求助，最值问题

笨笨

mathe

这个论坛里面应该出现过，利用了
https://www.wolframalpha.com/inp ... here+x+from+0+to+pi

markfang2050

解答见附件。
由`f(x)`单调性知\[
\begin{split}\sin^2x\sin2x≤\frac{3\sqrt3}8=\left(\frac34\right)^{\frac32}\end{split}\\
\begin{split}\sin^22x\sin4x≤\frac{3\sqrt3}8=\left(\frac34\right)^{\frac32}\end{split}\\
\begin{split}\sin^24x\sin8x≤\frac{3\sqrt3}8=\left(\frac34\right)^{\frac32}\end{split}\\
.........\\
\begin{split}\sin^22^{n-1}x\sin2^nx≤\frac{3\sqrt3}8=\left(\frac34\right)^{\frac32}\end{split}\\
\]以上各式相乘得\[
g(x)=\sin^2x\sin^32x\sin^34x\cdot\cdots\cdot\sin^32^{n-1}x\sin^22^nx≤\left(\frac34\right)^{\frac32n}\]故\[
\begin{split}F(x)=\sin^3x\sin^32x\sin^34x\cdot\cdots\cdot\sin^32^{n-1}\sin^32^n=g(x)\sin x\sin2^nx≤g(x)≤\left(\frac34\right)^{\frac32n}\end{split}\]\[
F(x)^{\frac23}=\sin^2x\sin^22x\sin^24x\cdot\cdots\cdot\sin^22^{n-1}x\sin^22^nx≤\left(\frac34\right)^n\]原目标式\[
\lim_{n\to\infty}(\sin^2x\sin^22x\sin^24x\cdot\cdots\cdot\sin^22^{n-1}x\sin^22^nx)^{\frac1n}≤\frac34\]

markfang2050

解答1.见附件