The inequality $$3 + 4 \cos \theta + \cos 2 \theta \geq 0$$ plays a key role in the proof of the classical zerofree region of the Riemann zeta function. Are there other inequalities of the form $$\sum_{i=0}^k a_i \cos b_i \theta \geq 0,\;\;\;\;\;a_\geq 0$$ such that $a_{i_0} = \sum_{i\ne i_0} a_i$ for some $0\leq i_0\leq k$ and $a_{0} < \frac{3}{4} a_{i_0}$, $b_0=0$?

3$\begingroup$ $1+\cos \theta \ge 0$; $99+100 \cos \theta +\cos 2\theta \ge 0$; $98+100\cos \theta +\cos (2\theta) + \cos (4\theta) \ge 0$ (didn't check the last one, but surely it's correct). Depends on what you're after? For zero free regions this is not the relevant criterion (one coefficient being sum of the others)  presumably you've already looked at Kadiri, Stechkin ... $\endgroup$ – Lucia Feb 5 at 18:58

$\begingroup$ I should have added that $a_0<a_{i_0}$. What would be useful would be $a_0< 3 a_{i_0}/4$, actually. $\endgroup$ – H A Helfgott Feb 5 at 19:12

$\begingroup$ apologies, but I do not understand what difference the condition you added makes; in the example I gave in the answer box I have $k=i_0=3$, $a_0,a_1,a_2=1,2,3$, $b_0,b_1,b_2=1,2,3$, and $a_{3}=6$, $b_3=0$, so $a_0=1<3a_{i_0}/4=9/2$. $\endgroup$ – Carlo Beenakker Feb 5 at 19:29

$\begingroup$ I should have said $b_0=0$. $\endgroup$ – H A Helfgott Feb 5 at 19:29

$\begingroup$ @Lucia: what I have in mind is actually just to improve explicit bounds on $1/\zeta(\sigma+it)$ for $\sigma>1$ (so as to improve explicit bounds on $1/\zeta(1+it)$). Or would a better zerofree region necessarily follow from an inequality such as the one I request? $\endgroup$ – H A Helfgott Feb 5 at 19:46
Assuming the $b_i$ are all distinct (or at least nonzero for $i \neq 0$), this is not possible. (Otherwise there are trivial examples, e.g. $1 + 2 \cos(0 \theta)+ \cos(0 \theta) \geq 0$ or $1 + 4 \cos \theta + \cos(2\theta) + 2 \cos(0 \theta) \geq 0$.)
Suppose that $\sum_{i=0}^k a_i \cos b_i \theta \geq 0$. Since $a_{i_0} = \sum_{i \neq i_0} a_i$, this implies that whenever $\cos b_{i_0} \theta = 1$, one must have $\cos b_i \theta = +1$ for all other $i$. In particular, the other $b_i$ must be integer multiples of $2b_{i_0}$. We now have
$$ a_0 + a_{i_0} \cos b_{i_0} \theta + \sum_{i \neq 0, i_0} a_i \cos b_i\theta \geq 0$$
with the $b_i$ in the sum nonzero integer multiples of $2b_{i_0}$. Performing a Taylor expansion around $\theta = \pi / b_{i_0}$ to second order, we conclude that
$$  a_{i_0} \frac{b_{i_0}^2}{2} + \sum_{i \neq 0, i_0} a_i \frac{b_i^2}{2} \geq 0$$
and hence (since $b_i^2 \geq 4 b_{i_0}^2$ and $b_{i_0} \neq 0$)
$$ \sum_{i \neq 0, i_0} a_i \leq \frac{1}{4} a_{i_0}$$
or equivalently
$$ a_0 \geq \frac{3}{4} a_{i_0}.$$
Thus one cannot have $a_0 < \frac{3}{4} a_{i_0}$. This argument also shows that up to rescaling and other trivial rearrangements, Mertens' inequality $3 + 4 \cos(\theta)+\cos(2\theta) \geq 0$ is the unique inequality that attains $a_0 = \frac{3}{4} a_{i_0}$.
At a more metamathematical level, if there were a variant of Mertens' trigonometric inequality that gave superior numerical results towards the classical zero free region, I would imagine that this would already have been noticed by now. :)

1$\begingroup$ Nice. I had a different and much more modest possible application in mind, but, as I made my question more precise, I started seeing that it could give a numerically better zerofree region (and hence the answer was almost certainly "no", for metamathematical reasons, as you put it). $\endgroup$ – H A Helfgott Feb 6 at 8:12
_{ this is an answer to the question as originally posed, without the additional conditions on $a_0$ and $b_0$ }
for example, $$6+\cos \theta+2 \cos 2 \theta+3 \cos 3 \theta\geq 0,$$ or more generally $$\tfrac{1}{2}k(k+1)+\sum_{n=1}^k n \cos n\theta\geq 0.$$