1531

已知 $x > 0$ , $y > 0$ , $x^{2} + y^{2} = x y (x^{2} y^{2} + 2)$ , 求 $\frac{1}{x} + \frac{1}{y}$ 的最小值

解

将条件整理为 $(x + y)^{2} = x y (x^{2} y^{2} + 4) \Rightarrow x + y = \sqrt{x y (x^{2} y^{2} + 4)},$ 从而 $\frac{1}{x} + \frac{1}{y} = \frac{x + y}{x y} = \sqrt{\frac{x^{2} y^{2} + 4}{x y}} \geq \sqrt{\frac{4 x y}{x y}} = 2.$