4. If positive real numbers $x, y$ satisfy $y>2 x$, then the minimum value of $\frac{y^{2}-2 x y+x^{2}}{x y-2 x^{2}}$ is

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答军 4 .
解析 令 $t=\frac{x}{y} \in\left(0, \frac{1}{2}\right)$, 由 $\frac{y^{2}-2 x y+x^{2}}{x y-2 x^{2}}=\frac{1-2 t+t^{2}}{t-2 t^{2}}=-\frac{1}{2}+\frac{3 t-2}{2\left(2 t^{2}-t\right)}$.令 $f(t)=-\frac{1}{2}+\frac{3 t-2}{2\left(2 t^{2}-t\right)}, t \in\left(0, \frac{1}{2}\right)$, 则 $f^{\prime}(t)=-\frac{3 t^{2}-4 t+1}{\left(2 t^{2}-t\right)^{2}}=-\frac{(3 t-1)(t-1)}{\left(2 t^{2}-t\right)^{2}}$.令 $f^{\prime}(t)=0$ 得 $t=\frac{1}{3}$ 或 $t=1$ (舍).
当 $0<t<\frac{1}{3}$ 时, $f^{\prime}(t)<0$, 当 $\frac{1}{3}<t<\frac{1}{2}$ 时, $f^{\prime}(t)>0$, 当 $t=\frac{1}{3}$ 时, $f(t)$ 取得最小值 $f\left(\frac{1}{3}\right)=4$.

Answer 4.
Analysis: Let $t=\frac{x}{y} \in\left(0, \frac{1}{2}\right)$, then $\frac{y^{2}-2 x y+x^{2}}{x y-2 x^{2}}=\frac{1-2 t+t^{2}}{t-2 t^{2}}=-\frac{1}{2}+\frac{3 t-2}{2\left(2 t^{2}-t\right)}$. Let $f(t)=-\frac{1}{2}+\frac{3 t-2}{2\left(2 t^{2}-t\right)}, t \in\left(0, \frac{1}{2}\right)$, then $f^{\prime}(t)=-\frac{3 t^{2}-4 t+1}{\left(2 t^{2}-t\right)^{2}}=-\frac{(3 t-1)(t-1)}{\left(2 t^{2}-t\right)^{2}}$. Let $f^{\prime}(t)=0$ to get $t=\frac{1}{3}$ or $t=1$ (discard).
When $0<t<\frac{1}{3}$, $f^{\prime}(t)<0$, when $\frac{1}{3}<t<\frac{1}{2}$, $f^{\prime}(t)>0$, when $t=\frac{1}{3}$, $f(t)$ reaches its minimum value $f\left(\frac{1}{3}\right)=4$.