We are given the equations $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x}$. We want to find possible solutions for $x, y, z$.

1. Problem Description

We are given the equations $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x}$. We want to find possible solutions for $x, y, z$.

2. Solution Steps

From the given equations, we have two equations:

$x + \frac{1}{y} = y + \frac{1}{z}$ and $y + \frac{1}{z} = z + \frac{1}{x}$.

From the first equation, we have:

$x - y = \frac{1}{z} - \frac{1}{y}$

$x - y = \frac{y - z}{yz}$

From the second equation, we have:

$y - z = \frac{1}{x} - \frac{1}{z}$

$y - z = \frac{z - x}{xz}$

Substituting the second equation into the first, we have:

$x - y = \frac{\frac{z - x}{xz}}{yz}$

$x - y = \frac{z - x}{xyz^2}$

If $x = y$, then $x + \frac{1}{y} = x + \frac{1}{x} = y + \frac{1}{z}$. Since $x = y$, we have $x + \frac{1}{z} = z + \frac{1}{x}$. This implies $x = z$.

Thus, $x = y = z$ is a solution.

If $x, y, z$ are not all equal, then $x - y \ne 0$, $y - z \ne 0$, and $z - x \ne 0$.

From $x - y = \frac{z - x}{xyz^2}$, we get:

$(x - y)xyz^2 = z - x$

Similarly, we can write:

$y + \frac{1}{z} = z + \frac{1}{x}$ implies

$y - z = \frac{1}{x} - \frac{1}{z} = \frac{z - x}{xz}$, so $(y - z)xz = z - x$

$z + \frac{1}{x} = x + \frac{1}{y}$ implies

$z - x = \frac{1}{y} - \frac{1}{x} = \frac{x - y}{xy}$, so $(z - x)xy = x - y$

Multiplying the three equations, we get:

$(x - y)(y - z)(z - x) (xyz)^2 = (z - x)(z - x)(x - y)$

$(x - y)(y - z)(z - x) (xyz)^2 = (x - y)(z - x)^2$.

Since $x, y, z$ are not all equal, $(x - y)(y - z)(z - x) \ne 0$.

So, $(xyz)^2 = 1$.

$xyz = \pm 1$.

If $xyz = 1$, then

$x - y = \frac{z - x}{z/xy z^2}=\frac{y-z}{yz}$

$(x - y) = \frac{z - x}{xyz} z = z - x$ so we get $xyz^2 = 1$

Since $(y - z)xz = z - x$, and $z - x = (x - y)xyz = x - y$

$(y - z)xz = x - y$ which gives us :

$xyz=1,$ and $(x-y) = z-x$.

We have $(y - z)xz = z - x$. If $xyz = 1$, then $z = \frac{1}{xy}$.

$x + \frac{1}{y} = y + \frac{1}{z}$

$x + \frac{1}{y} = y + xy$

$x - y = xy - \frac{1}{y}$

$x - y = \frac{xy^2 - 1}{y}$

$y + \frac{1}{z} = z + \frac{1}{x}$

$y + xy = \frac{1}{xy} + \frac{1}{x}$

$y + xy = \frac{1 + y}{xy}$

$xy^2 + x^2 y^2 = 1 + y$

When $x = 1, y = 1, z = 1$ (or $x = y = z = 1$), the equations hold.

When $x = 1, y = -1, z = -1$ $xyz=1$, equations hold.

$xyz=-1,$ then we get $(x-y) = -z+x$.

If we let $x=1, y = 2$.

$1 + \frac{1}{2} = 2 + \frac{1}{z}$ then $\frac{3}{2} = 2 + \frac{1}{z}$ means $\frac{1}{z} = \frac{3}{2} - 2 = -\frac{1}{2}$.

So $z = -2$. Then $xyz = 1 \cdot 2 \cdot (-2) = -4$. Not $\pm 1$. So, no result there.

However, when $x = y = z$, the equation $x + \frac{1}{x} = x + \frac{1}{x} = x + \frac{1}{x}$ holds.

3. Final Answer

$x=y=z$ is a solution.