Consider the ball $ B(0, R) := \{ x | ||x|| \le R \} $ and consider a point $x$ outside of the ball, that is $||x|| > R$. Now i construct another ball of radius $\frac{1}{2}(||x|| - R)$ around $x$ and i want to show that these two balls have no points in common. For this consider a $x'$ which lies in the second ball, then the following hold:
- $|x' - x| \le \frac{1}{2}(||x|| - R)$
- $||x|| > R$
- $||x' - x|| + ||x|| \ge ||x'||$ (triangle inequality)
from this i want to deduce that $||x'|| > R$ but i am stuck. Any help?