Based on some computational experiments, it seems that, if $x_{0}, x_{1}, \dotsc, x_{n} \in \mathbb{Z}_{\ge 1}$ are such that \begin{equation*} \left(\sum_{i=1}^{n} x_{i}\right)^{2} = x_{0}\prod_{i=1}^{n}x_{i}, \end{equation*} then $x_{0} \le n^{2}$.
Is this bound known to be true? And (to ask a less precise question) whether or not this bound is true, where should one get started to learn methods for proving bounds like this, assuming no background beyond elementary number theory?
On
Some examples to get you started; $n=3,$ where I am printing only the fundamental solutions.
Amusing: for fundamental with $x_1 \geq x_2 \geq .... \geq x_n \geq 1,$ apparently $x_1 \leq 5n + 10.$ Maybe provable, maybe not. The gound solution with maximal $x_1$ comes out $a=1$ and $x_1 = 5n+10, x_2 = 4n + 8; x_3 = 5, x_4 = x_5 = x_6 = ... = x_n = 1$ Your equation becomes $100 (n+2)^2 = 100 ( n+2)^2 $
With $x \geq y \geq z \geq 1$ and $(x+y+z)^2 = a xyz,$ a fundamental solution also satisfies $ayz \geq 2(x+y+z).$ Each fundamental solution should lead to a tree of solution triples, but this needs proof. In Hurwitz (1907) this is Satz 5. Meanwhile, the earliest illustration is the tree of Markov Triples, https://en.wikipedia.org/wiki/Markov_number
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$ For $n=3.$ Just the fundamental solutions.
jagy@phobeusjunior:~$ ./mse
x: 1 y: 1 z: 1 a: 9 coprime
x: 2 y: 1 z: 1 a: 8 coprime
x: 3 y: 2 z: 1 a: 6 coprime
x: 3 y: 3 z: 3 a: 3 gcd(x,y,z): 3
x: 4 y: 2 z: 2 a: 4 gcd(x,y,z): 2
x: 5 y: 4 z: 1 a: 5 coprime
x: 6 y: 4 z: 2 a: 3 gcd(x,y,z): 2
x: 8 y: 4 z: 4 a: 2 gcd(x,y,z): 4
x: 9 y: 6 z: 3 a: 2 gcd(x,y,z): 3
x: 9 y: 9 z: 9 a: 1 gcd(x,y,z): 9
x: 16 y: 8 z: 8 a: 1 gcd(x,y,z): 8
x: 18 y: 12 z: 6 a: 1 gcd(x,y,z): 6
x: 25 y: 20 z: 5 a: 1 gcd(x,y,z): 5
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$n=4.$ Here we begin to see that the smallest variable, here $w,$ must be very small. In Hurwitz, the Tabelle on page 194, shows that $x_j = 1$ for $j \geq 5.$ That's an important part of how he made the table, hand calculations.
