NMO Selection Tests for the Junior Balkan Mathematical Olympiad - Second Selection Test

From the given condition, we have $xyz \le 4 \left(x - \frac{1}{y}\right)$, which is equivalent to $4x \ge xyz + \frac{4}{y}$. By AM-GM, we get $4x \ge xyz + \frac{4}{y} \ge 2\sqrt{xyz \cdot \frac{4}{y}} = 4\sqrt{xz}$, so $x \ge z$.

Analogously, from $xyz \le 4 \left(y - \frac{1}{z}\right)$ and $xyz \le 4 \left(z - \frac{1}{x}\right)$, we get $y \ge x$ and $z \ge y$, so necessarily $x = y = z$.

Now the requirement holds if and only if $x^3 \le 4 \left(x - \frac{1}{x}\right)$, i.e. $(x^2 - 2)^2 \le 0$, which leads to $x = \sqrt{2}$, so $x = y = z = \sqrt{2}$ is the only solution.