Let $\mathbf{x},\mathbf{y},\mathbf{z}$ be pairwise orthonormal vectors from $\mathbb{R}^n$ where $n \geq 3$. For any $1 \leq i,j,k \leq n$ define $$p_{ijk} = \left| \begin{array}{ccc} x_i^2 & x_iy_i & x_iz_i \newline x_jy_j & y_j^2 & y_jz_j \newline x_kz_k & y_kz_k & z_k^2 \end{array} \right| = x_i y_j z_k \left| \begin{array}{ccc} x_i & y_i & z_i \newline x_j & y_j & z_j \newline x_k & y_k & z_k \end{array} \right|.$$ Notice that $p_{ijk}$ might attain negative values, but at the same time $\sum_{i,j,k} p_{ijk} = 1$ by orthonormality. It can thus be regarded as a discrete quasi-probability distribution on {$1,\ldots,n$}$^3$.
Extensive numerical experiments suggest that the following Cauchy-Schwarz-like inequality is satisfied for any symmetric matrix $(a_{ij})$ with non-negative entries: $$\left| \sum_{i,j,k} a_{ik}a_{jk}p_{ijk} \right| \leq \sqrt{\sum_{i,j,k} a_{ik}^2 p_{ijk}}\sqrt{\sum_{i,j,k} a_{jk}^2 p_{ijk}}.$$
The sums under the square roots are nonnegative, because one has that $\sum_{i,j,k} a_{ik}^2 p_{ijk} = \tfrac{1}{2} \sum_{i,k} a_{ik}^2 (x_iz_k - x_kz_i)^2$ and similarly $\sum_{i,j,k} a_{jk}^2 p_{ijk} = \tfrac{1}{2} \sum_{j,k} a_{jk}^2 (y_jz_k - y_kz_j)^2$, where in the calculations one again uses the orthonormality of the three vectors and the symmetricity of $(a_{ij})$.
The question is, how to (dis)prove this inequality? The proofs of the standard Cauchy-Schwarz inequality are not applicable here because of $p_{ijk}$ being negative sometimes. Any hint would be much appreciated!