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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?

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    It would help to give a different name to the center of the second ball, maybe $c$. You use $x$ in more than one context.2012-11-09

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$\def\norm#1{\left\|#1\right\|}$Use the 'other' triangle inequality, your version gives upper bounds for $\norm{x'}$ which aren't of any help to deduce lower bounds. We have (by the triangle inequality $\norm{x} \le \norm{x'} + \norm{x-x'}$ and your other two inequalities) \begin{align*} \norm{x'} &\ge \norm{x} - \norm{x-x'}\\ &\ge \norm{x} - \frac 12(\norm x - R)\\ &= \frac 12\norm x + \frac 12R\\ &> \frac 12 R + \frac 12R\\ &= R. \end{align*}