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.)



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


