[tex]P=\frac{9}{y}+\frac{18}{x}+\frac{x}{6}-\frac{5y}{12}+2019=\frac{9}{y}+\frac{1}{4}y+\frac{18}{x}+\frac{1}{2}x-\frac{1}{3}(x+2y)+2019\geq 2\sqrt{\frac{9}{y}.\frac{1}{4}y}+2\sqrt{\frac{18}{x}.\frac{1}{2}x}-\frac{1}{3}.18+2019=3+6-9+2019=2019[/tex]
Dấu "=" xảy ra tại [tex]x=y=6[/tex]