# Disjoint balls on Euclidean space

I'm doing some self-study and I'm stuck on a proof. Prove that if two open balls on $N$-dimensional Euclidean space are disjoint then $d(x,x') \ge r + r'$

$x$ and $x'$ are the centers of the balls, $r$ and $r'$ are the radii, and the distance between $x$ and $x'$ is given by:

$d(x,x') = \sqrt{ (x_1-x'_1)^2 + (x_2-x'_2)^2 + ...(x_N - x'_N)^2 }$

Also, $x$ and $x'$ are distinct points.

• If they are disjoint then why $\ge$ sign? will it be just $>$? – Marso Nov 4 '12 at 17:09
• They're open balls. I changed the question to be more specific. – ncRubert Nov 4 '12 at 17:21

HINT: Look at the line segment $S$ with endpoints $x$ and $x'$: $$S=\Big\{\alpha x+(1-\alpha)x':\alpha\in[0,1]\Big\}\;.$$ When $\alpha=0$, you get $x'$; as $\alpha$ increases towards $1$, the point moves along $S$ towards $x$, reaching $x$ at $\alpha=1$. When $\alpha=\frac12$, the point is midway between $x$ and $x'$. Let $p_\alpha=\alpha x+(1-\alpha)x'$; show that
$$d(p_\alpha,x')=\alpha d(x,x')$$ and $$d(p_\alpha,x)=(1-\alpha)d(x,x')\;.$$
Now suppose that $d(x,x')<r+r'$. Show that there is an $\alpha\in[0,1]$ such that $d(x,p_\alpha)<r$ and $d(x',p_\alpha)<r'$, and conclude that $p_\alpha\in B(x,r)\cap B(x',r')$.