62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

From the statement, we have $(xy+1)(y+1)=(yz+1)(x+1)$, so $x{y}^2+xy+y=xyz+x+yz$. Similarly $y{z}^2+yz+z=xyz+y+zx$, and also $z{x}^2+zx+x=xyz+z+xy$. After adding these three equations, we get $x{y}^2+y{z}^2+z{x}^2=3xyz$, 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 $x=y=z$.