vì x>= 2y
=> P=\(\frac{2\cdot x^2+y^2-2\cdot x\cdot y}{xy}>=\frac{2\cdot\left(2y\right)^2+y^2-2\cdot\left(2y\right)\cdot y}{2y\cdot y}\)=\(\frac{8\cdot y^2+y^2-4y^2}{2y^2}=\frac{5y^2}{2y^2}=\frac{5}{2}\)
Vậy min P=5/2
Dấu = khi x=2y
\(P=\frac{2\left(\frac{x}{y}\right)^2-2.\frac{x}{y}+1}{\frac{x}{y}}=\frac{2t^2-2t+1}{t}\text{ }\left(t=\frac{x}{y}\ge2\right)\)
\(=2t+\frac{1}{t}-2=\frac{t}{4}+\frac{1}{t}+\frac{7t}{4}-2\ge2\sqrt{\frac{t}{4}.\frac{1}{t}}+\frac{7}{4}.2-2=\frac{5}{2}\)
Dấu bằng xảy ra khi t=2 hay x=2y