TITLE: Inequality with four positive integers looking for upper bound
QUESTION [8 upvotes]: 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

REPLY [6 votes]: 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.