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Suppose $a,b$ are positive numbers, $a\neq b$. Then, the relationship $$ \frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)>\frac{1}{a+b} $$ is true because we can rewrite it as $$ a+b>4\frac{ab}{a+b} $$ or $$ (a-b)^2>0 $$ My question: Is there a similar relationship for p.d. matrices (of suitable dimension so that addition and multiplication work), i.e., can one establish that $$ \frac{1}{4}\left(A^{-1}+B^{-1}\right)-(A+B)^{-1}>0? $$ With similar steps to the scalar case, I write the claim as $$ A+B-4B(A+B)^{-1}A>0, $$ from which the analogous steps to the scalar case do not directly go through anymore. Using, e.g., Inverse of the sum of matrices did not help me proceed, either.

For context, the assertion would help me establish the general case here (I believe/hope that specific application has no properties that I do not mention in my question here): https://stats.stackexchange.com/questions/588398/help-partitioned-samples-efficiency-in-ols-compared-to-one-sample-regression

Christoph Hanck
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    For what it's worth, a numerical search suggests that the claim is true for $2\times 2$ matrices. – user7530 Sep 13 '22 at 09:47
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    Are you aware that your first equation can be written $\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)>\frac{2}{a+b}$, otherwise said $\frac{1}{H}>\frac{1}{A}$, itself equivalent to $H<A$ where $A$ is the arithmetic mean of $a,b$ and $H$ their harmonic mean ? – Jean Marie Sep 13 '22 at 13:09
  • Thanks, very useful indeed! – Christoph Hanck Sep 13 '22 at 13:45

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Yes, because $\frac{1+x^{-1}}{4}-(1+x)^{-1}>0$ for all positive $x$ and $$ \frac{1}{4}\left(A^{-1}+B^{-1}\right)-(A+B)^{-1} =A^{-1/2}\left[\frac{I+X^{-1}}{4}-(I+X)^{-1}\right]A^{-1/2} $$ where $X=A^{-1/2}BA^{-1/2}$.

user1551
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  • Thanks, I suppose this is the/a way to go. It may be considered hand holding at this point, but since the original claim was not clear to me, nor is that the term in square brackets is p.d... – Christoph Hanck Sep 18 '22 at 10:51