9. (14 points) For two real numbers $a, b$, $\min \{a, b\}$ represents the smaller of $a, b$. Find all non-zero real numbers $x$ such that
$$
\min \left\{x+\frac{4}{x}, 4\right\} \geqslant 8 \min \left\{x, \frac{1}{x}\right\} .
$$
When $x>0$, $x+\frac{4}{x} \geqslant 2 \sqrt{x \cdot \frac{4}{x}}=4$; when $x>0$; \\ 
$x+\frac{4}{x}, & x>1$; \\ 
$x, & x \leqslant-1 \text { or } 0<1$; \\ 
(1) When $x>1$, the original inequality is $4 \geqslant \frac{8}{x} \Rightarrow x \geqslant 2$. 
At this time, $x \in[2,+\infty)$. 
(2) When $0<x \leqslant 1$, the original inequality is 
$4 \geqslant 8 x \Rightarrow x \leqslant \frac{1}{2}$. 
At this time, $x \in\left(0, \frac{1}{2}\right]$. 
(3) When $-1<x<0$, the original inequality is $x+\frac{4}{x} \geqslant \frac{8}{x} \Leftrightarrow x^{2} \leqslant 4$. 
At this time, $x \in(-1,0)$. 
(4) When $x \leqslant-1$, the original inequality is 
$x+\frac{4}{x} \geqslant 8 x \Leftrightarrow x^{2} \geqslant \frac{4}{7}$. 
At this time, $x \in(-\infty,-1]$. 
In summary, the range of $x$ that satisfies the problem is $(-\infty, 0) \cup\left(0, \frac{1}{2}\right] \cup[2,+\infty)$.