CaptainBlack, could you post the solution? I seem to be the only one working on it, and I've more or less given up.

I'm sure there's a 3-line solution that's going to have me slapping my forehead...

Prove inductively:

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

Now, assume it is true for some value: n = k
f(2) = 2^k + a_1*2^(k-1) + ... + a_{k-1}*2 + 1 >= 3^k

Show that it is true for the next term: n = k + 1
f(2) = 2^(k+1) + a_1*2^k + ... + a_{k}*2 + 1 >= 3^(k+1)

We have:
2^(k+1) + a_1*2^k + ... + a_{k}*2 + 1
= 2*[2^k + a_1*2^(k-1) + ... + a_{k-1}*2 + a_{k}] + 1


It just occurred to me that I can go no futher until I know how to use the fact that there are n real roots. I know that is important to the proof (because it limits the values of a_1 to a_n), so I need to figure out how that effects the problem before I can go on.

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!

As f(x)>0 for all x>=0 all roots are negative. Let these be -b1, -b2, .. -bn,
with all the b's >0.

Then:

f(x) =(x+b1)(x+b2) ..(x+bn)

Also as the constant term is 1 we know that b1.b2. .. bn = 1.

Now 2 + bk = 1 + 1 + bk >= 3 cuberoot(1.1.bk), by the Arithmetic-Geometric
mean inequality. So (2+bk) >= 3cuberoot(bk)

So for x=2:

f(2) = (2+b1)(2+b2) ...(2+bn) >= 3^n cuberoot(b1.b2. .. bn) = 3^n

RonL

Was my proof wrong?

That was the thing I needed. I suspected it was true, but I didn't know it was a theorem. Good problem.

Yes, because the inductive step does not hold as you have the product
of the first k r's equal to 1, the k+1 st r must be 1. So this proof only works if
all the roots are -1, but this is in general not true.

RonL

I wasn't arguing that the r_{x} roots must be 1 (that is that every root must be 1), but that if the roots for the n = k case are r_1, r_2, ..., r_k, where r_1, r_2, ..., r_k are real numbers, and that the first "k" roots of the n = k + 1 case are r_1, r_2, ... , r_k, where r_1, r_2, ..., r_k are the same values from the n = k case, then the r_{k + 1} root must be 1.

[Edit] There is a problem with this logic that just occurred to me: The roots of n = k + 1 don't have to be the same as those of n = k.