jagy@phobeusjunior:~$ ./mse
x: 1 y: 1 z: 1 w: 1 a: 16 coprime
x: 2 y: 2 z: 1 w: 1 a: 9 coprime
x: 2 y: 2 z: 2 w: 2 a: 4 gcd(x,y,z,w): 2
x: 3 y: 1 z: 1 w: 1 a: 12 coprime
x: 4 y: 2 z: 1 w: 1 a: 8 coprime
x: 4 y: 3 z: 3 w: 2 a: 2 coprime
x: 4 y: 4 z: 3 w: 1 a: 3 coprime
x: 4 y: 4 z: 4 w: 4 a: 1 gcd(x,y,z,w): 4
x: 5 y: 2 z: 2 w: 1 a: 5 coprime
x: 6 y: 2 z: 2 w: 2 a: 3 gcd(x,y,z,w): 2
x: 6 y: 3 z: 2 w: 1 a: 4 coprime
x: 6 y: 4 z: 1 w: 1 a: 6 coprime
x: 6 y: 6 z: 3 w: 3 a: 1 gcd(x,y,z,w): 3
x: 8 y: 4 z: 2 w: 2 a: 2 gcd(x,y,z,w): 2
x: 8 y: 5 z: 5 w: 2 a: 1 coprime
x: 9 y: 6 z: 2 w: 1 a: 3 coprime
x: 10 y: 5 z: 4 w: 1 a: 2 coprime
x: 10 y: 8 z: 1 w: 1 a: 5 coprime
x: 10 y: 10 z: 9 w: 1 a: 1 coprime
x: 12 y: 6 z: 4 w: 2 a: 1 gcd(x,y,z,w): 2
x: 12 y: 8 z: 3 w: 1 a: 2 coprime
x: 15 y: 10 z: 3 w: 2 a: 1 coprime
x: 18 y: 9 z: 8 w: 1 a: 1 coprime
x: 21 y: 14 z: 6 w: 1 a: 1 coprime
x: 30 y: 24 z: 5 w: 1 a: 1 coprime
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$n=5$
x: 1 y: 1 z: 1 w: 1 t: 1 a: 25 coprime
x: 2 y: 1 z: 1 w: 1 t: 1 a: 18 coprime
x: 2 y: 2 z: 2 w: 1 t: 1 a: 8 coprime
x: 3 y: 3 z: 1 w: 1 t: 1 a: 9 coprime
x: 3 y: 3 z: 2 w: 2 t: 2 a: 2 coprime
x: 4 y: 1 z: 1 w: 1 t: 1 a: 16 coprime
x: 4 y: 3 z: 2 w: 2 t: 1 a: 3 coprime
x: 4 y: 3 z: 3 w: 1 t: 1 a: 4 coprime
x: 4 y: 4 z: 4 w: 2 t: 2 a: 1 gcd: 2
x: 5 y: 2 z: 1 w: 1 t: 1 a: 10 coprime
x: 5 y: 5 z: 3 w: 1 t: 1 a: 3 coprime
x: 6 y: 2 z: 2 w: 1 t: 1 a: 6 coprime
x: 6 y: 3 z: 1 w: 1 t: 1 a: 8 coprime
x: 7 y: 4 z: 1 w: 1 t: 1 a: 7 coprime
x: 7 y: 7 z: 3 w: 3 t: 1 a: 1 coprime
x: 8 y: 2 z: 2 w: 2 t: 2 a: 2 gcd: 2
x: 8 y: 4 z: 2 w: 1 t: 1 a: 4 coprime
x: 8 y: 5 z: 5 w: 1 t: 1 a: 2 coprime
x: 9 y: 3 z: 3 w: 2 t: 1 a: 2 coprime
x: 9 y: 4 z: 3 w: 1 t: 1 a: 3 coprime
x: 9 y: 6 z: 1 w: 1 t: 1 a: 6 coprime
x: 9 y: 8 z: 4 w: 2 t: 1 a: 1 coprime
x: 10 y: 5 z: 2 w: 2 t: 1 a: 2 coprime
x: 11 y: 11 z: 9 w: 1 t: 1 a: 1 coprime
x: 12 y: 4 z: 4 w: 3 t: 1 a: 1 coprime
x: 12 y: 6 z: 2 w: 2 t: 2 a: 1 gcd: 2
x: 12 y: 6 z: 4 w: 1 t: 1 a: 2 coprime
x: 12 y: 8 z: 2 w: 1 t: 1 a: 3 coprime
x: 14 y: 7 z: 4 w: 2 t: 1 a: 1 coprime
x: 15 y: 10 z: 3 w: 1 t: 1 a: 2 coprime
x: 15 y: 12 z: 1 w: 1 t: 1 a: 5 coprime
x: 16 y: 9 z: 9 w: 1 t: 1 a: 1 coprime
x: 18 y: 12 z: 3 w: 2 t: 1 a: 1 coprime
x: 20 y: 10 z: 8 w: 1 t: 1 a: 1 coprime
x: 21 y: 12 z: 7 w: 1 t: 1 a: 1 coprime
x: 24 y: 16 z: 6 w: 1 t: 1 a: 1 coprime
x: 35 y: 28 z: 5 w: 1 t: 1 a: 1 coprime
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$n=6. $ There is still a single fundamental solution with gcd 2. All the others have $x_6=1$
x: 1 y: 1 z: 1 w: 1 t: 1 u: 1 a: 36 coprime
x: 2 y: 2 z: 1 w: 1 t: 1 u: 1 a: 16 coprime
x: 3 y: 2 z: 2 w: 2 t: 2 u: 1 a: 3 coprime
x: 3 y: 3 z: 2 w: 2 t: 1 u: 1 a: 4 coprime
x: 4 y: 3 z: 2 w: 1 t: 1 u: 1 a: 6 coprime
x: 4 y: 4 z: 1 w: 1 t: 1 u: 1 a: 9 coprime
x: 4 y: 4 z: 2 w: 2 t: 2 u: 2 a: 1 gcd: 2
x: 4 y: 4 z: 4 w: 2 t: 1 u: 1 a: 2 coprime
x: 5 y: 1 z: 1 w: 1 t: 1 u: 1 a: 20 coprime
x: 5 y: 5 z: 4 w: 4 t: 1 u: 1 a: 1 coprime
x: 6 y: 2 z: 1 w: 1 t: 1 u: 1 a: 12 coprime
x: 6 y: 3 z: 3 w: 3 t: 2 u: 1 a: 1 coprime
x: 6 y: 6 z: 3 w: 1 t: 1 u: 1 a: 3 coprime
x: 7 y: 2 z: 2 w: 1 t: 1 u: 1 a: 7 coprime
x: 8 y: 2 z: 2 w: 2 t: 1 u: 1 a: 4 coprime
x: 8 y: 4 z: 1 w: 1 t: 1 u: 1 a: 8 coprime
x: 8 y: 6 z: 6 w: 2 t: 1 u: 1 a: 1 coprime
x: 8 y: 8 z: 3 w: 3 t: 1 u: 1 a: 1 coprime
x: 9 y: 3 z: 2 w: 2 t: 1 u: 1 a: 3 coprime
x: 9 y: 3 z: 3 w: 1 t: 1 u: 1 a: 4 coprime
x: 9 y: 8 z: 2 w: 2 t: 2 u: 1 a: 1 coprime
x: 9 y: 8 z: 4 w: 1 t: 1 u: 1 a: 2 coprime
x: 10 y: 5 z: 2 w: 1 t: 1 u: 1 a: 4 coprime
x: 12 y: 4 z: 3 w: 2 t: 2 u: 1 a: 1 coprime
x: 12 y: 6 z: 2 w: 2 t: 1 u: 1 a: 2 coprime
x: 12 y: 8 z: 1 w: 1 t: 1 u: 1 a: 6 coprime
x: 12 y: 12 z: 9 w: 1 t: 1 u: 1 a: 1 coprime
x: 14 y: 7 z: 2 w: 2 t: 2 u: 1 a: 1 coprime
x: 14 y: 7 z: 4 w: 1 t: 1 u: 1 a: 2 coprime
