929

设实数 ${\displaystyle x,y}$ 满足 ${\displaystyle {\displaystyle x>\frac{3}{2},y>3}}$，不等式 ${\displaystyle k(2x-3)(y-3)\le 8x^3+y^3-12x^2-3y^2}$ 恒成立，求实数 ${\displaystyle k}$ 的最大值.

记 ${\displaystyle 2x-3=m>0,y-3=n>0}$, 则原不等式等价于$$\begin{array}{rl}& kmn\le 4x^2(2x-3)+y^2(y-3)=m(m^2+6m+9)+n(n^2+6n+9)\\ \Rightarrow & k\le \frac{m^2+6m+9}{n}+\frac{n^2+6n+9}{m}.\end{array}$$
而$$\begin{array}{r}\frac{m^2+6m+9}{n}+\frac{n^2+6n+9}{m}\ge \frac{12m}{n}+\frac{12n}{m}\ge 24,\text{“=”}⟺m=n=3,\end{array}$$
故 ${\displaystyle k}$ 的最大值为 ${\displaystyle 24}$.