6. Given the system of equations $\left\{\begin{array}{l}x-2 y=z-2 u, \\ 2 y z=u x .\end{array}\right.$ For each set of positive real solutions $(x, y, z, u)$, where $z \geqslant y$, there exists a positive real number $M$ such that $M \leqslant \frac{z}{y}$. Then the maximum value of $M$ is ( ).
A. 1
B. $3+2 \sqrt{2}$
C. $6+4 \sqrt{2}$
D. $3-2 \sqrt{2}$
6. $\mathrm{C}$

Let the set of all positive real solutions $(x, y, z, u)$ of the system of equations be $S$, then $M \leqslant \min _{(x, y, z, k) \in:} \frac{z}{y}$, so $M_{\max }=\min _{(1, y, z, u) \in \leq} \frac{z}{y}$.
Since $x-2 y=z-2 u$, we have $x+2 u=z+2 y$.
Thus, $(x+2 u)-8 x u=(x-2 u)^{2} \geqslant 0$.
Therefore, $(z+2 y)^{2}-16 z y \geqslant 0$.
Thus, $\left(\frac{z}{y}\right)^{2}-12\left(\frac{z}{y}\right)-4 \geqslant 0$, and $\frac{z}{y} \geqslant 1$.
Hence, $\frac{z}{y} \geqslant 6+4 \sqrt{2}$.
When $x=2, u=1, z=2+\sqrt{2}, y=\frac{2-\sqrt{2}}{2}$, we have $\frac{z}{y}=6+4 \sqrt{2}$. Therefore, $\min _{(x, y, x, u) \in,} \frac{z}{y}=$ $6+4 \sqrt{2}$.
Thus, $M_{\max }=6+4 \sqrt{2}$.