If $x,y\in \Bbb{R}^n,\|x+y\|=\|x\|+\|y\|$ then show there is a $\lambda \in \Bbb{R}$ such that $x=\lambda y$. If $x,y\in \Bbb{R}^n,\|x+y\|=\|x\|+\|y\|$ then show there is a $\lambda \in \Bbb{R}$ such that $x=\lambda y$.
 A: Recall $\| x\| = \sqrt{\langle x, x\rangle} $ for $x\in \mathbb{R}^n$

We know that $$\| x + y\|^2 = \|x\|^2 + \| y \|^2 + 2\| x\|\|y \| $$
Then, we have that $$\langle x, x \rangle + \langle y, y \rangle + 2\sqrt{\langle x, x \rangle \cdot \langle y, y \rangle} $$ $$= \langle x + y, x + y \rangle = \langle x , x + y \rangle + \langle y, x + y \rangle = \langle x, x \rangle + \langle y , y\rangle + 2 \langle x, y \rangle   $$
By Cauchy Schwartz:
$$ 2|\langle x , y\rangle | \leqslant 2\sqrt{\langle x, x\rangle \cdot \langle y, y \rangle}  $$
Then it's the equality case of inequality. i.e., $x = \lambda y~$ for some $\lambda \in \mathbb{R}.$ Hence done!
A: If both $x,y$ are zero then $x=0 y$.
Suppose $y \neq 0$, and write $x=s y + z$, where $z \bot y$.
Then $\|x+y\|^2 = (s+1)^2 \|y\|^2 + \|z\|^2$, $\|x\|^2 = s^2 \|y\|^2 + \|z\|^2$
and $(\|x\|+\|y\|)^2 = s^2 \|y\|^2 + \|z\|^2 + \|y\|^2+2 |s| \|y\| \sqrt{s^2 \|y\|^2 + \|z\|^2}$.
Equating & simplifying yields $2s\|y\|^2 = 2 \|y\| \sqrt{s^2 \|y\|^2 + \|z\|^2}$.
This tells us that $s \ge 0$, and squaring both sides gives
$s^2 \|y\|^4 = s^2 \|y\|^4 + \|y\|^2 \|z\|^2$ from which we see that $z=0$.
Hence $x=sy$ with $s\ge 0$.
