# Bounding Euler products (or almost) by products of zeta functions

Let $s_1, s_2 \in (1/2,1\rbrack$. I would like to bound the product

$$A=\prod_p \left(1 + \frac{p^{-s_1} p^{-s_2}}{(1-p^{-s_1}+p^{-1}) (1-p^{-s_2}+p^{-1})}\right)$$

Now, I am almost positive that $$A\leq \frac{\zeta(s_1+s_2) \zeta(2 s_1+ s_2) \zeta(s_1+2s_2)}{\zeta(s_1+2) \zeta(s_2+2) \zeta(4)}$$ Is this (or results like this) known? Is there an elegant way to show this?

(It does seem to be the case that every term in the infinite product on the left is less than the corresponding term in the (implicit) infinite product on the right (where we expand the zeta function as its Euler product), but of course that remains to be proven.)

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I have edited to improve readability. I think maybe an $s_1$ in the first display ought to be an $s_2$, but I'll leave it to OP to edit that in, if warranted. – Gerry Myerson Feb 24 '12 at 11:07
I've just corrected that :*) . – H A Helfgott Feb 24 '12 at 14:07
What I would do is just use a computer program to expand out the infinite series on each side (for each prime $p$) and subtract. Presumably the lower-order terms vanish. Then hopefully the first nonvanishing term is patently nonnegative and the tail can be bounded by this first nonvanishing term. – Matt Young Feb 24 '12 at 14:46
If you want to prove the inequality for each Euler factor, then setting $x_i=p^{-s_i}$, isn't it just a matter of checking whether some (complicated) polynomial $P(x_1,x_2)$ is $\geq 0$ on $[1/p,1/\sqrt{p}]^2$ ? – François Brunault Feb 24 '12 at 15:59
Regarding showing a degree ten-polynomial non-negative on $x<y<\sqrt{x}$. I asked a similar question a while back at mathoverflow.net/questions/1493/… The answer was QEPCAD which is a software that can do quantifier elimination for the theory of real closed fields. The software is not polished, but is usable. It might, or might not choke on your polynomial (time and space complexity is large), but it is worth a try. – Boris Bukh Feb 24 '12 at 23:03

Actually, I now have a follow-up question. How on earth do I list QEPCAD in the bibliography for the answer it provided?

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[1] QEPCAD version ..., available at ... (BTW I would insert a follow-up question in the original question.) – GH from MO Mar 9 '12 at 13:26

Following up on Boris's suggestion, let me tell of my mostly happy experience with QEPCAD.

First of all - QEPCAD seems to crash on three variables (at least for the slightly hairy expressions we are dealing with here). So we have to start by reducing our problem to a two-variable problem by means of human.

The inequality that $A\leq ζ(s1+s2)ζ(2s1+s2)ζ(s1+2s2)/ζ(s1+2)ζ(s2+2)ζ(4)$ would naturally rest on turns out to be false; no QEPCAD needed here (though QEPCAD caught this when fed a special value for one of the variables). If this strong inequality is true, it's doubtful it has a nice proof.

Now for the slightly weaker inequality (call it inequality B; it is neither the strongest nor the weakest one) that I mentioned above, namely: $A\leq \frac{\zeta(s_1+s_2)ζ(2 s_1+s_2)ζ(s_1+2 s_2)}{ζ(3)ζ(3)ζ(4)}$; this, as you can easily check, follows if we show that $1 + \frac{y_1 y_2}{(1-y_1+x) (1-y_2+x)} \leq \frac{(1-x^3)^2 (1-x^4)}{(1-y_1 y_2) (1-y_1 y_2^2) (1- y_1^2 y_2)}$ for $0\leq x\leq 1/2$ and $x\leq y_1, y_2 \leq \sqrt{x}$.

QEPCAD chokes on this. However, this human realized that, if we change variables to $x$, $s = y_1 + y_2$ and $r = y_1 y_2$, we get that we must show that a polynomial quadratic on $s$ with positive leading coefficient adopts only non-positive values within a range. Hence it is enough to check for $s$ extremal given $r$ - and this happens when either $y_1=y_2$ ($s$ is then minimal) or $y_i = x$ or $y_i=\sqrt{x}$ for some $i=1,2$ (so that $s$ is maximal).

QEPCAD proves the inequality very quickly in the first two cases. For $y_i = \sqrt{x}$, defining $x$ as $y_i^2$ (so that we have a polynomial) gives us a polynomial of degree too large for QEPCAD to handle. Inputting a stronger inequality of lower degree (with $(1-3 x^3)$ instead of $(1-x^3)^2 (1-x^4)$) makes QEPCAD give an affirmative answer, thus proving inequality B.

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Hey! I liked the comment! – H A Helfgott Feb 28 '12 at 22:23
Dear Harald, I am glad that you deepened the intimate connection between machine and human. – GH from MO Mar 9 '12 at 13:22