Conditions for $AB+BA$ posdef, given $A,B$ are symmetric and posdef?

Question

Let $A,B$ be two symmetric and positive definite (posdef) real matrices.

Under what conditions is

$$AB+BA$$

also positive (semi-)definite?

Note: By posdef, I mean that $x'Ax > 0$ for any non-zero vector $x$.

Answer 1

A nice case is when $A,B$ commute. Then $A,B$ can be diagonalised simultaneously:

$$A=VSV',\quad B=VDV'$$

where $S,D$ are diagonal and positive, and $V$ is an orthogonal matrix with the common set of eigenvectors. See Do commuting matrices share the same eigenvectors?.

It follows that

$$AB=BA=VSDV'$$

Then

$$\frac{1}{2} x'(AB+BA)x = x' VSDV' x = y' SD y > 0$$

for all non-zero $x$ because $SD$ is positive, where $y=V'x$.

Question: Are there other cases where $AB+BA$ is posdef but $A,B$ do not commute?