Let $$H_i = \{x\in \mathbb{R}^n\mid A_i^T x - b_i = 0\}$$ for all $i\in{1, \ldots, m}$ with $m>n$.
Let $D = \{x \in \mathbb{R}^n\mid A_i^T x - b_i \leq 0, \forall i\in \overline{1,m}\}$. Let $f:D \rightarrow \mathbb{R},$ $f(x) = C^T \cdot x$ where $C \in \mathbb{R}^n$. Let $\hat{A} = \begin{bmatrix}A_1^T\\ \vdots \\ A_n^T \end{bmatrix}$ and $\hat{B} = \begin{bmatrix}b_1\\ \vdots\\ b_n \end{bmatrix}$.
Suppose $x^* = \bigcap_{k=1}^n H_k \in D$ that is $x^{*} = \hat{A}^{-1}\cdot \hat{B} $ (the matrix $\hat{A}$ is invertible) such that if any line in matrix $\hat{A}$ is replaced with $A_i^T$ with $i\in \overline{n+1,m}$ (same story for $\hat{B}$) one at a time, thus obtaining $\hat{A_i}$ and $\hat{B_i}$, (and if $\hat{A_i}$ is invertible) the resulting point $x_i = \hat{A_i}^{-1} \hat{B_i}$ (if in $D$) has a lower value, i.e $f(x^*)\geq f(x_i)$.
Question: How can I prove (or disprove) that in this conditions $x^*$ is actually a global maximum ...