[tex]3(x\sqrt{y-9}+y\sqrt{x-9})=xy\Leftrightarrow \frac{3\sqrt{y-9}}{y}+\frac{3\sqrt{x-9}}{x}=1[/tex]
Ta có: [tex]3\sqrt{x-9}\leq \frac{x-9+9}{2}=\frac{x}{2}\Rightarrow \frac{3\sqrt{x-9}}{x}\leq \frac{1}{2}[/tex]
Tương tự thì [tex]\frac{3\sqrt{y-9}}{y}\leq \frac{1}{2}\Rightarrow \frac{3\sqrt{x-9}}{x}=\frac{3\sqrt{y-9}}{y}\leq 1[/tex]
Dấu "=" xảy ra khi x = y = 18. Khi đó S = 2.