Let $\mathbb{S}_n$ denote the set of $n \times n$ symmetric positive semidefinite matrices. I am trying to figure out whether $k: \mathbb{S}_n \times \mathbb{S}_n \to \mathbb{R}_+$ defined as:

$$k(A, B) = \text{Tr}[(A^{1/2} B A^{1/2})^{1/2}]$$

is a positive definite kernel. Can anyone find a counterexample showing it is not? Or prove it is indeed a positive definite kernel?

**Some analysis so far:** Having failed to find a counterexample numerically, I am trying to show:
$$\sum_{ij} k(A_i, A_j) x_i x_j \geq 0$$
holds for any $A_1, \dots, A_m \in \mathbb{S}_n$ and $x_1, \dots, x_m \in \mathbb{R}$.
Without loss of generality we can assume that $x_1, \dots, x_m \in \{-1, +1\}$ since the change of $x_i \mapsto \text{sign}(x_i)$ and $A_i \mapsto x_i^2 A_i$ preserves the value on the left hand side above and all matrices remain positive semidefinite after this change of variables.

Let $\mathcal{P}$ denote the set of indices where $x_i x_j = +1$ and let $\mathcal{N}$ denote the set of indices where $x_i x_j = -1$. Then we just need to show the following inequality holds: $$\sum_{i=1}^m \text{Tr}[A_i] + 2 \sum_{(i,j)\in\mathcal{P}} k(A_i, A_j) \geq 2 \sum_{(i,j)\in\mathcal{N}} k(A_i, A_j)$$ This is where I'm stuck, or maybe a different approach is needed.

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