distance and correlations

Question

For two random variables $X_i$ and $X_j$, I define: $d(X_i,X_j)$ = $\sqrt{2(1-\rho_{i,j})}$ where $\rho_{i,j}$ is the Pearson correlation coefficient. I want to show that d is a pseudometric.

The four axioms can be found here: https://en.wikipedia.org/wiki/Metric_(mathematics)

The axioms of nonnegativity and of symmetry are trivial. If I am not mistaken, the identity of indiscernibles does not hold because if $X_i = a X_j$, $ \ d(X_i, X_j) = 0$ but $X_i \ne X_j$, that is why I want to show that it is "only" a pseudometric.

My main question is about the triangle inequality. How can I check it holds? If $cor(X,Y) = a$, $cor(Y,Z) = b$ and $cor(X,Z) = c$, by positive semidefiniteness of the correlation matrix you can say that $b \in [ac-\sqrt{(1-a^2)(1-c^2)}, \ ac+\sqrt{(1-a^2)(1-c^2)}]$ (the determinant is positive).

I want to show that $\sqrt{2(1-a)}+\sqrt{2(1-b)} \geq \sqrt{2(1-c)}$

In the "worst" case, $b = ac+\sqrt{(1-a^2)(1-c^2)}$

I tried to study the functions: a $\mapsto \sqrt{2(1-c)} - \sqrt{2(1-a)} - \sqrt{2(1-b)}$ with b replaced by its upper bound - and the same function of c - to show that they are always negative on $[-1,1]^2$ but it soon gets intractable. I could only check the result numerically. I imagine there are much better ways to solve the problem. Do you know any of those?

PS: I have found the thread Is triangle inequality fulfilled for these correlation-based distances? and the answer of ttnphns but I could not find a proof of the fact that if the correlation matrix is PSD then $d$ is Euclidean. Isn't it at odd with the identity of indiscernibles?