If $|u+v| = |u| + |v|$ then $u = \lambda v$. How do I prove $\lambda \ge 0$? I'm trying to prove that if $|u+v| = |u| + |v|$ implies $u = \lambda v$ for $\lambda \ge 0$.
To this end I have $|u + v|^2 = (|u| + |v|)^2 \Rightarrow  |u|^2 + |v|^2 + 2|u||v| = |u|^2 + |v|^2 + 2u\cdot v \Rightarrow |u||v| = u \cdot v \Rightarrow u = \lambda v$, where the last implication follows by Cauchy-Schwarz.
However, how do I prove that $\lambda \ge 0$ ?

If $\lambda <0$ then for $v\neq 0$
$$u\cdot v=\lambda v\cdot v=\lambda |v|^2<0.$$
Can you take it from here?
