I'm pretty sure that the inequality $$ \vert {\rm tr}(A^*B) \vert \leq {\rm tr}( |A| |B| ) $$ holds whenever $A$ and $B$ are Hilbert-Schmidt, just by using the polar decompositions of $A$ and $B$.
If this is the case, then we'd have
$$ \eqalign{ {\rm tr} ((A+B)^*(A+B)) & = {\rm tr}(A^*A) + {\rm tr}(A^*B) + {\rm tr}(B^*A) + {\rm tr}(B^*B) \\ & \leq {\rm tr}(\vert A\vert^2) + {\rm tr}(\vert A\vert \vert B\vert) + {\rm tr}(\vert B\vert \vert A\vert) + {\rm tr}(\vert B\vert^2) = {\rm tr}((\vert A\vert + \vert B\vert)^2) } $$ which implies that you can get away with $c=1$. Taking $A=B$ to be positive shows that this is sharp.