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

,

,

( nc)=ac(1+mn))
and
 + (ma)c = ac(m+n))
Let's assume that

, then


(n-1) \geq 0)
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.