9. (NET 2) $)^{\mathrm{IMO} 4}$ Find all solutions in positive real numbers $x_{i}(i=$ $1,2,3,4,5)$ of the following system of inequalities:  
$$ \begin{aligned} & \left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right) \leq 0, \\ & \left(x_{2}^{2}-x_{4} x_{1}\right)\left(x_{3}^{2}-x_{4} x_{1}\right) \leq 0, \\ & \left(x_{3}^{2}-x_{5} x_{2}\right)\left(x_{4}^{2}-x_{5} x_{2}\right) \leq 0, \\ & \left(x_{4}^{2}-x_{1} x_{3}\right)\left(x_{5}^{2}-x_{1} x_{3}\right) \leq 0, \\ & \left(x_{5}^{2}-x_{2} x_{4}\right)\left(x_{1}^{2}-x_{2} x_{4}\right) \leq 0 . \end{aligned} $$
9. Clearly $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$ is a solution. We shall show that this describes all solutions. Suppose that not all $x_{i}$ are equal. Then among $x_{3}, x_{5}, x_{2}, x_{4}, x_{1}$ two consecutive are distinct: Assume w.l.o.g. that $x_{3} \neq x_{5}$. Moreover, since $\left(1 / x_{1}, \ldots, 1 / x_{5}\right)$ is a solution whenever $\left(x_{1}, \ldots, x_{5}\right)$ is, we may assume that $x_{3} < x_{5}$. Then $x_{5}^{2} > x_{1} x_{3}$, which together with (iv) gives $x_{4}^{2} \leq x_{1} x_{3} < x_{5} x_{3}$, a contradiction. Consider next the case $x_{1} > x_{2}$. We infer from (i) that $x_{1} \geq \sqrt{x_{3} x_{5}} > x_{3}$ and $x_{2} \leq \sqrt{x_{3} x_{5}} < x_{2}$. Second solution.  $$ \begin{aligned} 0 & \geq L_{1}=\left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right)=x_{1}^{2} x_{2}^{2}+x_{3}^{2} x_{5}^{2}-\left(x_{1}^{2}+x_{2}^{2}\right) x_{3} x_{5} \\ & \geq x_{1}^{2} x_{2}^{2}+x_{3}^{2} x_{5}^{2}-\frac{1}{2}\left(x_{1}^{2} x_{3}^{2}+x_{1}^{2} x_{5}^{2}+x_{2}^{2} x_{3}^{2}+x_{2}^{2} x_{5}^{2}\right) \end{aligned} $$  and analogously for $L_{2}, \ldots, L_{5}$. Therefore $L_{1}+L_{2}+L_{3}+L_{4}+L_{5} \geq 0$, with the only case of equality $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$.