If $A>0$ and $B>0$ and $A^{-1}, then can we claim that $x^T(AB)x\geq x^Tx$? ($A$ and $B$ are real symmetric matrices).
I know these facts (from Matrix Mathematics book) (It seemed to me these might help but I haven't been able to use them in my advantage!),
Let $A,B\in\mathbb{F}^{n\times n}$ (real or complex matrix), and assume that $A$ and $B$ are positive semi-definite. Then, $0\leq A if and only if $\rho(AB^{-1})<1$.
Let $A,B\in\mathbb{N}^{n\times n}$ (positive semi-definite matrix). Then AB is semi simple, and every eigenvalue of $AB$ is nonnegative. If in addition $A$ and $B$ are positive definite, then every eigenvalue of $AB$ is positive.