WLOG $x>=z$ now from $x+\frac{1}{y} = z$ we get $\frac{1}{y} < 0$ so we have $y<0$. But if $x, y, z$ is the answer then so is $-z, -y, -x$ so $y$ can be greater than zero, contradiction.

Not only is there no real solutions, but no complex solutions either. This is because clearing the denominators gives $xy+1=yz$, $yz+1=xz$, $xz+1=xy$ and adding the three equations then gives $xy+xz+yz+3=xy+xz+yz$ and cancellation gives $3=0$, a contradiction.

Complete_quadrilateral wrote: WLOG $x>=y>=z$ now from $x+\frac{1}{y} = z$ we get $\frac{1}{y} < 0$ so we have $y<0$. But if $x, y, z$ is the answer then so is $-z, -y, -x$ so $y$ can be grater than zero, contradiction. cool solution

Complete_quadrilateral wrote: WLOG $x>=y>=z$ now from $x+\frac{1}{y} = z$ we get $\frac{1}{y} < 0$ so we have $y<0$. But if $x, y, z$ is the answer then so is $-z, -y, -x$ so $y$ can be grater than zero, contradiction. We can't assume $x\geq y\geq z$ since the expressions are not symmetric.

pggp wrote: Determine whether there exist real numbers $x$, $y$, $z$, such that \[x+\frac{1}{y}=z,\quad y+\frac{1}{z}=x,\quad z+\frac{1}{x}=y.\] Multiplying the equations with $y,z,x$ reespectivly gives us \[ xy+1=zx,\; yz+1=zx,\; zx+1=xy \]Summing these three equations up we get \[ xy+yz+zx+3=xy+yz+zx \]which is clearly impossible.

The inequality solution is easily fixed though: W.l.o.g. $x,y$ have the same sign (otherwise permute). By swapping signs, w.l.o.g. $x,y>0$. But then also $z=x+\frac{1}{y}>0$. But then the first equation yields $x>z$, the second $y>x$ and the third $z>y$, which leaves us with $x>z>y>x$, which is a fair contradiction.

pi_quadrat_sechstel wrote: Complete_quadrilateral wrote: WLOG $x>=y>=z$ now from $x+\frac{1}{y} = z$ we get $\frac{1}{y} < 0$ so we have $y<0$. But if $x, y, z$ is the answer then so is $-z, -y, -x$ so $y$ can be grater than zero, contradiction. We can't assume $x\geq y\geq z$ since the expressions are not symmetric. Thank you, I'll fix it now.

$\color{blue} \boxed{\textbf{SOLUTION}}$ These are equivalent to, $xy+1=yz \implies yz-xy=1$ $yz+1=xz \implies xz-yz=1$ $xz+1=xy \implies xy-xz=1$ Summing these gives, $0=3$ So, no real solution for these equations $\blacksquare$