# Seeking diagonal $D$ such that $AD^2B$ is positive definite (A and B are symmetric and p.d.)

Let A,B be two real symmetric, non-commuting positive definite matrices. Then the product AB is not positive definite (in the sense that ${\bf x}^TAB{\bf x}$ may be anything).

I'm trying to find conditions on a diagonal matrix D such that ${\bf x}^TAD^2B{\bf x}\geq 0$ for any nonzero real vector ${\bf x}$.

I'm particularly interested in allowing $D$ to contain zeros. Then $AD^2B$ would be singular.

I know that this is equivalent to proving that $AD^2B+BD^2A$ is symmetric and positive definite, but cannot get any further. Thanks for any help!