Determine the real numbers $x$, $y$, $z > 0$ for which
$xyz \leq \min\left\{4(x - \frac{1}{y}), 4(y - \frac{1}{z}), 4(z - \frac{1}{x})\right\}$
1. We start with the given inequality:
   \[
   xyz \leq \min\left\{4\left(x - \frac{1}{y}\right), 4\left(y - \frac{1}{z}\right), 4\left(z - \frac{1}{x}\right)\right\}
   \]

2. Consider the first part of the minimum:
   \[
   xyz \leq 4\left(x - \frac{1}{y}\right)
   \]
   Rearrange this inequality:
   \[
   xyz \leq 4x - \frac{4}{y}
   \]
   Multiply both sides by \( y \):
   \[
   xy^2z \leq 4xy - 4
   \]
   Rearrange to isolate terms involving \( z \):
   \[
   xy^2z + 4 \leq 4xy
   \]
   \[
   xy^2z \leq 4xy - 4
   \]
   Divide both sides by \( xy \):
   \[
   yz \leq 4 - \frac{4}{x}
   \]

3. Similarly, consider the second part of the minimum:
   \[
   xyz \leq 4\left(y - \frac{1}{z}\right)
   \]
   Rearrange this inequality:
   \[
   xyz \leq 4y - \frac{4}{z}
   \]
   Multiply both sides by \( z \):
   \[
   xyz^2 \leq 4yz - 4
   \]
   Rearrange to isolate terms involving \( x \):
   \[
   xyz^2 + 4 \leq 4yz
   \]
   \[
   xyz^2 \leq 4yz - 4
   \]
   Divide both sides by \( yz \):
   \[
   xz \leq 4 - \frac{4}{y}
   \]

4. Finally, consider the third part of the minimum:
   \[
   xyz \leq 4\left(z - \frac{1}{x}\right)
   \]
   Rearrange this inequality:
   \[
   xyz \leq 4z - \frac{4}{x}
   \]
   Multiply both sides by \( x \):
   \[
   x^2yz \leq 4xz - 4
   \]
   Rearrange to isolate terms involving \( y \):
   \[
   x^2yz + 4 \leq 4xz
   \]
   \[
   x^2yz \leq 4xz - 4
   \]
   Divide both sides by \( xz \):
   \[
   xy \leq 4 - \frac{4}{z}
   \]

5. From the inequalities derived:
   \[
   yz \leq 4 - \frac{4}{x}, \quad xz \leq 4 - \frac{4}{y}, \quad xy \leq 4 - \frac{4}{z}
   \]
   We can infer that \( x = y = z \). Let \( x = y = z \).

6. Substitute \( x = y = z \) into the original inequality:
   \[
   x^3 \leq 4\left(x - \frac{1}{x}\right)
   \]
   Simplify the right-hand side:
   \[
   x^3 \leq 4x - \frac{4}{x}
   \]
   Multiply both sides by \( x \):
   \[
   x^4 \leq 4x^2 - 4
   \]
   Rearrange to form a polynomial inequality:
   \[
   x^4 - 4x^2 + 4 \leq 0
   \]
   Let \( u = x^2 \), then the inequality becomes:
   \[
   u^2 - 4u + 4 \leq 0
   \]
   Factorize the quadratic:
   \[
   (u - 2)^2 \leq 0
   \]
   Since a square is always non-negative, the only solution is:
   \[
   u - 2 = 0 \implies u = 2 \implies x^2 = 2 \implies x = \sqrt{2}
   \]

7. Therefore, the only solution is:
   \[
   x = y = z = \sqrt{2}
   \]

The final answer is \( \boxed{ x = y = z = \sqrt{2} } \).