I'll take an approach similar to slobone:

Let f(x) = (x + r_1)(x + r_2)*...*(x + r_{n-1})(x + r_n)
Where r_1*r_2*...*r_n = 1

Now, I'll try proving it inductively:

First, show it is true for the first term: n = 1, which means r_1 = 1
f(2) = (2 + 1) = 3 >= 3^1 = 3

Now, assume it is true for some term: n = k
f(2) = (2 + r_1)(2 + r_2)*...*(2 + r_k) >= 3^k
Where r_1*r_2*...*r_k = 1

Show that it works for the next term: n = k + 1
f(2) = (2 + r_1)(2 + r_2)*...*(2 + r_k)(2 + r_{k+1}) >= 3^(k+1)

Recall that r_1*r_2*...*r_k = 1, but r_1*r_2*...*r_k*r_{k+1} MUST also equal 1, so r_{k+1} MUST equal 1.

Therefore, we have:
(2 + r_1)(2 + r_2)*...*(2 + r_k)(2 + 1) >= 3^k*(2 + 1) = 3*3^k = 3^(k+1)

Q.E.D.

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Notes:

The coefficient of all the "x" terms in f(x) MUST be 1 (or MUST multiply to equal one - but the way I set up this proof allowed for me to avoid this) because the original polynomial form of f(x) began with 1*x^n, so the expansion of the factored form of f(x) must have the same. Note also that r_1 through r_n take care of the coefficients of the constants, a_1 through a_{n-1}, of the remaining "x"s in the expanded form.

In the factored form of f(x), the product of the roots, r_1*r_2*...*r_n, MUST equal 1 because the constant term of the polynomial form of f(x) was 1, and so expanding the factored form of f(x) MUST also result in a constant term of 1.

I LOVE INDUCTIVE PROOFS!