# Why in this proof we get $\alpha \geq 0$?

I've solved the following problem: "Let $u,v \in \mathbb{R}^n$ with $u \neq 0$ be such that $|u+v|=|u|+|v|$ (euclidean norm), show that there's $\alpha \in \mathbb{R}$ with $\alpha \geq 0$ such that $v = \alpha u$".

My approach was the following: By definition of the euclidean norm, we have that $$|u+v|^2=\left\langle u+v,u+v\right\rangle=|u|^2+2\left\langle u,v\right\rangle+|v|^2,$$ but by hypothesis $|u+v|^2=(|u|+|v|)^2=|u|^2+2|u||v|+|v|^2$ so that equating those two we must have $\left\langle u,v\right\rangle = |u||v|$ and so $|\left\langle u,v\right\rangle| =|u||v|$ so that by the Cauchy-Schwarz inequality there must be $\alpha \in \mathbb{R}$ such that $v=\alpha u$.

Since I've not shown that $\alpha \geq 0$ in this proof, I thought it was wrong, but the answer in the book gives the same proof and says that all of this implies $\alpha \geq 0$. I've gone again through the proof of Cauchy-Schwarz Inequality and it doesn't seem to be there the reason for this. Indeed, if $x,y \in \mathbb{R}^2$ are simply $x=(1,1)$ and $y=(-2,-2)$ then $y=-2x$ and indeed $|\left\langle x,y\right\rangle| = |x||y|$, so by this example we see that Cauchy-Schwarz Inequality doesn't imply that the scalar that multiply one vector to get the other must be positive.

So, how this $\alpha$ is shown to be greater or equal to zero?

EDIT: I think I've found the answer. I've shown that $\left\langle u,v\right\rangle =|u||v|$ and thus $|\left\langle u,v\right\rangle| = |u||v|$ by Cauchy-Schwarz Inequality we must have $v=\alpha u$, but since $\alpha = \left\langle u,v\right\rangle /|u|^2$, since the inner product itself is a positive number (because it's a product of two norms), then $\alpha$ must be positive. Is this right?
You are correct: $\alpha \ge 0$ should be shown independently.