Let $(a_{ij})$ be a collection of non-negative numbers indexed by integers $1\le i,j \le N$ where $N$ is some fixed integer. Let $(c_{ij})$ be another collection of real numbers also indexed by integers $1\le i,j \le N$ which satisfies the inequality $$\sum_{1\le i,j \le N}c_{ij} \cdot a_{ik} a_{jl} \ge 0$$ for all $1\le k,l \le N$. Does it imply that $$\sum_{1\le i,j \le N}c_{ij} \cdot a_{ij} \ge 0 \ \ \ ?$$
I would already be happy if one can prove this inequality under the additional assumption that $(a_{ij})$ is a symmetric matrix whose corresponding quadratic form has signature $(1, N-1)$. Thanks!
The given condition means that $A^TCA\ge0$ (entrywise) and the desired condition means that the sum of all entries of the Hadamard product $C\circ A$ is nonnegative, or formally, $e^T(C\circ A)e\ge0$, where $e$ denotes the all-one vector. There isn't any apparent relationship between the two conditions. It's implausible that the first will entail the second. Here is a random counterexample: $$ A=\pmatrix{1&5\\ 5&4},\ C=\pmatrix{1&-2\\ -2&4}. $$ $A$ is a $2\times2$ real symmetric matrix that has a positive trace but a negative determinant. So it has exactly one positive eigenvalue and exactly one negative eigenvalue. However, $$ A^TCA=\pmatrix{81&27\\ 27&9}\ge0,\quad e^T(C\circ A)e=-3<0. $$