If $A>0$ and $B>0$ and $A^{-1}If $A>0$ and $B>0$ and $A^{-1}<B$, 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, $0\leq A<B$ 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.

Since $B=A^{-1}+C$ with $C$ symmetric and positive, one asks whether $x^TACx\geqslant0$ for every vector $x$ and every symmetric positive matrices $A$ and $C$. 
Consider $A=\begin{pmatrix}5 & 2 \\ 2 & 1\end{pmatrix}$ and $C=\begin{pmatrix}1 & -2 \\ -2 & 10\end{pmatrix}$. Then $A$ and $C$ are symmetric positive, $A^{-1}=\begin{pmatrix}1 & -2 \\ -2 & 5\end{pmatrix}$, $B=\begin{pmatrix}2 & -4 \\ -4 & 15\end{pmatrix}$ and $AC=\begin{pmatrix}1 & 10 \\ 0 & 6\end{pmatrix}$. Hence $x^TACx\lt0$ for $x^T=(1,-1)$. 
Thus, the result above does not hold in full generality.
