I am trying to figure out if given the matrix C and any vector $\vec{p}$ with $p_i\in[0,1]$, and all elements summing up to 1, there exists a vector $\vec{x}$ such that the following inequality holds: $$(C\vec{p}).\vec{x} \gt a$$
Is there any general strategy to solve this for a given C and $a\in\mathbb{R_+}$? I'm specifically trying to solve this problem for the following matrix C: $$C=\begin{pmatrix} 1&-1&0&0\\ 0&0&0&0\\ 0.5&-0.5&0&0\\ 0&0&0&0\\ 0&0&1&-1\\ 0&0&0.5&-0.5\\ 0.5&-0.5&0&0\\ 0&0&0.5&-0.5\\ 0.25&-0.25&0.25&-0.25\\ \end{pmatrix}$$ and $\vec{p}$ is a 4-dim vector and $\vec{x}$ a 9-dim vector. The give some context on why I want to solve this: The elements of the vector $\vec{p}$ are the probabilities for every choice, hence the range restriction and summing up to 1 restriction. $(C\vec{p})\cdot{x}$ is a function of which I want to set the lower bound. So I want to find the function that gives me a certain lower bound, hence I need to find to vector $\vec{x}$ that gives me the correct lower bound. The lower bound doesn't have to be a just larger than a.