This is in the real numbers.
And n>=2, ai>0 (i=1...n).
I'm not sure if I understand what you're asking, but here goes.
I think here:
You meant to say: (1+a1)(1+a2)
...as that would make a lot more sense.
n >=2, so let's start with the minimum case: n=2.
(1+a1)(1+a2) = 1 + a1 + a2 + a1a2
1+(a1+a2) = 1 + a1 + a2
The first expression > the second, because there's that extra a1a2 term in there.
This will be true for any n >= 2! This is because the expanded multiplication will always have all the terms of the simple addition, plus some additionals that are combinations of the a's.
That proof probably wouldn't satisfy a mathematician, but it sure works for me.
El que pega primero pega dos veces.