Originally Posted by slobone

POSSIBLE OUTLINE OF PROOF

We can also write

f(x) = (x-r_1)(x-r_2)...(x-r_n)

(because f is monic) where r_1, r_2, ..., r_n are the n real roots.

Step 1: show that each of the r_i must be <= 0.

Proof by contradiction: For convenience, arrange the subscripts so that r_1 >= r_2 >=...>= r_n. By assumption, r_1 > 0.

If r_1 > r_2, let x be such that r_1 > x > r_2. Then exactly one of (x-r_1), (x-r_2),... (x-r_n) is < 0, so f(x) < 0, contradiction.

If r_1 = r_2, we have a problem. We could keep going down the list of roots until we find a number that's less than an odd number of the roots, but what if there isn't any such number? That is, r_1 = r_2, r_3 = r_4, and so on, and n is even. In this case I don't know what to do -- perhaps you could make some kind of combinatorial argument using the binomial theorem to show that if all the coefficients are positive, all the roots must be negative.

Step 2. Let's make a notational switch and define R_j = ABS (r_j) [absolute value]. Then f(2) = (2+R_1)(2+R_2)...(2+R_n), where all the R's are positive. Since the constant term of f is 1, it follows that

R_1R_2...R_n = 1.

I believe that for any set of R_j's with this property, it must be true that f(2) >= 3^n.

For example, (2 + 1/3)(2 + 1/2)(2 + 6) = 280/6 > 3^3.

Unfortunately this is the hard part of the proof.

[I apologize if my notation is horrific, but I'm not used to doing math on the Internet.]