Let $ z = ( z_1, z_2, \dots, z_m) \in \mathbb{C}^m$ and $z_j>0$ for all $ 1 \leq j \leq m.$ Then $\| \sum_{j} z_j \|_2 = \sum_j \| z_j \|_2$ if and only if $ z_j = \alpha_j z_1$ for some $ \alpha_j>0.$
This side is clear: If $ z_j = \alpha_j z_1$ for $ \alpha_j>0,$ then $ \| \sum_{j} z_j \|_2 = \| z_1 + \alpha_2 z_1 + \dots + \alpha_m z_1 \|_2 = \| ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m )z_1\|_2 = ( 1 + \alpha_2 + \alpha_3 + \dots + \alpha_m ) \|z_1\|_2 = \|z_1\|_2 + \|z_2\|_2 + \dots + \| z_m\|_2 = \sum_{j} \|z_j\|_2.$
For the other side of the proof, I know that I need to use the case of equality in the Cauchy-Bunyakovskii-Schwarz(CBS) Inequality, which says that: For all $x,y \in \mathbb{C}^{m \times 1 }$ we have \begin{align} |x^ * y| = \|x\| \|y\| \text{, where $x^*$ is the conjugate transpose of x,} \end{align} if and only if $ y = \alpha x$ for $ \alpha = \dfrac{x^ * y}{x^ * x}.$
Any directions or help with the proof?