Continue the sequence $x . y, z, w$ satisfying $x y+2 y z+z w \leqslant A\left(x^{2}+y^{2}+z^{2}+w^{2}\right)$, then the minimum value of the real number $A$ is
9. $\frac{1+\sqrt{2}}{2}$.

Let $\lambda, \mu>0 (\lambda, \mu$ to be determined $)$, then $2 x y \leqslant \lambda x^{2}+\frac{1}{\lambda} y^{2}, 2 y z \leqslant y^{2}+z^{2}, 2 z w \leqslant \ldots z^{2}+\frac{1}{\mu} z^{2} \cdot$ F is $x y+2 y z+z w \leqslant-\frac{\lambda}{2} x^{2}+\left(\frac{1}{2 \lambda}+1\right) y^{2}+\left(1+\frac{\mu}{2}\right) z^{2}+\frac{1}{2 \mu} w^{2}$. Let $\frac{\lambda}{2}=\frac{1}{2 \lambda}+1=1+\frac{\mu}{2}=\frac{1}{2 \mu}$, then $\lambda=\sqrt{2}+1, \mu=\sqrt{2}-1$. At this point, $x y+2 y z+z w \leqslant \frac{1+\sqrt{2}}{2}\left(x^{2}+y^{2}+z^{2}+w^{2}\right)$, equality holds when $x=$ $w=\sqrt{2}-1, y=z=1$. Therefore, the minimum value of $A$ is $\frac{1+\sqrt{2}}{2}$.