Consider $ (R^p, \| \cdot \|)$ a norm space and let $x$ and $y$ be vectors such that $\| x + y \| = \| x \| + \| y \|$. Now show that for any $\alpha$, $\beta$ $\geq 0$, $\|\alpha x + \beta y\| = \|\alpha x \| + \|\beta y\|$.
This is with regard to general norms, i.e.
$$ ||\bar{x}|| \geq 0 $$ $$ ||\alpha\bar{x}|| = |\alpha | ||\bar{x}|| $$ $$ ||\bar{x} + \bar{y} || \leq ||\bar x|| + ||\bar y|| $$ $$ ||\bar{x}||= 0 \Leftrightarrow \bar x = 0 $$
With $|| \cdot ||_2$ the proof of this is simple, it implies colinearity of the vectors. For a general norm, I think it implies a somewhat weaker notion of "colinearity", in the hand-wavey sense. This is a homework question and I don't really want the answer---that will not be of much help.
My assessment of the problem is to solve it by making strict use of the four properties of a norm that I've listed above but I am missing some key insight that tells me my maths toolbox is lacking. I would appreciate some problem solving guidance that will eventually lead me to figure this out.
First, you may assume that $a > 0$ and $b > 0$. Then, without loss of generality that $a \geq b$. Set $\lambda := b/a \in (0,1]$. It suffices to show that $||x+ \lambda y||= ||x|| + ||\lambda y||$. By the triangle inequality, "$\leq$" holds. For the remaining inequality, write $$ ||x|| + ||\lambda y|| = ||x+ y|| - (1-\lambda)||y|| = ||x+y|| - ||(1-\lambda)y|| \leq ||x+ y - (1-\lambda)y||\,, $$ using the assumption in the first equal sign.