x: 15 y: 6 z: 5 w: 2 t: 1 u: 1 a: 1 coprime
x: 15 y: 10 z: 2 w: 1 t: 1 u: 1 a: 3 coprime
x: 16 y: 8 z: 4 w: 2 t: 1 u: 1 a: 1 coprime
x: 18 y: 12 z: 3 w: 1 t: 1 u: 1 a: 2 coprime
x: 18 y: 14 z: 7 w: 1 t: 1 u: 1 a: 1 coprime
x: 20 y: 16 z: 1 w: 1 t: 1 u: 1 a: 5 coprime
x: 21 y: 14 z: 3 w: 2 t: 1 u: 1 a: 1 coprime
x: 22 y: 11 z: 8 w: 1 t: 1 u: 1 a: 1 coprime
x: 27 y: 18 z: 6 w: 1 t: 1 u: 1 a: 1 coprime
x: 40 y: 32 z: 5 w: 1 t: 1 u: 1 a: 1 coprime
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
On
Here is the list of fundamental solutions for $n=9.$ For the first time (lowest $n$), all have $ \gcd( x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9) =1.$ For you, the point of the list is that the coefficient (I call it $a$ ) occurs as the $a$ value for fundamental solutions. In time, enough inequalities may (probably will) show that all $a \leq n^2.$ With my order $x_1 \geq x_2 \geq x_3 \geq \cdots \geq x_9 \geq 1$ the other reasonable conjecture is that $x_1 \leq 5n+10.$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
1 x1: 11 x2: 11 x3: 3 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 12 x2: 5 x3: 5 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 15 x2: 15 x3: 9 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 16 x2: 11 x3: 11 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 16 x2: 4 x3: 4 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 16 x2: 9 x3: 3 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 18 x2: 16 x3: 8 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 18 x2: 6 x3: 4 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 20 x2: 10 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 20 x2: 8 x3: 5 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 22 x2: 11 x3: 4 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 27 x2: 12 x3: 9 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 28 x2: 14 x3: 8 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 30 x2: 20 x3: 3 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 36 x2: 24 x3: 6 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 49 x2: 45 x3: 5 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 55 x2: 44 x3: 5 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 5 x2: 5 x3: 5 x4: 5 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 6 x2: 6 x3: 4 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 8 x2: 4 x3: 3 x4: 3 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
1 x1: 9 x2: 4 x3: 2 x4: 2 x5: 2 x6: 2 x7: 1 x8: 1 x9: 1 a: 1
2 x1: 10 x2: 9 x3: 5 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 15 x2: 5 x3: 3 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 18 x2: 6 x3: 6 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 18 x2: 9 x3: 2 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 20 x2: 10 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 27 x2: 18 x3: 3 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 2 x2: 2 x3: 2 x4: 2 x5: 2 x6: 2 x7: 2 x8: 1 x9: 1 a: 2
2 x1: 3 x2: 3 x3: 3 x4: 3 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 5 x2: 5 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 8 x2: 6 x3: 3 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 8 x2: 7 x3: 7 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
2 x1: 9 x2: 4 x3: 4 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
3 x1: 12 x2: 2 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
3 x1: 15 x2: 5 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
3 x1: 24 x2: 16 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
3 x1: 4 x2: 3 x3: 3 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
3 x1: 8 x2: 6 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
3 x1: 9 x2: 9 x3: 3 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
