9. (NET 2) ${ }^{\text {IMO4 }}$ Find all solutions in positive real numbers $x_{i}(i=$ $1,2,3,4,5)$ of the following system of inequalities:
$$
\begin{array}{l}
\left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right) \leq 0, \\
\left(x_{2}^{2}-x_{4} x_{1}\right)\left(x_{3}^{2}-x_{4} x_{1}\right) \leq 0, \\
\left(x_{3}^{2}-x_{5} x_{2}\right)\left(x_{4}^{2}-x_{5} x_{2}\right) \leq 0 \\
\left(x_{4}^{2}-x_{1} x_{3}\right)\left(x_{5}^{2}-x_{1} x_{3}\right) \leq 0 \\
\left(x_{5}^{2}-x_{2} x_{4}\right)\left(x_{1}^{2}-x_{2} x_{4}\right) \leq 0 .
\end{array}
$$
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 $x_{6}$ are equal. Then among $x_{3}, x_{5}, x_{2}, x_{4}, x_{1}$ two consecutive are distinct: Assume w.lo.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 $x_{3}x_{3}$. 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 $x_{2} \leq \sqrt{x_{35}}x_{2}$.
Serond 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 $L_{2}, \ldots, L_{3}$. Therefore $L_{1}+L_{2}+L_{3}+L_{4}+L_{5} \geq 0$, with the only case of equality $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$.