9. 15 Determine all positive real number tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ that satisfy the following inequalities:
$$\begin{array}{l}
\left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right) \leqslant 0 \\
\left(x_{2}^{2}-x_{4} x_{1}\right)\left(x_{3}^{2}-x_{4} x_{1}\right) \leqslant 0 \\
\left(x_{3}^{2}-x_{5} x_{2}\right)\left(x_{4}^{2}-x_{5} x_{2}\right) \leqslant 0 \\
\left(x_{4}^{2}-x_{1} x_{3}\right)\left(x_{5}^{2}-x_{1} x_{3}\right) \leqslant 0 \\
\left(x_{5}^{2}-x_{2} \dot{x}_{4}\right)\left(x_{1}^{2}-x_{2} x_{4}\right) \leqslant 0
\end{array}$$
[Solution] Let $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ be the required set of solutions and $x_{1}=\min \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}, x_{2}=\min \left\{x_{2}, x_{3}, x_{4}, x_{5}\right\}$. Since $x_{1}^{2} \leqslant x_{3} x_{5}, x_{2}^{2} \leqslant x_{3} x_{5}$ and $\left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right) \leqslant 0$, we have
$$x_{1} \leqslant x_{2}=x_{3}=x_{5} \leqslant x_{4}$$

Thus, $x_{4}^{2} \geqslant x_{1} x_{3}, x_{5}^{2} \geqslant x_{1} x_{3}$, and from $\left(x_{4}^{2}-x_{1} x_{3}\right)\left(x_{5}^{2}-x_{1} x_{3}\right) \leqslant 0$ we get
$$x_{1}=x_{2}=x_{3}=x_{5} \leqslant x_{4} .$$

Furthermore, by $\left(x_{5}^{2}-x_{2} x_{4}\right)\left(x_{1}^{2}-x_{2} x_{4}\right) \leqslant 0$, we have
$$x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$$

Other cases can be similarly proven (1) holds.
On the other hand, for any positive real number $x$, it is clear that $(x, x, x, x, x)$ is a required set of solutions. Therefore, all positive real number tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ that satisfy (1) are the solutions sought.