4 x1: 12 x2: 3 x3: 2 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
4 x1: 16 x2: 8 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
4 x1: 4 x2: 2 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
4 x1: 5 x2: 5 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
5 x1: 35 x2: 28 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 5
6 x1: 12 x2: 4 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 6
6 x1: 21 x2: 14 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 6
6 x1: 6 x2: 3 x3: 3 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 6
8 x1: 14 x2: 7 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 8
8 x1: 4 x2: 4 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 8
8 x1: 9 x2: 8 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 8
9 x1: 7 x2: 7 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 9
10 x1: 10 x2: 2 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 10
10 x1: 8 x2: 5 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 10
11 x1: 11 x2: 4 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 11
15 x1: 5 x2: 3 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 15
18 x1: 2 x2: 2 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 18
18 x1: 9 x2: 2 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 18
24 x1: 3 x2: 2 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 24
32 x1: 8 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 32
36 x1: 4 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 36
50 x1: 2 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 50
81 x1: 1 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 81
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
S = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9
P = x1 * x2 * x3 * x4 * x5 * x6 * x7 * x8 * x9
S^2 = a * P
fundamental:
x1 >= x2 >= x3 >= x4 >= x5 >= x6 >= x7 >= x8 >= x9 >= 1 AND
a*P >= 2*x1 * S
x1: 1 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 81
x1: 2 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 50
x1: 2 x2: 2 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 18
x1: 2 x2: 2 x3: 2 x4: 2 x5: 2 x6: 2 x7: 2 x8: 1 x9: 1 a: 2
x1: 3 x2: 2 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 24
x1: 3 x2: 3 x3: 3 x4: 3 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 4 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 36
x1: 4 x2: 2 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
x1: 4 x2: 3 x3: 3 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
x1: 4 x2: 4 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 8
x1: 5 x2: 3 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 15
x1: 5 x2: 5 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 5 x2: 5 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
x1: 5 x2: 5 x3: 5 x4: 5 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 6 x2: 3 x3: 3 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 6
x1: 6 x2: 6 x3: 4 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 7 x2: 7 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 9
x1: 8 x2: 1 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 32
x1: 8 x2: 4 x3: 3 x4: 3 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 8 x2: 5 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 10
x1: 8 x2: 6 x3: 3 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 8 x2: 6 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
x1: 8 x2: 7 x3: 7 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 9 x2: 2 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 18
x1: 9 x2: 4 x3: 2 x4: 2 x5: 2 x6: 2 x7: 1 x8: 1 x9: 1 a: 1
x1: 9 x2: 4 x3: 4 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 9 x2: 8 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 8
x1: 9 x2: 9 x3: 3 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
