Frobenius Norm Triangle Inequality

You can use the fact that it's a norm from inner product : $$\langle A,B \rangle = \text{Trace}(A^TB)$$ Then: $$\|A+B\|^2=\|A\|^2+\|B\|^2+2\langle A,B\rangle$$ using Cauchy-Schwarz inequalitie we have : $$\langle A,B\rangle \leq \|A\| \|B\|$$ and this gives: $$\|A+B\|^2 \leq \|A\|^2+\|B\|^2+2 \|A\| \|B\|= (\|A\| + \|B\|)^2$$

It suffices to show $$\sqrt{\sum_{i=1}^nx_i^2}+\sqrt{\sum_{i=1}^ny_i^2}\geqslant\sqrt{\sum_{i=1}^n(x_i+y_i)^2}.$$Sqaure it, it's equivalent to show$$\sqrt{\sum_{i=1}^nx_i^2}\sqrt{\sum_{i=1}^ny_i^2}\geqslant\sum_{i=1}^nx_iy_i,$$which is the Cauchy's Inequality.

It also works in $\mathbb{C}$ with slight modification.