It's pretty basic and I'm sure I'm missing something dumb here, but I'd like to know why $||A+B||_F \leq ||A||_F+||B||_F$
The way I understand it,
$||A+B||^2_F=tr((A+B)^T(A+B))=tr((A^T+B^T)(A+B))=tr(A^TA+A^TB+B^TA+B^TB)$
Now using the property the trace is linear:
$tr(A^TA+A^TB+B^TA+B^TB) = tr(A^TA)+tr(B^TB)+tr(A^TB+B^TA)=||A||^2_F+||B||^2_F+tr(A^TB+B^TA)$
Now if we were to prove that $2||A||_F||B||_F \geq tr(A^TB+B^TA)$ that would solve the question. But I don't see how that's trivial, and generally multiplying $\sum$s together is something I avoid like the plague. Is this indeed the way? would someone help me with this last step?
Notice that $\text{tr}(A^TB)=\sum_{i,j}a_{ij}b_{ij}$, so by the Cauchy-Schwarz applied to $(a_{ij})$ and $(b_{ij})$ treated as vectors $|\text{tr}(A^TB)|\leq\text{tr}(A^TA)^{1/2}\text{tr}\,(B^TB)^{1/2}=\|A\|_F\|B\|_F$, and switching $|\text{tr}(B^TA)|\leq\text{tr}(B^TB)^{1/2}\text{tr}\,(A^TA)^{1/2}=\|A\|_F\|B\|_F$. Therefore, $$|\text{tr}(A^TB+B^TA)|\leq|\text{tr}(A^TB)|+|\text{tr}(B^TA)|\leq 2\|A\|_F\|B\|_F.$$