For the absolute value $|C|=(C^*C)^\frac{1}{2}$ and the Hilbert-Schmidt norm $\parallel C\parallel_{HS}=(trC^*C)^\frac{1}{2}$ of the operator $C$. The following inequlityinequality is shown by Araki et al in 'An Inequality for Hilbert-Schmidt Norm, Commun. Math. Phys. 81, 89-96 (1981)'
For any two bounded linear operators $A$ and $B$ on a Hilbert space $\mathbb{H}$, $$\parallel |A|-|B| \parallel_{HS}\le \sqrt{2}\parallel A-B \parallel_{HS},$$ and the factor $\sqrt{2}$ is best possible.
For any two bounded linear operators $A$ and $B$ on a Hilbert space $\mathbb{H}$, $$\parallel |A|-|B| \parallel_{HS}\le \sqrt{2}\parallel A-B \parallel_{HS},$$ and the factor $\sqrt{2}$ is best possible. What is the best constant factor $c$ such that $$\parallel |A|+|B| \parallel_{HS}\ge c\parallel A+B \parallel_{HS}.$$$$\parallel |A|+|B| \parallel_{HS}\ge c\parallel A+B \parallel_{HS}?$$