LOL I just used the rearrangement inequality on Aopswiki to solve it.
But that's a bit lame I guess, I'll try to prove it
Hmmm let's consider a simpler case of two numbers {a,b}, {c,d},
,
.
In fact, let
,
,
and
Let's assume that
, then
And since
are both greater than 1, this is must be true, thus
■
Now consider {a,b,c}, {d,e,f}.
,
How could we pick combinations to maximimise their products?
If we picked
, that cannot be the maximum, because
So something bigger will be
. But this cannot be the maximum, since
So something bigger will be
. This is the maximum since this inequality can't be applied anymore.
This argument can be extended to sets of any size.
This means that with sets
,
, with
,
,
The maximum product is
Hehe soz this isn't really a proof, at least it's more like a 'rationale' for my answer. The mathematicians in the audience will probably be laughing.