We are given a normed space ($\mathbb{R^p}, ||.||$),
a) For all $\mathbf{x}$, $\mathbf{y}$ where $\mathbf{x}\neq\mathbf{y}$, $||\mathbf{x}||=||\mathbf{y}||$ implies $||\mathbf{x}+\mathbf{y}||<2$.
b) For all $\mathbf{x}$, $\mathbf{y}$ in $\mathbb{R^p}$, $\mathbf{y}\neq 0$ and $||\mathbf{x}+\mathbf{y}||=||\mathbf{x}||+||\mathbf{y}||$ implies $\mathbf{x} = a\mathbf{y}$ for a $\ge 0$
We are then asked to prove that a implies b by contradiction.
They tell us we should assume $\frac{\mathbf{x}}{||\mathbf{x}||}\neq\frac{\mathbf{y}}{||\mathbf{y}||}$ as a hint and work from there.
I have abolutely no clue how to proceed. Any advice will be appreciated.
Suppose $y \neq 0$ and $||x+y||=||x||+||y||$. Consider two cases: 1) $\frac x {||x||} =\frac y {||y||}$ and 2) $\frac x {||x||} \neq \frac y {||y||}$. In case 1) we already have $x=ay$ with $a=\frac {||x||} {||y||} \geq 0$. In case 2) we have $||\frac {x+y} {||x||} +\frac y {||y||} -\frac y {||x||}|| \geq ||\frac {x+y} {||x||}|| - ||y(\frac 1 {||x||} -\frac 1 {||y||})||= \frac {||x||+||y||} {||x||}-||y||(\frac {||y||-||x||} {||x||||y||})=2$ if $||y|| \geq ||x||$. We have proved that $||\frac x {||x||} +\frac y {||y||}||=2$ which contradicts 1). A similar argument can be given if $||y|| \leq ||x||$.