x1: 10 x2: 2 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 10
x1: 10 x2: 9 x3: 5 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 11 x2: 4 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 11
x1: 11 x2: 11 x3: 3 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 12 x2: 2 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
x1: 12 x2: 3 x3: 2 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
x1: 12 x2: 4 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 6
x1: 12 x2: 5 x3: 5 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 14 x2: 7 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 8
x1: 15 x2: 5 x3: 3 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 15 x2: 5 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
x1: 15 x2: 15 x3: 9 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 16 x2: 4 x3: 4 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 16 x2: 8 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 4
x1: 16 x2: 9 x3: 3 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 16 x2: 11 x3: 11 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 18 x2: 6 x3: 4 x4: 3 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 18 x2: 6 x3: 6 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 18 x2: 9 x3: 2 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 18 x2: 16 x3: 8 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 20 x2: 8 x3: 5 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 20 x2: 10 x3: 2 x4: 2 x5: 2 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 20 x2: 10 x3: 4 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 21 x2: 14 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 6
x1: 22 x2: 11 x3: 4 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 24 x2: 16 x3: 2 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 3
x1: 27 x2: 12 x3: 9 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 27 x2: 18 x3: 3 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 2
x1: 28 x2: 14 x3: 8 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 30 x2: 20 x3: 3 x4: 2 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 35 x2: 28 x3: 1 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 5
x1: 36 x2: 24 x3: 6 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 49 x2: 45 x3: 5 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
x1: 55 x2: 44 x3: 5 x4: 1 x5: 1 x6: 1 x7: 1 x8: 1 x9: 1 a: 1
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Here is a solution. Thank you to Will Jagy for suggesting Vieta jumping. I had been trying Vieta jumping, but I guess I needed a bit more encouragement to push it through.
Let $x_{0} \in \mathbb{Z}_{\ge 1}$ be such that \begin{equation} \left(\sum_{i=1}^{n} x_{i}\right)^{2} = x_{0}\prod_{i=1}^{n}x_{i} \label{eq:1} \end{equation} for some positive integers $x_{1} \le \dotsb \le x_{n}$. Using Vieta jumping, we may assume, without loss of generality, that $x_{1} \le \dotsb \le x_{n} \le \sum_{i=1}^{n-1} x_{i}$. For, Equation \ref{eq:1} is equivalent to \begin{equation*} x_{n}^{2} - \left(\prod_{i=0}^{n-1}x_{i} - 2 \sum_{i=1}^{n-1}x_{i}\right) x_{n} + \left(\sum_{i=1}^{n-1}x_{i}\right)^{2} = 0. \end{equation*} Letting $x_{n}'$ be the other root of the polynomial $t^{2} - \left(\prod_{i=0}^{n-1}x_{i} - 2 \sum_{i=1}^{n-1}x_{i}\right) t + \left(\sum_{i=1}^{n-1}x_{i}\right)^{2} \in \mathbb{Z}[t]$, we have that $x_{n}x_{n}' = \left(\sum_{i=1}^{n-1}x_{i}\right)^{2}$. If $x_{n}$ is already the smaller of these roots, then we are done. Otherwise, we have a solution $(x_{0}, x_{1}, \dotsc, x_{n-1}, x_{n}')$ to Equation \ref{eq:1} with $x_{n}' < \sum_{i=1}^{n-1}x_{i} < x_{n}$. Reordering $x_{1}, \dotsc, x_{n-1}, x_{n}'$ in increasing order, and then reapplying this reduction step a finite number of times, we eventually arrive at a solution of the desired form.
