No. 10, Dec., 1974, (Page 1086-1095) or https://prase.cz/kalva/putnam/putn73.html

Answer 1

Hint:

Let us assume it holds for $n=k$, that is: $|U_k| \leq \left(\frac{\sqrt{3}}{2}\right)^k$

Now what we want to show for the product for $n=k+1$ reduces to $\leq \left(\frac{\sqrt{3}}{2}\right)^{k-1} \sin(2^{k}x)\sin(2^{k+1}x)$, right? If we multiply it with $\cos(2^{k}x)$, what happens?

Answer 2

Easy to see that $|U_1(x)|\leq\frac{\sqrt3}{2}$ and $|U_2(x)|\leq\frac{3}{4}$ for every $x$.

Assume that $|U_k(x)|\leq\left(\frac{\sqrt3}{2}\right)^k$ for every real $x$ and every $k\leq n$, for some $n\ge2$.

• If $|\sin{x}|\leq\frac{\sqrt3}{2}$, then $$|U_{n+1}(x)|=|\sin x|\cdot|U_n(2x)|\leq\frac{\sqrt3}{2}\cdot\left(\frac{\sqrt3}{2}\right)^n=\left(\frac{\sqrt3}{2}\right)^{n+1}$$

• If $|\sin{x}|\geq\frac{\sqrt3}{2}$, then $|\cos{x}|\leq\frac{1}{2}$ and $|\sin{x}\sin2x|=|2(1-\cos^2x)\cos{x}|\leq\frac{3}{4}$. Thus, $$|U_{n+1}(x)|=|\sin{x}\sin2x|\cdot|U_{n-1}(4x)|\leq\frac{3}{4}\cdot\left(\frac{\sqrt3}{2}\right)^{n-1}=\left(\frac{\sqrt3}{2}\right)^{n+1}$$

By induction on $n\ge1$, we are done.