Minimizing a sum of hyperbolic cosine. My problem is as follows:

Given independent unit vectors $a_1,\ldots,a_n$ in $\mathbb R^n$, I want to find a lower bound on the function
$$f(x)=\sum_{i=1}^n \cosh(\left \langle a_i,x\right \rangle)$$
for $\| x \|=1$.

A trivial lower bound would be $n$, but I am looking for something as tight as possible, hence the desire to find the minimum of the function.
I tried solving this problem geometrically thinking that by renaming the $a_i$ into $-a_i$ as necessary we can ensure that $a_1,\ldots,a_n$ will be in a cone of minimum angle (among all such renamings) and the barycenter of these $a_i$ should (intuitively) maximize $f$, from which I hoped the $x$ that minimizes $f$ would be defined as some vector orthogonal to this barycenter. Still, I could not see how to prove either of those things. I also considered solving this problem by computing the derivative of $f$ and solving a system but to no avail...
I am convinced that there is a lower bound that can be expressed as a function of the inner products $\left \langle a_i,a_j \right \rangle$ but I was unable to find it.
EDIT:
As noted by @uniquesolution the maximum and the minimum are very close when we consider the $a_i$ to be an orthonormal basis. The difference is still important to me because what I need to lower bound is not exactly $f(x)$ for $||x||=1$ but for $||x||=r$ for some fixed r. In which case the difference between both extrema can be large. I will leave the problem stated as is because I think it is easier to represent geometrically when all points are on the unit sphere and the minimum will be reached at the same point when projected on the sphere. Moreover, if we are looking for a lower bound that depends on the inner products $\left \langle a_1, a_j \right \rangle$ then the orthonormal case will not be conclusive.
 A: It seems to me that in general you cannot improve on the trivial $n$ lower bound. Take $a_1=e_1$, the first unit vector, and the rest are infinitesimal perturbations of $e_1$, arbitrarily close to it but still linearly independent. Now take $x=e_2$. Then all the entries $\langle a_i,x\rangle$ will be either zero (for $i=1$) or arbitrarily close to zero, and so the value of $f$ at $e_2$ will be arbitrarily close to $n$.
You are probably correct that a more precise lower bound will somehow depend on the position of the $a_i$'s on the unit sphere. A simple exercise (using Lagrange's multipliers) is to show that if the $a_i$'s are an orthonormal basis, then the maximum of $f(x)$ over the sphere occurs when $x$ coincides with one of the $a_i$'s, in which case the value is $\cosh(1)+n-1$, as pointed out by Andy Walls in the comments above, but the minimal value will occur at the vertices of the cube $n^{-1/2}[-1,1]^n$, and the value of the function at such points is $n\cosh(n^{-1/2})$, which is at least $n+1/2$.
It is interesting to observe that for every $n$ the distance between the maximum $\cosh(1)+n-1$ and the minimum $n\cosh(n^{-1/2})$ is smaller than $0.05$.
