Showing that $\sin^2x\cdot\sin^22x\cdot\sin^24x\cdot\sin^28x\cdots\sin^22^nx\leq\frac{3^n}{4^n}$

Solution 1:

We first prove that

$$ (\sin x)^4(\sin 2x)^2 \leq \left(\frac{3}{4}\right)^3. $$

Indeed, applying the double angle formula $\sin 2x = 2\sin x\cos x$ and substituting $t = \sin^2 x$, we have

$$ (\sin x)^4(\sin 2x)^2 = 4t^3(1-t) $$

and the right-hand side is maximized at $t = \frac{3}{4}$ with the value $(3/4)^3$ as desired. Now, returning to the original problem, the above inequality yields

\begin{align*} &(\sin x)^2 (\sin 2x)^2 \dots (\sin 2^n x)^2 \\ &= \Biggl[ (\sin x)^2 (\sin 2^n x)^4 \prod_{k=0}^{n-1} (\sin 2^k x)^4 (\sin 2^{k+1}x)^2 \Biggr]^{1/3} \\ &\leq \Biggl[ \prod_{k=0}^{n-1} \left(\frac{3}{4}\right)^3 \Biggr]^{1/3} \\ &= \left(\frac{3}{4}\right)^n \end{align*}

as required.