Understanding an equality with open balls

Question

In order to understand a proof I want to know why the following is true:

Let X be a banach space and $x\in X$ with $x\in B(x_0,r)$ (open ball around $x_0$ with radius $r$), then $\frac{r}{2}x-x_0 \in B(x_0,r)$

That should be quite easy but I still can't understand it.

So I hope someone can explain why it is true. Maybe with a good explanation, but I would like to see a proof involving $d(\frac{r}{2}-x_0,x_0)\leq....\leq r$

Thanks!

Edit: The assumption also says that $||x||=1$

Answer 1

it is not true, say on the real line take $x_0=3$, $r=2$, $x=1.2$, then $\dfrac r 2 x-x_0= -1.8\not\in B(x_0,r)$.

Hmm, edit, now the assumption says $||x||=1$.

On the real line take $x_0=2.5$, $r=2$, $x=1$, then $\dfrac r 2 x-x_0= -1.5\not\in B(x_0,r)$.