Let $z\geq x,z \geq y$ $\frac{z}{xy} \geq \frac{y}{zx} \to x+z \geq z+y \to x \geq y$ $\frac{x}{yz} \geq \frac{y}{xz} \to y+x \geq z+y \to x \geq z \to x=z \to x=y=z=\frac{\sqrt{2}}{2}$

Well a different approach. Let $(\frac{x}{z},\frac{y}{x},\frac{z}{y})=(a,b,c)$ So $\frac{1}{(b+1)}\cdot (a+1)=\frac{1}{a}$ $\implies a^2+a=b+1$ Similarly $\frac{1}{(c+1)}\cdot (b+1)=\frac{1}{b}$ $\implies b^2+b=c+1$ and $\frac{1}{(a+1)}\cdot (c+1)=\frac{1}{c}$ $\implies c^2+c=a+1$ Adding all the equations, $a^2+b^2+c^2=3$ But as we know $abc=1$ and applying $A.M-G.M$ $a^2+b^2+c^2\geq 3\cdot {(abc)}^{\frac{2}{3}}=3$ As we have equality, $\implies a=b=c=1$ $\implies x=y=z$ Now putting values we get $x=y=z=\frac{1}{\sqrt{2}}$

Don't know whether the people checking the solution would consider this as a solution in RMO The system of equations becomes $\frac{1}{y}\left(\frac{1}{x^2}+1\right)=\frac{1}{z}\left(\frac{1}{y^2}+1\right)=\frac{1}{x}\left(\frac{1}{z^2}+1\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ Consider the first three equalities. We see, that the function $f(x)=\frac{1}{x}\left(\frac{1}{x^2}+1\right)$ is strictly decreasing in positive reals, and hence is injective. Thus, we conclude $x=y=z$. Putting back in the original equation, we get $x=y=z=\frac{1}{\sqrt{2}}\blacksquare$

No, it's not a correct solution

file:///C:/Users/arjun/Documents/IMG_20200930_0001.pdf check the above link to see my solution to the problem (i used nothing more than normal algebra)