7. $[20]$ Three positive reals $x, y$, and $z$ are such that
$$
\begin{array}{l}
x^{2}+2(y-1)(z-1)=85 \\
y^{2}+2(z-1)(x-1)=84 \\
z^{2}+2(x-1)(y-1)=89 .
\end{array}
$$

Compute $x+y+z$.
Answer: 18 . Add the three equations to obtain
$$
x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x-4 x-4 y-4 z+6=258,
$$
which rewrites as $(x+y+z-2)^{2}=256$. Evidently, $x+y+z=2 \pm 16$. Since $x, y$, and $z$ are positive, $x+y+z>0$ so $x+y+z=2+16=18$.