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.