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Let $\Lambda_1,\dots,\Lambda_n$ be strictly positive definite operators in the Euclidean space $\mathbb{R}^d$. By an operator arithmetic-harmonic mean inequality with weights $\Lambda_i$ I mean the following:

For all strictly positive definite operators $X_i$: $$\operatorname{tr} \left[ \sum_i \Lambda_i^{1/2} X_i^{-1} \Lambda_i^{1/2} \right]^{-1} \le \operatorname{tr} \sum_i \Lambda_i X_i$$

What are the necessary and sufficient conditions on $\Lambda_i$ for such an inequality to hold? In particular, is $\sum_i \Lambda_i \ge 1$ necessary or sufficient, as suggested by the scalar weight case $\Lambda_i = \lambda_i$?

Upd. As pointed out by Narutaka OZAWA in the comments, $\sum_i \Lambda_i = 1$ implies the stronger operator inequality: $\left[ \sum_i \Lambda_i^{1/2} X_i^{-1} \Lambda_i^{1/2} \right]^{-1} \le \sum_i \Lambda_i^{1/2} X_i \Lambda_i^{1/2}$. It follows that $\sum_i \Lambda_i \ge 1$ is sufficient for the above inequality with traces.

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  • $\begingroup$ Even for scalar weights, I don't see why this should be true for all strictly positive definite $X_i$... $\endgroup$
    – Yemon Choi
    Dec 13, 2014 at 23:55
  • $\begingroup$ @YemonChoi For scalar weights a stronger statement is known: $\left[ \sum_i \lambda_i X_i^{-1} \right]^{-1} \le \sum_i \lambda_i X_i$ in the operator sense. $\endgroup$ Dec 14, 2014 at 0:07
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    $\begingroup$ @YemonChoi: And it's actually fairly elementary, since the $n=2$ case $(\lambda X_1^{-1} + (1-\lambda) X_2^{-1})^{-1} \le \lambda X_1 + (1-\lambda) X_2$ is equivalent to the more obvious $(\lambda + (1-\lambda) X_1^{1/2} X_2^{-1} X_1^{1/2})^{-1} \le \lambda + (1-\lambda) X_1^{-1/2} X_2 X_1^{-1/2}$, and general $n$ follows by induction from $n=2$. $\endgroup$ Dec 14, 2014 at 0:17
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    $\begingroup$ Yes, $\sum\Lambda_i\geq1$ is necessary and sufficient. This follows from one of Jensen's operator inequalities and the fact that $t^{-1}$ is operator convex. See Theorem 2.1 in arxiv.org/abs/math/0204049 where it deals with the case $\sum\Lambda_i=1$. The rest is probably an exercise. $\endgroup$ Dec 14, 2014 at 0:21
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    $\begingroup$ @AlexanderShamov: In fact, I didn't think of it. I don't see either. $\endgroup$ Dec 14, 2014 at 2:41

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