Question

Cho x, y dương thỏa mãn : x ≥ 2y. Tìm GTNN của : \(\frac{x^2+y^2}{xy}\)

Answer 1

\(x\ge2y\Rightarrow\frac{x}{y}\ge2\)

\(P=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\frac{3x}{4y}+\frac{x}{4y}+\frac{y}{x}\ge\frac{3}{4}.2+2\sqrt{\frac{xy}{4xy}}=\frac{5}{2}\)

\(\Rightarrow P_{min}=\frac{5}{2}\) khi \(x=2y\)

Answer 2

@Nguyễn Việt Lâm ủng hộ cách khác

\(\frac{x^2+y^2}{xy}-\frac{5}{2}=\frac{2\left(x^2+y^2\right)-5xy}{2xy}\)

\(=\frac{x^2-4xy+4y^2+x^2-2y^2-xy}{2xy}\)

\(=\frac{\left(x-2y\right)^2+\left(x+y\right)\left(x-2y\right)}{2xy}\ge0\) (do \(x\ge2y\))

\(\Rightarrow\frac{x^2+y^2}{xy}\ge\frac{5}{2}."="\Leftrightarrow x=2y\)