下列选项正确的是（ ）
A．$\frac{\ln\pi}{\pi} < \frac{\ln 3}{3}$ B．$9\ln\frac{2}{3} < 8\ln\frac{3}{4}$ C．$\sin\frac{1}{2} < \frac{3}{2}\sin\frac{1}{3}$ D．$\sin\frac{1}{2} > \frac{\sqrt{2}}{\pi}$
解：A 可化为$\frac{\ln x}{x}$
设$f(x) = \frac{\ln x}{x}$，则$f'(x) = \frac{1 - \ln x}{x^{2}}$
$\therefore x > e$时$f(x)$单减，$0 < x < e$时$f(x)$单增
$\therefore f(\pi) > f(3)$，$f\left( \frac{2}{3} \right) < f\left( \frac{3}{4} \right)$
$\therefore \frac{\ln\pi}{\pi} > \frac{\ln 3}{3}$，故 A 错误
$\frac{\ln\frac{2}{3}}{\frac{2}{3}} < \frac{\ln\frac{3}{4}}{\frac{3}{4}}$，则$9\ln\frac{2}{3} < 8\ln\frac{3}{4}$，故 B 正确
C 选项可化为$\frac{\sin x}{x}$，$\sin\frac{1}{2} < \frac{3}{2}\sin\frac{1}{3}$
设$g(x) = \frac{\sin x}{x}$，则$g'(x) = \frac{x\cos x - \sin x}{x^{2}}$
设$t(x) = x\cos x - \sin x$，则$t'(x) = - x\sin x$
$\therefore t(x)$单减，$t(x) < t(0) = 0$
$\therefore g'(x) = \frac{t(x)}{x^{2}} < 0$
$\therefore 0 < x < \frac{\pi}{2}$时，$g(x)$单减
$\therefore g\left( \frac{1}{2} \right) < g\left( \frac{1}{3} \right)$
$\therefore \frac{\sin\frac{1}{2}}{\frac{1}{2}} < \frac{\sin\frac{1}{3}}{\frac{1}{3}}$，故 C 正确
D 选项可化为$\frac{\sin\frac{1}{2}}{\frac{1}{2}} > \frac{\frac{\sqrt{2}}{2}}{\frac{\pi}{4}} = \frac{\sin\frac{\pi}{4}}{\frac{\pi}{4}}$
$\because g\left( \frac{1}{2} \right) > g\left( \frac{\pi}{4} \right)$
$\therefore \frac{\sin\frac{1}{2}}{\frac{1}{2}} > \frac{\sin\frac{\pi}{4}}{\frac{\pi}{4}}$，故 D 正确
综上，故选：BCD.