Show the negative-definiteness of a squared Riemannian metric

Let $\Bbb{S}_{++}^n$ denote the space of symmetric positive definite (SPD) $n\times n$ real matrices.

The geodesic distance between $A,B\in\Bbb{S}_{++}^n$ is given by the following Riemannian metric
$$
d(A,B):= \Bigg(\operatorname{tr}\bigg(\ln^2\big(\sqrt{A^{-1}}B\sqrt{A^{-1}}\big)\bigg)\Bigg)^{\frac{1}{2}}.
$$

EDIT
It could also be defined as follows
$$
d(A,B):= \lVert\log(A^{-1}B)\rVert_{F}
$$

EDIT II I could show the negative-definiteness of $d^2$ if I could write $d$ as follows
$$
d(A,B)=\lVert P-Q\rVert_{F},
$$
where $P$ depends solely on $A$, and $Q$ depends only on $B$. So, is there any way of expressing $d$ as above?

I would like to prove that $d^2$ is a negative-definite function.

For the negative-definiteness of a function $f\colon\mathcal{X}\times\mathcal{X}\to\Bbb{R}$, it suffices to show that, for all $m\in\Bbb{N}$, $\{x_1,\ldots,x_m\}\subset\mathcal{X}$, $m\in\Bbb{N}$, $\{c_1,\ldots,c_m\}\subset\Bbb{R}$, $\sum_{i=1}^{m}c_i=0$, the following holds true
$$
\sum_{i,j=1}^{m}c_ic_jf(x_i,x_j)\leq0.
$$

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