Suppose $A$ and $B$ are positive definite (symmetric) real matrices.
In a previous post (An inequality on the root of matrix products) I asked whether
$(AB)^{1/2}+(BA)^{1/2} \geq A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}$ ?
where $S$ is a square contractive matrix (i.e. a square matrix that obeys $I-SS^T\geq 0$).
This was shown by counterexample to be false in some cases (it is true in others of course). I have since run numerous simulations on random matrices (obeying the stated properties) and found in those simulations that
$(AB)^{1/2}+(BA)^{1/2} - A^{1/2}SB^{1/2}-B^{1/2}S^TA^{1/2}$
always has at least one positive eigenvalue. In other words while the first inequality above may be false I have not been able to confirm that
$A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}\geq (AB)^{1/2}+(BA)^{1/2}$
is ever possible. Thus my question is: Does
$(AB)^{1/2}+(BA)^{1/2} - A^{1/2}SB^{1/2}-B^{1/2}S^TA^{1/2}$
always have at least one positive eigenvalue (or in other words is this construct either positive definite or indefinite for all $A,B,S$ obeying their respective property - i.e. is it never negative definite)?
NO!!!!!
Let $A=B=S=Id$
Clearly we get
$$(II)^{\frac{1}{2}}+ (II)^\frac{1}{2} - (I)^\frac{1}{2} I (I)^\frac{1}{2} -(I)^\frac{1}{2} I^T (I)^\frac{1}{2}$$
$$= \sqrt{I} + \sqrt{I} - I - I$$ $$=\mathbb{0}$$
And the zero $n$ square matrix has $n$ zero eigenvalues (not one positive).