Algebra integral solutions

Question

How many integer solutions are there to $$\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} = 6$$

I tried to bound it using Cauchy-Schwarz and got $$6 \geq \sqrt{xy}+\sqrt{yz}+\sqrt{xz}$$

Answer 1

Using AM GM inequality you will get

$$\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\ge 3 \sqrt[3]{xyz}$$

From where manual checking can be done.

Answer 2

I think the bounds assume $x,y,z\ge 0$ and OP did not specify this in the specification. Allowing negative numbers would perhaps increase the number of solutions.

One way to explore this is to assume $x,y,z\ne 0$ and let $x=z$ which yields $$2y+\frac{x^2}{y} = 6$$ and since $x^2>0$ we must have $y>0$ which forces $y \in \{1,2,3\}$, and that in turn yields $x^2 \in \{4,4,0\}$ resulting in the solutions $(\pm 2, 1, \pm 2)$ and $(\pm 2, 2, \pm 2)$. Since $x,y,z$ are symmetric, the first of those answers yields 6 solutions, and the second yields 3.

The only class left is all different non-zero $x,y,z$. One can try to argue (not sure how yet) that if you let the solutions differ, their range will not be beyond what grouping the solutions together does, which would limit your search to one of the variables being in $\{1,2,3\}$.