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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.

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    Just to put the definition of positive definite , that is, A>0, in terms of determinants, straight. It is not enough that \det(A)>0, what is needed is that all leading principal minors are positive, that is, \det(A_{ii})>0, for all $i$. This is the determinants off the submatrices consisting of first $i$ rows and columns.2012-09-02

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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.

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Do you think this is a counter example for this question?

We know that $\rho(AB)>1$. If we assume $AB=\begin{bmatrix}2 & -8\\ 0 &2\end{bmatrix}$ The eigenvalues of this matrix are $(2,2)$

$\frac{1}{2}(AB+(BA)')=\begin{bmatrix}2 & -4\\ -4 & 2\end{bmatrix}$ The eigenvalues of this matrix are $(-2,6)$.

Therefore $x^TABx=\frac{1}{2}x^T(AB+(BA)')x$ is not greater or less than $x^Tx$! $(AB+(BA)')$ is an indefinite matrix.

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    What are $A$ and $B$ in your example?2012-09-10