Recall that the absolute value of a bounded operator $T$ on a Hilbert space $H$ is the unique positive operator $|T|$ such that $$\||T|x\|=\|Tx\|$$ for all $x\in H$. It can be defined using the continuous functional calculus, or if you have square-roots of positive operators in hand, by $|T|=(T^*T)^{1/2}$. Likewise, this can be defined in any C*-algebra.

With respect to the usual ordering on self-adjoint elements, i.e., $S\leq T$ if $T-S$ is positive, does this absolute value behave like an absolute value? For instance, does it satisfy a triangle inequality like the one below? $$||T|-|S||\leq|T-S|\leq|T|+|S|$$

If $S$ and $T$ are normal and commute, then perhaps one could use the functional calculus on the commutative C*-algebra they generate to show this, but not for general operators.

How about other inequalities with respect to this ordering?