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. :-)

_{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 $$\tfrac{1}{2}k(k+1)+\sum_{n=1}^k n \cos n\theta\geq 0.$$