Let us check when $4 \le x + \frac{4}{x}$. If $x > 0$, then this inequality is equivalent to $4x \le x^2 + 4$ or $0 \le (x-2)^2$. This is always true. If $x < 0$, then we get $0 \ge (x-2)^2$. This is never the case, so $4 \ge x + \frac{4}{x}$. Hence,
$$\min \left\{ 4, x + \frac{4}{x} \right\} = \begin{cases} 4, & \text{if } x > 0, \\ x + \frac{4}{x}, & \text{if } x < 0. \end{cases}$$
For $x > \frac{1}{x}$ and $x > 0$ we get $x^2 > 1$, so $x > 1$. For $x > \frac{1}{x}$ and $x < 0$ we get $x^2 < 1$, so $-1 < x < 0$. Hence,
$$\min \left\{ x, \frac{1}{x} \right\} = \begin{cases} \frac{1}{x}, & \text{if } x > 1 \text{ or } -1 < x < 0, \\ x, & \text{if } x \le -1 \text{ or } 0 < x \le 1. \end{cases}$$
In the case of $x > 1$ the given inequality is equivalent to $4 \ge \frac{8}{x}$, which implies $x \ge 2$. The inequality holds for all numbers $x \in [2, \infty)$. If $0 < x \le 1$, then we have $4 \ge 8x$, so $\frac{1}{2} \ge x$. Thus, the inequality also holds for $x \in (0, \frac{1}{2}]$.

Let $-1 < x < 0$. We get $x + \frac{4}{x} \ge \frac{8}{x}$, or, equivalently, $x \ge \frac{4}{x}$. Multiplying by $x$ yields $x^2 \le 4$, and this holds for $-1 < x < 0$. The inequality holds for all $x \in (-1, 0)$.

Finally, let $x \le -1$. We get $x + \frac{4}{x} \ge 8x$ or $4 \le 7x^2$. Since $x \le -1$, we have $x^2 \ge 1$, so $7x^2 \ge 7 > 4$. The inequality holds for all $x \in (-\infty, -1]$ as well.

We have shown that the given inequality holds for all $x$ in $(-\infty, 0) \cup (0, \frac{1}{2}] \cup [2, \infty)$.