Show that if equality takes place in triangle inequality then one of the vectors is either null vector or they are linearly dependent

Question

Given $x,y \in \mathbb{C}^2$, where $x$ and $y$ - complex vectors and given equality $$\|x+y\| = \|x\| + \|y\|$$ show that either $x = 0$ or $ y = \lambda x, \lambda \in \mathbb{C}$

I tried to square the equality and see what follows but didn't come up with anything resembling a good result, how would one approach this ?

Answer 1

$$\|x+y\| = \|x\| + \|y\|$$

Squaring on both sides :

$$\|x+y\|^ 2= \|x\|^2 + \|y\|^2 + 2\|x\| \|y\|$$

But as $\|x+y\|^ 2=\|x\|^2 + \|y\|^2+2Re \langle x,y\rangle$,

We get :

$Re \langle x,y\rangle=\|x\| \|y\|$

Again as $Re \langle x,y\rangle \leq |\langle x,y\rangle|$ and by Cauchy-Schwarz, $|\langle x,y\rangle|\leq \|x\| \|y\|$,

We must have $|\langle x,y\rangle|= \|x\| \|y\|$, which holds iff $x = 0$ or $ y = \lambda x, \lambda \in \mathbb{C}$

Answer 2

Hint: Square both sides to get $$ \langle x,y \rangle = \|x\|^2 + \|y\|^2 \implies\\ \|x\|^2 + \|y\|^2 + 2 \operatorname{Re}\langle x,y \rangle = \|x^2\| + \|y\|^2 + 2 \|x\| \,\|y\| \implies\\ \operatorname{Re}[\langle x,y \rangle] = \|x\|\cdot\|y\|. $$