If x=y=z then $x=y=x=\pm \frac{1}{\sqrt{2}}$. If not all three of them are equal then x>y or y>z or z>x. Suppose x>y (other cases similar). Divide equation 2 by eq 3: $\frac{\frac{1}{yz}}{\frac{1}{zx}}=\frac{\frac{y}{x}+1}{\frac{z}{y}+1}$ which is equivalent to $\frac{x^{2}}{y^{2}}=\frac{y+x}{y+z}$ Well, x>y and therefore y+x>y+z, yielding x>z. Now repeat same reasoning with the first and second equation. This results in z<y. Same reasoning again with first and third equation gives us x<y which contradicts our assumption. Done.

Beat wrote: Solve in real numbers the system of equations: $\frac{1}{xy}=\frac{x}{z}+1$ (1) $\frac{1}{yz}=\frac{y}{x}+1$ (2) $\frac{1}{zx}=\frac{z}{y}+1$ (3) If x<0. If y>0 .(1)$\rightarrow{z>0}$;(3)$\rightarrow$ Left hand <0<Right Hand So y<0.(2)$\rightarrow{z<0}$.Thus x,y,z<0 If x>0,similarly ,we have x,y,z>0 We consider x,y,z>0.Suppose $x\geq{y}$ From (2),(3) we have :$\frac{x^{2}}{y^{2}}=\frac{x+y}{y+z}$ thus$x\geq{z}$ On the other hand ,from(1),(2),similarly,we have $\frac{z^{2}}{x^{2}}=\frac{x+z}{y+x}$ thus $y\geq{z}$ So $x\geq{y}\geq{z}$ From(1) and (3),we have $\frac{z^{2}}{y^{2}}=\frac{x+z}{y+z}$ Thus $x\leq{y}$.So x=y=z .We easily have :$x=y=z=\pm{\frac{1}{\sqrt{2}}}$ (Consider x,y,z<0)

If we examine the signs of \( x, y, z \) on L.H.S. and R.H.S., we can only have all \( x, y, z \) either positive or negative. i.) Say all \( x, y, z \) are positive and \( x \geq y \geq z \) (similarly it works for all permutations). From the first equation we get \( \frac{1}{xy} \geq 2 \implies xy \leq \frac{1}{2} \). Now from the second equation we get \( y \geq x \) as \( \frac{1}{2} \geq xy \geq yz \). This gives us \( x = y = z = \frac{1}{\sqrt{2}} \). ii.) If all \( x, y, z \) are negative, we put \( x = -x_0, y = -y_0, z = -z_0 \) to get the same system where \( x_0, y_0, \) and \( z_0 \) are positive reals. This gives us \( x = y = z = -\frac{1}{\sqrt{2}} \). So, the only solutions to the system are \[ \boxed{\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)} \text{ and } \boxed{\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)} \].

WLOG, $$ \begin{gathered} x≤y \leq z \\ \frac{1}{x y}=\frac{x}{z}+1 \\ \Rightarrow \frac{z}{x y}=x+z-(i) \end{gathered} $$ Similarly, $$ \begin{aligned} & \rightarrow \frac{y}{z x}=z+y-(ii) \\ & \Rightarrow \frac{x}{y z}=x+y-(iii) \end{aligned} $$ Considering $(i)+(ii)+(iii)$, we get, $$ \begin{aligned} & \frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}=2(x+y+z) \\ \Rightarrow & \frac{x^2+y^2+z^2}{2 x y z}=x+y+z \\ \Rightarrow & \frac{x^2+y^2+z^2}{2 x y z}=x+\frac{y}{z x} \\ \Rightarrow & x^2-y^2+z^2=2 x^2 y z \\ \Rightarrow & x^2=\frac{z^2-y^2}{2 z y-1} \end{aligned} $$ Now, $$ \begin{aligned} & x^2 \geqslant 0 \\ & z-y \geqslant 0 \end{aligned} $$ Similarly, $$ y-x \geqslant 0 $$ But, $$ x-z \leq 0 $$ Hence all must be equal $$ \begin{aligned} & \frac{1}{x x}=\frac{x}{x}+1 \\ & \Rightarrow x=±\frac{1}{\sqrt{2}}=y=z \end{aligned} $$

The red thing is WLOG, x≤y≤z z/xy=x+z-(i)...then continue