Answer:
5272

Solution 1: Let $p=x y z$ and $q=(x+1)(y+1)(z+1)$. Then, we get
$$
p q=[x(1+y)] \cdot[y(1+z)] \cdot[z(1+x)]=(1-p)(2-p)(4-p) .
$$

Additionally, note that
$$
q-p=x y+y z+z x+x+y+z+1=(x+x y)+(y+y z)+(z+x z)+1=8-3 p .
$$

Therefore, we have $q=8-2 p$. Substituting this into our earlier equation gives us
$$
p(8-2 p)=(1-p)(2-p)(4-p) .
$$

We can rearrange this to get $(4-p)\left(2-5 p+p^{2}\right)=0$. Solving this gives us $p=4, \frac{5 \pm \sqrt{17}}{2}$. Thus, our maximum solution is $\frac{5+\sqrt{17}}{2}$, which yields an answer of 5272 . To show that such a solution exists, see Solution 2.

Solution 2: Let $r=x y z-1$. Observe that
$$
r x=x(y+y z+x y z)-(x+x y+x y z)=2 x-1 \Longleftrightarrow x=\frac{1}{2-r} .
$$

Similarly, $y=\frac{2}{4-r}$ and $z=\frac{4}{1-r}$. Therefore $8=(1+r)(1-r)(2-r)(4-r)$. This factors as $r(r-3)\left(r^{2}-3 r-2\right)=0$, so the maximum possible value for $r$ is $\frac{3+\sqrt{17}}{2}$.
Now let's check that this yields a valid solution for $x, y, z$. Let $r=\frac{3+\sqrt{17}}{2}$ and let $x=\frac{1}{2-r}, y=\frac{2}{4-r}, z=$ $\frac{4}{1-r}$. Then $x y z-1=\frac{8}{(2-r)(4-r)(1-r)}-1=1+r-1=r$. Now, we may do our above computations in reverse to get
$$
2 x-1=2 x-(2-r) x=r x=x^{2} y z-x=x(y+y z+x y z)-(x-x y+x y z) .
$$

Repeating the same thing for $y$ and $z$ yields that
$$
\left(\begin{array}{ccc}
-1 & x & 0 \\
0 & -1 & y \\
z & 0 & -1
\end{array}\right)\left(\begin{array}{l}
1 \\
2 \\
4
\end{array}\right)=\left(\begin{array}{ccc}
-1 & x & 0 \\
0 & -1 & y \\
z & 0 & -1
\end{array}\right)\left(\begin{array}{l}
x+x y+x y z \\
y+y z+x y z \\
z+x z+x y z
\end{array}\right) .
$$

However, since $x y z-1 \neq 0$, the determinant of the matrix is nonzero, so we may multiply by its inverse to find that
$$
\left(\begin{array}{l}
1 \\
2 \\
4
\end{array}\right)=\left(\begin{array}{l}
x+x y+x y z \\
y+y z+x y z \\
z+x z+x y z
\end{array}\right)
$$

Therefore this construction is valid.
10. Real numbers $x, y, z$ satisfy
$$
x+x y+x y z=1, \quad y+y z+x y z=2, \quad z+x z+x y z=4 .
$$

The largest possible value of $x y z$ is $\frac{a+b \sqrt{c}}{d}$, where $a, b, c, d$ are integers, $d$ is positive, $c$ is square-free, and $\operatorname{gcd}(a, b, d)=1$. Find $1000 a+100 b+10 c+d$.