9.8 Find the region composed of $(x, y)$ that satisfies the inequality $\log _{x} y \geqslant \log _{y} \frac{x}{y}(x y)$.
［Solution］ Clearly, we have $x>0, y>0, x \neq 1, x \neq y$. Let
$$u=\log _{x} y$$

Then $u \neq 1$, and
$$\log _{y}(x y)=\frac{\log _{x}(x y)}{\log _{x} \frac{x}{y}}=\frac{1+u}{1-u}$$

Clearly, $u \neq \frac{1+u}{1-u}$, and
$$u>\frac{1+u}{1-u} \Leftrightarrow u-\frac{1+u}{1-u}=\frac{u^{2}+1}{u-1}>0 \Leftrightarrow u>1$$

Therefore, the region formed by $(x, y)$ satisfying $\log _{x} y \geqslant \log _{\frac{x}{y}}(x y)$ is
$$\{(x, y) \mid 01 \text { and } y>x\} \text { . }$$