Find all non-zero real numbers $x$ such that
\[\min \left\{ 4, x+ \frac 4x \right\} \geq 8 \min \left\{ x,\frac 1x\right\} .\]
We need to find all non-zero real numbers \( x \) such that
\[
\min \left\{ 4, x+ \frac{4}{x} \right\} \geq 8 \min \left\{ x, \frac{1}{x} \right\}.
\]

To solve this, we will consider different cases based on the values of \( x \) and analyze the inequality in each case.

### Case 1: \( x \leq -1 \)
In this case, \( \min \left\{ 4, x + \frac{4}{x} \right\} = x + \frac{4}{x} \) and \( \min \left\{ x, \frac{1}{x} \right\} = x \) because \( x \leq -1 \).

The inequality becomes:
\[
x + \frac{4}{x} \geq 8x.
\]

Rearranging terms, we get:
\[
x + \frac{4}{x} - 8x \geq 0 \implies -7x + \frac{4}{x} \geq 0.
\]

Multiplying both sides by \( x \) (note \( x \) is negative, so the inequality sign flips):
\[
-7x^2 + 4 \leq 0 \implies 7x^2 \geq 4 \implies x^2 \geq \frac{4}{7}.
\]

Since \( x \leq -1 \), we have:
\[
x \in (-\infty, -1].
\]

### Case 2: \( -1 \leq x < 0 \)
In this case, \( \min \left\{ 4, x + \frac{4}{x} \right\} = x + \frac{4}{x} \) and \( \min \left\{ x, \frac{1}{x} \right\} = \frac{1}{x} \) because \( x < 0 \).

The inequality becomes:
\[
x + \frac{4}{x} \geq \frac{8}{x}.
\]

Rearranging terms, we get:
\[
x + \frac{4}{x} - \frac{8}{x} \geq 0 \implies x - \frac{4}{x} \geq 0.
\]

Multiplying both sides by \( x \) (note \( x \) is negative, so the inequality sign flips):
\[
x^2 - 4 \leq 0 \implies x^2 \leq 4.
\]

Since \( -1 \leq x < 0 \), we have:
\[
x \in [-1, 0).
\]

### Case 3: \( 0 < x \leq 1 \)
In this case, \( \min \left\{ 4, x + \frac{4}{x} \right\} = 4 \) and \( \min \left\{ x, \frac{1}{x} \right\} = x \) because \( 0 < x \leq 1 \).

The inequality becomes:
\[
4 \geq 8x.
\]

Rearranging terms, we get:
\[
x \leq \frac{1}{2}.
\]

Since \( 0 < x \leq 1 \), we have:
\[
x \in (0, \frac{1}{2}].
\]

### Case 4: \( 1 \leq x \leq 2 \)
In this case, \( \min \left\{ 4, x + \frac{4}{x} \right\} = 4 \) and \( \min \left\{ x, \frac{1}{x} \right\} = \frac{1}{x} \) because \( 1 \leq x \leq 2 \).

The inequality becomes:
\[
4 \geq \frac{8}{x}.
\]

Rearranging terms, we get:
\[
4x \geq 8 \implies x \geq 2.
\]

Since \( 1 \leq x \leq 2 \), the only solution in this range is:
\[
x = 2.
\]

### Case 5: \( x \geq 2 \)
In this case, \( \min \left\{ 4, x + \frac{4}{x} \right\} = 4 \) and \( \min \left\{ x, \frac{1}{x} \right\} = \frac{1}{x} \) because \( x \geq 2 \).

The inequality becomes:
\[
x + \frac{4}{x} \geq \frac{8}{x}.
\]

Rearranging terms, we get:
\[
x + \frac{4}{x} - \frac{8}{x} \geq 0 \implies x - \frac{4}{x} \geq 0.
\]

Multiplying both sides by \( x \) (note \( x \) is positive):
\[
x^2 - 4 \geq 0 \implies x^2 \geq 4.
\]

Since \( x \geq 2 \), we have:
\[
x \in [2, +\infty).
\]

Combining all the cases, the solution set is:
\[
x \in (-\infty, 0) \cup (0, \frac{1}{2}] \cup [2, +\infty).
\]

The final answer is \(\boxed{x \in (-\infty, 0) \cup (0, \frac{1}{2}] \cup [2, +\infty)}\).