Note that the quantity inside the outer parentheses is the arithmetic mean of the three terms being summed. Now, the arithmetic mean is always greater than the geometric mean of the same set of numbers, with equality applying only when all of the values in the set are equal to one another. So regardless of the value of y, we know that this expression is greater than:

lim y->0  (2^y*2^2y*2^3y)^1/3y =

lim y->0 (2^(y+2y+3y))^1/3y = lim y->0 2^(6y/3y) = 2^2 = 4

when y finally "reaches" 0 (see? I said this wan't super rigorous) the approximation of geometric for arithmetic mean is exact as all three values (2^y, 2^2y, 2^3y) are equal to one another (and = 1) and so the limit is 4

There's nothing particularly special about these three values, so we can generalize:

lim y->0 ((a^y+b^y+c^y)/3)^1/y = (abc)^1/3 as long as a,b,c are >0

So for the bonus, where a=2, b=6, c=18, abc = 216 and (abc)^1/3 = 6, which is consistent with computer explorations below.

I'd love to see the transition from a lower bound to equality as y->0 fleshed out in more detail, but I'm convinced that the result is accurate after some computer exploration of my own.