$\frac{xy+1}{yz+1}=\frac{x+1}{y+1} \to \frac{y(x-z)}{yz+1}=\frac{x-y}{y+1}$ same way $\frac{z(y-x)}{zx+1}=\frac{y-z}{z+1},\frac{x(z-y)}{xy+1}=\frac{z-x}{x+1}$ If we multiply it, then we get $ \frac{xyz(x-z)(y-x)(z-y)}{(xy+1)(yz+1)(zx+1)}= \frac{(x-y)(y-z)(z-x)}{(x+1)(y+1)(z+1)}$ or $(x-y)(y-z)(z-x) \Big( \frac{xyz}{(xy+1)(yz+1)(zx+1)}+\frac{1}{(x+1)(y+1)(z+1)} \Big)=0 \to (x-y)(y-z)(z-x)=0$ If $x=y$ then $x=z$ so all they have to be equal

After multiplying we get : xyz + xy + z + 1 = x^2*z + xz + x + 1 xyz + yz + x + 1 = x*y^2 + xy + y + 1 xyz + xy + y + 1 = y * z^2 + yz + z + 1 Summing all equations and after canceling we get: 3*xyz = z* x^2 + x*y^2 + y*z^2 Since x, y and z are positive real numbers we have: z*x^2 + x*y^2 + y*z^2 >= 3* xyz ( By AM - GM inequality ), so x = y = z.

$$xyz + yz + x + 1 = xy^2 + xy + y + 1, xyz + xy + y + 1 = y z^2 + yz + z + 1,xyz + xy + z + 1 =z x^2 + zx + x + 1$$Summing all equations and after canceling we get $$ xy^2 + yz^2+zx^2=3xyz$$By AM - GM , we have $$xy^2 + yz^2+zx^2\geq 3xyz$$So $ x = y = z.$