How much does the absolute value of an operator behave like an absolute value? 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?
 A: While the above inequalities may not be valid for all operators, it is true that if  and T are normal and commute, then $|T-S| \leq|T|+|S|$, see Corollary 3.6 in Mortad - on the absolute value of the product and the sum of linear operators, which also proves other related inequalities
A: The first inequality for example is false but
https://projecteuclid.org/download/pdf_1/euclid.pja/1195519395
shows that a slightly weaker inequality with logarithmic factors does hold for the spectral radii when the operators are selfadjoint. I don't know if it can be extended to a relaxed inequality for the Loewner order.
A: The answer is no. Loewner partial order is subtle. There are some simple examples given in R. Bhatia,  F. Kittaneh, The matrix arithmetic–geometric mean inequality
revisited, Linear Algebra and its Applications 428 (2008) 2177–2191
A: In general, if $p_1$, $p_2$ are non-commuting rank $1$ projectors, then
$$p_1 + p_2 \not \succeq |p_1 - p_2|$$
Indeed, may assume $p_1 = \left(\matrix{ 1 &0\\0&0}\right)$, $p_2 = \left(\matrix{ \cos^2 \theta &\cos \theta \sin \theta\\\cos \theta \sin \theta&\sin^2 \theta}\right)$, then one checks $(p_1-p_2)^2 = \sin^2\theta \cdot I_2$, and so 
$$p_1 + p_2 - |p_1 - p_2| =  \left(\matrix{ 1+ \cos^2 \theta - |\sin \theta| &\cos \theta \sin \theta\\\sin \theta \cos \theta&\sin^2 \theta - | \sin \theta|}\right)$$
