Consider the space $X$ of all scalar products on $\mathbb{R}^n$. For a scalar product $s$ and a base $B:=b_1\ldots,b_n$ let $M_{s,B}$ denote the matrix, whose $(i,j)$-th entry is $(s(b_i,b_j))$ . Given two scalar products $s,s'$ one can find by PCA a orthonormal basis $B$ of $s$ such that $M_{s',B}$ is a diagonal matrix. By positive definiteness, all diagonal entries $\lambda_,\ldots,\lambda_n$ are positive. Let $d(s,s'):=\sqrt{\sum_{i=1}^n\log(\lambda_i)^2}$. I want to show, that this defines a metric on the set of all scalar products. Symmetry and Definiteness are clear. But why does it satisfy the triangular inequality ?

To be honest I already know, that it is a metric. This distance function comes from a Riemannian metric on the set of all scalar products. But I am looking for a simpler way (without computing the Levi-Civita connection and showing, that the geodesics satisfy the ODE and so on).

Furthermore the resulting space should be a $CAT(0)$-space. It would be nice if one could show the CAT(0) inequality directly. The geodesic from $s$ to $s'$ is given by

$$[0;1]\rightarrow X\qquad t\mapsto s_t,\mbox{ where } s_t(b_i,b_j)=\begin{cases}\lambda_i^t&i=j \\\ 0& i\neq j\end{cases}$$