∵正实数x，y，z满足x+2y+z=1，
∴

1

x+y

+

9(x+y)

y+z

=

x+y+y+z

x+y

+

9(x+y)

y+z

=1+

y+z

x+y

+

9(x+y)

y+z

≥1+2

y+z x+y × 9(x+y) y+z

=7，当且仅当

y+z

x+y

＝

9(x+y)

y+z

，x+y+y+z=1，即x+y＝

1

4

，y+z＝

3

4

时，取等号．
∴则

1

x+y

+

9(x+y)

y+z

的最小值为7．
故答案为7．