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.