Find words containing every vowel

GolfScript, 19 characters

n%{'aeiouy'&,6=},n*

Usage:

golfscript vowels.gs < wordlist.txt

Output:

abstemiously
authoritatively
behaviourally
[...]
uninformatively
unintentionally
unquestionably
unrecognisably

If you also want to output the count at the end you can use

n%{'aeiouy'&,6=},.n*n@,

which is four characters longer.

Python - 46 characters

filter(lambda x:set(x)>=set("AEIOUY"),open(f))

Readable version: It's already pretty readable :-)

Ruby 38

Edit 34: Best so far (from @O-I):

a.split.select{|w|'aeiouy'.count(w)>5} 

Edit 1: I just noticed the question asked for 'y' to be included among the vowels, so I've edited my question accordingly. As @Nik pointed out in a comment to his answer, "aeiouy".chars is one character less than %w[a e i o u y], but I'll leave the latter, for diversity, even though I'm risking nightmares over the opportunity foregone.

Edit 2: Thanks to @O-I for suggesting the improvement:

s.split.select{|w|'aeiouy'.delete(w)==''}

which saves 11 characters from what I had before.

Edit 3 and 3a: @O-I has knock off a few more:

s.split.select{|w|'aeiouy'.tr(w,'')==''}

then

s.split.reject{|w|'aeiouy'.tr(w,'')[1]}

and again (3b):

a.split.select{|w|'aeiouy'.count(w)>5} 

I am a mere scribe!

Here are two more uncompettive solutions:

s.split.group_by{|w|'aeiouy'.delete(w)}['']       (43)
s.gsub(/(.+)\n/){|w|'aeiouy'.delete(w)==''?w:''} (48)

Initially I had:

s.split.select{|w|(w.chars&%w[a e i o u y]).size==6}

s is a string containing the words, separated by newlines. An array of words from s that contains all five vowels is returned. For readers unfamiliar with Ruby, %w[a e i o u y] #=> ["a", "e", "i", "o", "u", "y"] and & is array intersection.

Suppose

s = "abbreviations\nabduction\n"
s.split #=> ["abbreviations", "abduction"] 

In the block {...}, initially

w             #=> "abbreviations"
w.chars       #=> ["a", "b", "b", "r", "e", "v", "i", "a", "t", "i", "o", "n", "s"]
w.chars & %w[a e i o u y] #=> ["a", "e", "i", "o"]
(w.chars & %w[a e i o u y]).size == 6 #=> (4 == 6) => false,

so "abbreviations" is not selected.

If the string s may contain duplicates, s.split.select... can be replaced by s.split.uniq.select... to remove duplicates.

Just noticed I could save 1 char by replacing size==6 with size>5.