Suppose $A$ and $B$ are positive definite (symmetric) real matrices.
Is it true that $(AB)^{1/2}+(BA)^{1/2} \geq A^{1/2}B^{1/2}+B^{1/2}A^{1/2}$ ?
EDIT: Shown to be false below. Extension sought: Consider now a square strictly contractive matrix $S$ (such that $I-SS^T>0$) and consider the inequality $(AB)^{1/2}+(BA)^{1/2}\geq A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}$ ? Does this contraction change things, i.e. is the inequality true?
A related question but "in reverse" is asked here: An inequality on the root of matrix products (part 2 - the reverse case)
It's not true. If it's true, by continuity, it's also true for positive semidefinite matrices. Yet we have a counterexample in the latter case: \begin{align*} &A=X^2,\ B=Y^2\ \text{ where }\ X=\sqrt{\frac85}\pmatrix{1&2\\ 2&4},\ Y=\pmatrix{1&0\\ 0&\tfrac12},\ \\ &AB=\pmatrix{8&4\\ 16&8}=R^2,\ \text{ where }\ R=\pmatrix{2&1\\ 4&2},\\ &R+R^T=\pmatrix{4&5\\ 5&4},\\ &XY+YX=\sqrt{\frac85}\pmatrix{2&3\\ 3&4}. \end{align*} Neither is $R+R^T\succeq XY+YX$ (the last entry of $R+R^T$ is smaller than the last entry of $XY+YX$) and nor is $XY+YX\succeq0$ (its determinant is negative).
It should be easy to generate by computer a random counterexample for the positive definite case, but obtaining one with nice matrix entries may require a bit more work.