Suppose that $A$ and $B$ be symmetric positive matrix. I know that $B^{-1/2} A B^{1/2}$ is similar to $A$, so the eigenvalue of $B^{-1/2} A B^{1/2}$ is positive. I am thinking whether $x^T B^{-1/2} A B^{1/2} x > 0$ for all $x$.
I feel that this is wrong because $B^{-1/2} A B^{1/2}$ is not symmetric.
Let $B^{1/2} = C$ then $C$ is also symmetric positive definite and admits the eigen decomposition $C = Q^\top \Lambda Q$. Then we have $$ x^\top B^{-1/2} A B^{1/2} x = x^\top Q^\top \Lambda^{-1} (QAQ^\top) \Lambda Q x = y^\top \Lambda^{-1} D \Lambda y. $$ Let $D = \begin{pmatrix}2&1\\1&1\end{pmatrix}, \Lambda = \begin{pmatrix}3&0\\0&1\end{pmatrix}, y = \begin{pmatrix}-1&1\end{pmatrix}^\top$ then $y^\top \Lambda^{-1} D \Lambda y = -\dfrac 1 3 <0$. Therefore this is not true in general.