The inequality $$3 + 4 \cos \theta + \cos 2 \theta \geq 0$$ plays a key role in the proof of the classical zero-free 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 $a\_{0} < \frac{3}{4} a\_{i\_0}$, $b\_0=0$?



Better trigonometrical inequalities for $\zeta(s)$?
Assuming the $b\_i$ are all distinct (or at least non-zero 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. :-)


