Proving an equality associated with symmetric positive definite matrices

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Let $Q$ be an $n \times n$ symmetric positive definite matrix, $\vec{a}, \vec{b} \in \Bbb R^n$ be two random vectors. Prove that $a^TQ(ba^T-ab^T)Qb$ is non-positive.


Since $Q$ has $n$ independent eigenvectors with positive eigenvalues, I've tried expressing $\vec{a}$ and $\vec{b}$ as linear combinations of those eigenvectors, but it didn't work out. Really appreciate any help.

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By Cauchy-Schwarz $$|\langle \sqrt{Q}b,\sqrt{Q}a\rangle|^2=|a^TQb|^2\le \langle \sqrt{Q}b,\sqrt{Q}b\rangle \langle \sqrt{Q}a,\sqrt{Q}a\rangle=b^TQba^TQa$$