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 $\| \alpha x+ \beta y\|=\alpha \|x\|+\beta \|y\|$ 
 A: 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:
$$(\alpha-\beta)\|y\| + \|\alpha x + \beta y\| = \alpha(\|x\|+\|y\|) .$$
A: I'm not sure what you mean by "normed linear space".  The usual definition includes, for example, the requirement that $\| \alpha x \| = |\alpha| \| x \|$, which equals $\alpha \| x\|$ if $\alpha \geq 0$, from which the claim follows trivially.
Perhaps your definition does not have that condition.  Any function $\|\cdot\|$ satisfying $\|x+y\| = \|x\| + \|y\|$ for all $x,y$ satisfies $\|mx\| = m\|x\|$ for $m\in \{1,2,\dots\}$ by induction.  This implies that $\|rx\| = r\|x\|$ for $r$ a positive rational number, by division.  If you know your function to be continuous, the equation for $r$ real follows.
Actually, note that if $\|x+y\| = \|x\| + \|y\|$ for all $x,y\in X$, then $\|x\| = \|x+0\| = \|x\| + \|0\|$, and so by subtracting $\|0\| = 0$.  Then $0 = \|0\| = \|x + (-x)\| = \|x\| + \|-x\|$, and so $\|-x\| = -\|x\|$, and the formula $\|rx\| = r\|x\|$ follows for all (possibly negative) rational $r$, and all real $r$ if $\|\cdot\|$ is continuous.
However, the usual definition of "norm" requires instead that $\|rx\| = |r|\|x\|$.  For $r$ negative, we conclude that $r\|x\| = -r\|x\|$, from which it follows that $\|x\| = 0$.  If your definition includes the condition $\forall x((\|x\| = 0) \Rightarrow (x = 0))$, then your vector space $X$ is the zero vector space.
So for the problem to be nontrivial, my conclusion is that either you are using funny definitions, or leaving out some quantifiers.
A: 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 $\beta$ $\geq0$.
