Method for finding a maximum of a convex function

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I have the following maximization problem:

\begin{align} \max_{v_i} \quad & \left(1-\frac{c}{2}\right) \sum_{i=1}^n v_i -\frac{ \left( k -\delta \right)}{\sum^k_{i=1}\frac{1}{v_i}} \label{The OP} \\ \textrm{s.t.} \quad & 1\leq v_i\leq n-1 \quad &\forall i \in [1,n], \label{integers}\\ \textrm{ } \quad & \sum_{i=1}^y v_i \leq y(y-1) + \sum_{i=y+1}^n \min (v_i,y) \quad &\forall y \in [1,n], \label{EG conditions O}\\ \textrm{ } \quad & \frac{k-\delta-1}{\sum_{ j \in K \setminus \{i\}} \frac{1}{v_j}} - v_i \leq 0 \quad &\forall i \in [1,k], \textrm{ } \textrm{ (*)} \\ \textrm{ } \quad & - \frac{ \left( k -\delta \right)}{\sum^k_{i=1}\frac{1}{v_i}} + v_j \leq 0 \quad &\forall j \in (k,n], \label{not attacked nodes} \end{align}

Observe that the objective function is convex. Originally, it was an integer maximization problem related to optimal network formation.

Observe also that constraint $(*)$ is concave. Thus this is a problem of convex function maximization on a non-convex set.

Therefore, additionally, I consider a Convex Hull of the set instead of the set itself. A tight convex hull can be achieved by replacing condition $(*)$ with a set of linear constraints, which spans through the extreme points of the concave function. After this, I applied Bauer maximum principle and started to search for extremes. I managed to show analytically that only three potential integer sequences $\vec{v}$ are at the extremes of a convex hull.

Then I verified numerically and realized that those three sequences happen to coincide with real maximizers (At least for $n<11$). I performed numerical verification by iterating the real degree sequences of networks (taken from the Wolfram database) for $n$ up to $11$, revealing the sequences that yield the maximum for a given $c$.

I am not very familiar with optimization. So the only question I have is whether it is a sound approach to this particular maximization problem. I could not find much literature about the Bauer maximum principle used for problems like that.