Suppose $X,Y,Z$ are real-valued random variables, with non-negative covariances: $$ \text{Cov}(X,Y),\text{Cov}(Y,Z),\text{Cov}(X,Z)\geq0. $$ Is it true that $$ \text{Var}(X-Y)+\text{Var}(Y-Z)\geq \text{Var}(X-Z)? $$ If this inequality is true then it has a nice consequence: $\text{Var}(X-Y)$ defines a metric (on a space of random variables with non-negative covariances).
It feels like, if this inequality is true, there must be a simple proof. However I am struggling. So far I have only managed to find an equivalent inequality: $$ \text{Cov}(X,Z)+\text{Var}(Y)\geq \text{Cov}(X,Y)+\text{Cov}(Y,Z). $$ Any thoughts much appreciated.
The inequality is false. Here is a counter-example. Consider independent real-valued random variables $U,V$, with $\text{Var}(U)\geq\text{Var}(V)>0$; and set $Y=U$, $X=U+V$, $Z=U-V$.
Then the pairs of $X,Y,Z$ see positive covariances. Meanwhile $$ \text{Var}(X-Y)+\text{Var}(Y-Z)=2\text{Var}(V)<4\text{Var}(V)=\text{Var}(Z-X). $$