每日一题[1574]递推证明

设 $n\in\mathbb N^{\ast}$，$\theta\in\mathbb R$．求证：$\left|\sin\theta\cdot \sin 2\theta\cdots \sin 2^n\theta\right|\leqslant \left(\dfrac{\sqrt 3}2\right)^n$．

解析    设 $f(x)=|\sin x|\cdot |\sin 2x|^{\frac 12}$，不等式左边为 $m$，则 $$f(x)=\sqrt[4]{4\sin^6x\cos^2x}=\sqrt[4]{\dfrac 43\cdot\sin^2x\cdot\sin^2x\cdot\sin^2x\cdot 3\cos^2x}\leqslant \sqrt[4]{\dfrac 43\left(\dfrac 34\right)^4}=\left(\dfrac {\sqrt 3}2\right)^{\frac 32},$$ 因此 $$m^{\frac 32}\leqslant |\sin\theta|\cdot \left|\sin 2\theta\cdots\sin2^{n-1}\theta\right|^{\frac 32}\cdot \left|\sin2^n\theta\right|^{\frac 12}=\prod_{k=0}^{n-1}f\left(2^k\theta\right)\leqslant \left(\dfrac {\sqrt 3}2\right)^{\frac {3n}2},$$ 从而原命题得证．