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 inequality 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.

What is the best constant factor $c$ such that
$$\parallel |A|+|B| \parallel_{HS}\ge c\parallel A+B
\parallel_{HS}?$$