You could use a regular expression as follows:

In [33]: bool(re.match('he', 'Hello', re.I))
Out[33]: True 

In [34]: bool(re.match('el', 'Hello', re.I))
Out[34]: False 

On a 2000-character string this is about 20x times faster than lower():

In [38]: s = 'A' * 2000

In [39]: %timeit s.lower().startswith('he')
10000 loops, best of 3: 41.3 us per loop

In [40]: %timeit bool(re.match('el', s, re.I))
100000 loops, best of 3: 2.06 us per loop

If you are matching the same prefix repeatedly, pre-compiling the regex can make a large difference:

In [41]: p = re.compile('he', re.I)

In [42]: %timeit p.match(s)
1000000 loops, best of 3: 351 ns per loop

For short prefixes, slicing the prefix out of the string before converting it to lowercase could be even faster:

In [43]: %timeit s[:2].lower() == 'he'
1000000 loops, best of 3: 287 ns per loop

Relative timings of these approaches will of course depend on the length of the prefix. On my machine the breakeven point seems to be about six characters, which is when the pre-compiled regex becomes the fastest method.

In my experiments, checking every character separately could be even faster:

In [44]: %timeit (s[0] == 'h' or s[0] == 'H') and (s[1] == 'e' or s[1] == 'E')
1000000 loops, best of 3: 189 ns per loop

However, this method only works for prefixes that are known when you're writing the code, and doesn't lend itself to longer prefixes.

You could convert the string to lowercase before checking.

variable.lower().startswith(('a', 'e', 'i', 'o', 'u'))

Alternatively, you could use a regular expression with the ignore case flag.

bool(re.match('(?i)[aeiou]', variable))