要使函数有意义需x+

1

x

＞0求得x＞0

∴x+

1

x

≥2

x• 1 x

=2

∴log

1

9

（x+

1

x

）≤log

1

9

2

∴y=3log

1

9

(x+

1

x

)=3-2 

log

(3

x+

1

x

)=(x+

1

x

)-2=

1

(x+ 1 x )2

≤

1

4

，当且仅当x=1时取等号．

故选A