Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:

*

*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.


*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2z^2 = 3xyz
$$


*Since (from 1.) the left hand side (LHS) is strictly positive, then $xyz \gt 0$


*Exploring the LHS of the equation from 1.:
$$
2x^2y^2+2x^2z^2+2y^2z^2 = (xy-xz)^2+(xy-yz)^2+(xz-yz)^2+2xyz(x+y+z)
$$
therefore
$$
x^2y^2+x^2z^2+y^2z^2 \geq xyz(x+y+z)
$$


*Replacing LHS of the last inequation by $3xyz$ (according to 2.) gives:
$$3xyz \geq xyz(x+y+z)$$  and since (from 3.) $xyz$ is positive, we can divide both  sides by $xyz$, obtaining: $$x+y+z \leq 3$$


*$xyz$ can be positive in two cases:
Case I.    $x, y, z$ are all positive. In this case $x \geq 1$, $y \geq 1$, $z \geq 1$,
so
$$
x+y+z \geq 3
$$
but according to 5. this must be an equation. Therefore all inequations for $x$, $y$ and $z$ must turn into equations, so $x=y=z=1$.
Case II.   One of $x, y, z$ is positive, two others are negative. Without loss of generality we can assume $x \gt 0$, $y \lt 0$,  $z \lt 0$
...
 A: ... then write $m = -y, n = -z$. Then $x,m,n$ are positive integers satisfying:
$$\frac{x(-m)}{-n} + \frac {x(-n)}{(-m)} + \frac {(-m)(-n)}{x} = 3$$
or
$$\frac {xm}{n} + \frac {xn}{m} + \frac {mn}{x} = 3$$
which is just our original equation. Using our previous analysis, $x=m=n=1$. Hence $y=z=-1$...
A shortcut to your analysis is to use the AM$\ge$GM inequality: for positive $x,y,z$,
$$\frac {xy}{z} + \frac {xz}{y} + \frac {yz}{x} \ge 3\sqrt[3]{xyz} \ge 3$$
as $xyz \ge1$. Equality holds iff $x=y=z=1$.
A: Hint: If $(x,y,z)$ is a solution, then show that so is $(x,-y,-z)$ and other cyclical permutations. This then severely restricts the values of $|x|, |y|, |z|$ can take after you have derived the inequality:
$$x+y+z\leq 3. $$
A: If $(x,y,z)$ is a solution then so is $(|x|,|y|,|z|)$, so without loss of generality $x,y,z>0$, and after rearranging we may assume without loss of generality that $x\geq y\geq z$. Then $\tfrac xy,\tfrac xz,\tfrac yz\geq1$, and hence
$$3=\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}> x\frac yz\geq x,$$
and similarly $y,z<3$, so $x,y,z\in\{1,2\}$. This leaves just four triplets $(x,y,z)$ to check.
