Umm. This comes from Diophantine quartic equation in four variables and will finish the most important part if it can be done.
Four positive integers $w,x,y,z.$ One equation and two inequalities $$ wxyz = (w+x+y+z)^2, $$ $$ w \geq x \geq y \geq z \geq 1, $$ $$ xyz \geq 2(w+x+y+z). $$
I am hoping for an upper bound. Since i made $w$ biggest, it would be an UPPER BOUND on $w.$ For example, I am running a computer program to find all such quadruples with $w \leq 1000.$
Sample question: is it true that $w \leq 1000?$
This is the method of Hurwitz 1907. I have a pdf. His techniques are almost right for this problem, to the point where i am already convinced that the answer to the question by hardmath comes out the exact same way.
EDIT, these imply easily that $$\color{green}{ x+y+z \geq w}. $$ Could be useful.
EDIT: almost forgot, these are what I believe to be all such quadruples:
w x y z xyz 2(w+x+y+z)
4 4 4 4 64 32
6 6 3 3 52 36
8 5 5 2 50 40
10 10 9 1 90 60
12 6 4 2 48 48
15 10 3 2 60 60
18 9 8 1 72 72
21 14 6 1 84 84
30 24 5 1 120 120
Indeed, one can get an upper bound and resolve this equation in general. The cheapest way is to denote $x+y+z=\alpha w$ and note that $\alpha\le 3.$ Plugging this into our equations we end up with:
$$xyz=(\alpha+1)^2w$$ and $$xyz\ge 2(\alpha+1)w.$$ Combining latter, we get $\alpha\ge 1.$ Recall that $x+y+z=\alpha w,$ so $$\frac{xyz}{x+y+z}=\frac{(\alpha+1)^2}{\alpha}\le 3+\frac{1}{3}+2=\frac{16}{3},$$ when $1\le \alpha\le 3.$ So we have the bound $xyz\le \frac{16}{3}(x+y+z)\le \frac{16}{3}(x+x+x)=16x.$ So $yz\le 16.$ One can then proceed and find all solutions.