Ukrainian National Mathematical Olympiad, Third Round, Second Tour

From the statement, we have $(xy + 1)(y + 1) = (yz + 1)(x + 1)$, so $xy^2 + xy + y = xyz + x + yz$. Similarly $yz^2 + yz + z = xyz + y + zx$, and also $zx^2 + zx + x = xyz + z + xy$. After adding these three equations, we get $xy^2 + yz^2 + zx^2 = 3xyz$, or $\frac{y}{z} + \frac{z}{x} + \frac{x}{y} = 3$. But for positive real numbers $x, y, z$, by inequality between the arithmetic mean and the geometric mean, we get $\frac{y}{z} + \frac{z}{x} + \frac{x}{y} \ge 3$, and the equality is achieved only when $\frac{x}{y} = \frac{y}{z} = \frac{z}{x} = 1$, implying $x = y = z$.