if $\|x+y\|=\|x\|+\|y\|$, then $\|\alpha x+\beta y\|=\alpha \|x\|+\beta \|y\|$ Let $X$ be a normed linear space. Assume that for $x,y \in X$, we have $||x+y||=||x||+||y||$. Show that $||\alpha x+\beta y||=\alpha ||x||+\beta ||y||$ for every $\alpha,\beta \geq 0$.

My attempt: Suppose $\beta \geq \alpha$. Then $\|\alpha x+\beta y\|=\|\alpha (x+y)-(\alpha-\beta)y\| \leq |\alpha|(\|x\|+\|y\|)+|\beta - \alpha|\|y\|=\alpha\|x\|+\beta\|y\|$. 

Since the inequality is true for all $\alpha,\beta \geq 0$ and $\beta \geq \alpha$, if we let $\alpha=\beta=1$, we have $||x+y|| \leq |x||+||y||$. If the inequality is a strict inequality, then this contradicts with $\|x+y\|=\|x\|+\|y\|$. Hence, we must have equality, which is the desired result.

Is my proof correct? 

EDIT: Suppose we have $\|x+y\|=\|x\|+\|y\|$ for some $x,y \in X$. For the same $x$ and $y$, prove that $\| \alpha x+ \beta y\|=\alpha \|x\|+\beta \|y\|$
W.l.o.g. $\alpha \ge \beta$.

$$ \alpha\|x+y\| = \|(\alpha-\beta)y + (\alpha x + \beta y) \| \le (\alpha-\beta)\|y\| + \|\alpha x + \beta y\| \le (\alpha-\beta)\|y\| + \alpha\|x\| + \beta \|y\| = \alpha(\|x\|+\|y\|) $$
Since the first-quantity = last-quantity, all the inequalities must be equalities.  In particular:
$$(\alpha-\beta)\|y\| + \|\alpha x + \beta y\| = \alpha(\|x\|+\|y\|) .$$
I believe this can be answered in a much more basic way.

If X is a normed linear space, given $x,y \in X$ we have that $\alpha x, \beta y \in X$, because $X$ is closed under scalar multiplication.

So if $||x+y|| = ||x|| + ||y||$ $\forall x,y \in X$ we must also have $||\alpha x + \beta y|| = ||\alpha x|| + ||\beta y||= |\alpha|||x|| + |\beta|||y||= \alpha||x|| + \beta||y||$

Since $\alpha$ and $\beta$ $\geq0$.