Are there any positive integers such that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1$? 
Are there any positive integers $x,y,z$ such that $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1?$$ Prove/Disprove.

I've plugged in random positive integers for $x,y,z$ and I have not been able to get the equation to equal $1$.
 A: Assume, that a solution exists. Notice, that in each of the summands in 
$$
\frac xy + \frac yz + \frac zx = 1
$$
has to lie in the open intervall $(0,1)$, because $x,y,z$ are required to be positive. Therefore, we have $\frac xy < 1$, i. e. $x < y$. Similarly, we conclude $y < z$ and $z < x$. Hence
$$
x < y < z < x,
$$
a contradiction. Thus, no solution exists.
A: Here is an alternative.
Multiply both sides by $xyz$
$$x^2z + xy^2 + yz^2 = xyz$$
Subtract the third term $yz^2$ from both sides
$$x^2z + xy^2 = (x - z)yz$$
The LHS is clearly positive so the RHS must be as well.  Hence $x > z$.  
Do the same for the other two terms gives: $z > y$ and $y > x$.  
These cannot all be true. 
A: By the AM-GM inequality, one has
 $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge 3\sqrt[3]{\frac{x}{y} \cdot\frac{y}{z} \cdot\frac{z}{x} }=3>1 $$
and hence it is impossible.
A: Rearrange the equality 
$x^2z+y^2x+z^2y=xyz$
Since $x,y,z>0$, we have
$x^2z+y^2x+z^2y>x^2z$
Multiply both sides with $\frac{y}{x}$
$y^2z>xyz$
And $y>x$
On the other hand,
$x^2z+y^2x+z^2y>y^2x$
Multiply both sides with $\frac{z}{y}$ and you get
$xz^2>xyz$, and $z>y$
Lastly, 
$x^2z+y^2x+z^2y>z^2y$ Multiply both sides with $\frac{x}{z}$ and get
$x^2y>xyz$ and $x>z$
So we got, $y>x, z>y, x>z$ which is clearly not possible when $x,y,z$ are positive.
A: Clearly, 2 among $x,y,z$ cannot be equal. Without loss of generality, we suppose:
$$z>y>x>0$$
Therefore, $\dfrac{z}{x}>1$, which contradicts $$\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=1$$
