Trace norm sum of tensor products For two arbitrary matrices $A, B$, it is fulfilled
$\|A+B\|\leq\|A\|+\|B\|$,
where $\|\cdot\|$ is the trace norm.
Is the following satisfied:
$\|A\otimes1+1\otimes B\|=\|A\|+\|B\|$?
As from the comments of user8675309 the above equality is probably not true, what about the following:
$\|\frac{1}{d_2} A\otimes1+ \frac{1}{d_1}1\otimes B\|=\|A\|+\|B\|$,
where $d_1/d_2$ ist the dimension of respectively the first/second  Hilbert space.
 A: The hoped for relation
$\big \Vert \frac{1}{d_2} A\otimes I+ \frac{1}{d_1}I\otimes B\big \Vert_{S_1}=\big \Vert A\big \Vert_{S_1} + \big \Vert B\big \Vert_{S_1}$
can't possibly true in general.  E.g. Set $A = I$ and $B:=-A$
then we have
$\big \Vert \frac{1}{d} I\otimes I+ \frac{1}{d}I\otimes (-I)\big \Vert_{S_1} = \big \Vert \frac{1}{d} I\otimes I- \frac{1}{d}I\otimes I\big \Vert_{S_1} =\big\Vert 0 \big\Vert_{S_1} = 0 \lt \big \Vert A\big \Vert_{S_1} + \big \Vert B\big \Vert_{S_1}$ 
where the RHS is $\gt 0$ by positive definiteness of the Schatten 1 norm (it is a norm afterall).  
now in general we do have
$\big \Vert \big(\frac{1}{d_2} A\big)\oplus \big(\frac{1}{d_1}B\big)\big \Vert_{S_1}$
$=\big \Vert \frac{1}{d_2} A\otimes I_{d_2}+ \frac{1}{d_1}I_{d_1}\otimes B\big \Vert_{S_1}$
$\leq \big \Vert \frac{1}{d_2} A\otimes I_{d_2} \big \Vert_{S_1} +\big\Vert \frac{1}{d_1}I_{d_1}\otimes B\big \Vert_{S_1}$
$=\big \Vert A\big \Vert_{S_1} + \big \Vert B\big \Vert_{S_1}$ 
by application of triangle inequality.  The final equality comes from the fact that Kronecker Products of two matrices, $X \otimes Y$, has singular values given by $\sigma(X\otimes Y)= \sigma(X)\otimes \sigma(Y)$ (which appears to be a homomorphism) 
