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About Euclidean space (n-dimensional)

dn(x,y) is Euclidean distance between x and y.

For any $a∈R^n$ and $r>0$ , $B(a,r):=\{x∈R^n:dn(a,x)<r\}$.

For any $b∈R^n$ and $R>0$ , $B(b,R):=\{x∈R^n:dn(b,x)<R\}$.

Does the following proposition hold?

Proposition;

If $\,dn(a,b)<r+R\,$ holds, $c =a+(b-a) \frac{r}{r+R} \in B(a,r) \cap B(b,R)$ holds (That is, $\,\,dn(a,c)<r\,\,$ and $\,\,dn(b,c)<R\,$ hold).

I drawn a picture and I thought that c is in $B(a,r)$ and $B(b,R)$.

However, I can't prove this proposition by using formula.I think triangle inequality may be useful,but

$dn(a,c)\\≦ dn(a,b) + dn(b,c) \\ < r+R + dn(b,c) \\ = r+R + dn(b,a+(b-a) \frac{r}{r+R})\\...$

I can't prove.

Does this proposition holds? And if so, how can I prove that?

daㅤ
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1 Answers1

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$$dn(a,c)=|a-c|=|\frac{r}{r+R}(b-a)|<r$$ $$dn(b,c)=|b-c|=|\frac{R}{r+R}(b-a)|<R$$ Plain calculation will do just fine.

Roderick H
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