You can use operator.itemgetter for that:

import operator
stats = {'a': 1000, 'b': 3000, 'c': 100}
max(stats.iteritems(), key=operator.itemgetter(1))[0]

And instead of building a new list in memory use stats.iteritems(). The key parameter to the max() function is a function that computes a key that is used to determine how to rank items.

Please note that if you were to have another key-value pair 'd': 3000 that this method will only return one of the two even though they both have the maximum value.

>>> import operator
>>> stats = {'a': 1000, 'b': 3000, 'c': 100, 'd': 3000}
>>> max(stats.iteritems(), key=operator.itemgetter(1))[0]
'b' 

If using Python3:

>>> max(stats.items(), key=operator.itemgetter(1))[0]
'b'

You can first determine the maximum with:

maximum = max(dic.values())

and then filter based on the maximum value:

result = filter(lambda x:x[1] == maximum,dic.items())

Example in the command line:

$ python2
Python 2.7.12 (default, Nov 19 2016, 06:48:10) 
[GCC 5.4.0 20160609] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> dic={0: 1.4984074067880424, 1: 1.0984074067880423, 2: 1.8984074067880425, 3: 2.2984074067880425, 4: 2.2984074067880425}
>>> maximum=max(dic.values())
>>> maximum
2.2984074067880425
>>> result = filter(lambda x:x[1] == maximum,dic.items())
>>> result
[(3, 2.2984074067880425), (4, 2.2984074067880425)]

Given you want to present the list of keys are a nice list and the value, you can define a function:

def maximum_keys(dic):
    maximum = max(dic.values())
    keys = filter(lambda x:dic[x] == maximum,dic.keys())
    return keys,maximum

which returns a tuple containing the list of keys and the maximum value:

>>> maximum_keys(dic)
([3, 4], 2.2984074067880425)