<s>[INST] Prove that $\vert\sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx)\vert < \left(\frac{\sqrt{3}}{2}\right)^n$ 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.

(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
 [/INST] 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.
</s>