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I was in an other proof, in that we have to find a positive lower bound for $|z-w|,$ where where $w$ lies in the boundary of the open ball $B(a,r)$ and $z\in B(a,r/2)$.

From the figure we can easily identify that $|z-w|\geq r/2$, my try is to use the result $B(a,r)=a+B(0,r)$, But I am not convinced, Is there any easy way to prove this?

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We have, using the Triangle inequality, \begin{align} |w-a| & \leq |w-z| + |z-a|, \\\implies |w-z| & \geq |w-a| - |z-a|. \end{align} We find from your statement that $|w-a|=r$ and $|z-a|\leq r/2$, therefore getting \begin{align} |w-z| \geq r-\frac{r}{2} = \frac{r}{2}. \end{align} I hope this helps!

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  • $\begingroup$ Yeah, it helps tq :-) $\endgroup$ – Madhan Kumar Sep 19 at 13:15
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$r=|a-w|\leq|a-z|+|z-w|$ by triangle inequality so that: $$|z-w|\geq r-|a-z|\geq r-\frac12r=\frac12r$$

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