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Suppose $A$ and $B$ are symmetric positive (semi-)definite, and $A<B$ in Loewner order, meaning $B-A$ is positive (semi-)definite. Is it true that, for a symmetric positive-definite $C$, we have $ACA < BCB$?

If it helps, $A$ and $B$ can be assumed to be invertible.

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  • $\begingroup$ Are you trying to show that $\forall A,B\exists C$ or $\forall A,B\forall C$? $\forall A,B\forall C$ can be easily disproven using computer calculations by testing random positive semidefinite matrices. $\endgroup$
    – Joseph Van Name
    Commented Sep 3, 2022 at 15:31
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    $\begingroup$ The function $X\mapsto X^2$ is not operator monotone (search for these keywords perhaps for more information), so this fails even when $C=1$. $\endgroup$
    – Christian Remling
    Commented Sep 3, 2022 at 16:26
  • $\begingroup$ Thanks Christian, that's an excellent counterexample. $\endgroup$
    – Athere
    Commented Sep 4, 2022 at 16:12

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