Now, given a solution $(x_{0}, x_{1}, \dotsc, x_{n}) \in \mathbb{Z}_{\ge 1}^{n-1}$ to Equation \ref{eq:1} with $x_{1} \le \dotsb \le x_{n} \le \sum_{i=1}^{n-1} x_{i}$, we solve for $x_{0}$ and find that \begin{align*} x_{0} &=% \frac{\left(\sum_{i=1}^{n} x_{i}\right)^{2}}{\prod_{i=1}^{n}x_{i}} \le % 4\frac{\left(\sum_{i=1}^{n-1} x_{i}\right)^{2}}{\prod_{i=1}^{n}x_{i}} \le % 4\cdot% \frac{\sum_{i=1}^{n-1} x_{i}}{x_{n}} \cdot \frac{\sum_{i=1}^{n-1} x_{i}}{\prod_{i=1}^{n-1}x_{i}} \\ &= % 4\cdot% \sum_{i=1}^{n-1} \frac{x_{i}}{x_{n}} \cdot \sum_{i=1}^{n-1} \frac{1}{\prod_{\substack{1 \le j \le n-1 \\ j \ne i}} x_{j}} \le % 4\cdot (n-1)\cdot \frac{n-1}{\prod_{j=1}^{n-2} x_{j}}. \end{align*} Thus, if at least two of the numbers $x_{1}, \dotsc, x_{n-2}$ are $\ge 2$, or at least one of the numbers $x_{1}, \dotsc, x_{n-2}$ is $\ge 4$, we have that $x_{0} \le (n-1)^{2}$, and we are done. Otherwise, $(x_{1}, \dotsc, x_{n})$ has either the form $(1, \dotsc, 1, 1, x_{n-1}, x_{n})$ with $x_{n-1} \le x_{n} \le n-2 + x_{n-1}$, or the form $(1, \dotsc, 1, 2, x_{n-1}, x_{n})$ with $2 \le x_{n-1} \le x_{n} \le n-1 + x_{n-1}$, or the form $(1, \dotsc, 1, 3, x_{n-1}, x_{n})$ with $3 \le x_{n-1} \le x_{n} \le n + x_{n-1}$.
In each of these cases, one can work out directly that $x_{0} \le n^{2}$. ETA: For example, if $(x_{1}, \dotsc, x_{n})$ has the form $(1, \dotsc, 1, 3, x_{n-1}, x_{n})$ with $3 \le x_{n-1} \le x_{n} \le n + x_{n-1}$, then \begin{align*} x_{0} &=% \frac{(n+x_{n-1}+x_{n})^{2}}{3x_{n-1}x_{n}} \\ &=% \frac{n^{2}}{3x_{n-1}x_{n}} + \frac{2}{3}\left(\frac{1}{x_{n-1}} + \frac{1}{x_{n}}\right)n + \frac{1}{3}\left(\frac{x_{n-1}}{x_{n}} + 2 + \frac{x_{n}}{x_{n-1}}\right) \\ &\le% \frac{n^{2}}{27} + \frac{2}{3} \cdot \frac{2}{3} \cdot n + \frac{1}{3}\left(1 + 2 + \frac{n + x_{n-1}}{x_{n-1}}\right) \\ &\le% \frac{n^{2}}{27} + \frac{4}{9} n + \frac{1}{3}\left(3 + \frac{n}{3} + 1\right) \\ &\le% \frac{n^{2}}{27} + \frac{5}{9} n + \frac{4}{3} \\ &\le% n^{2} \quad \text{for $n \ge 2$}. \end{align*}
Thank you again to Will Jagy for pointing out a missed case.