# Proving the inequality $r/2\leq|z-w|,$where $w$ lies in the boundary of the open ball $B(a,r)$and $z\in B(a,r/2)$

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?

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!

• Yeah, it helps tq :-) – Madhan Kumar Sep 19 at 13:15

$$r=|a-w|\leq|a-z|+|z-w|$$ by triangle inequality so that: $$|z-w|\geq r-|a-z|\geq r-\frac12r=\frac12r$$