Prove that if two open balls are disjoint then the distance between their centres is greater than the sum of their radii I was trying to prove the following obvious fact , but despite having the intuition on why this is true I could not prove it 
If , 
$ B(x,r)\cap B(x',r') = \emptyset $ 
then , 
$ d(x,x') \geqslant r + r'$
I tried forming the following equations 
 $ \forall X \in B(x,r) $
1.)$ d(x,X) < r $
2.) $ d(x',X) >r' $
Similarly , $\forall Y \in B(x',r')  $


*

*$d(x',Y) < r'$

*$ d(x,Y)> r $ 
Then I tried to add the equations to see if I can use the properties of the metric function to get the desired inequality but I could not,any help , hint would be appreciated . 
 A: Show the contrapositive instead. Define $s = r + r'$ and $\lambda = r/s$. Then $r = \lambda s,r' = (1-\lambda) s$. If $d(x,x') < r + r'$, define $y := (1-\lambda) x + \lambda x'$. Then
$$ d(y,x) = \|\lambda x' - \lambda x\| = \lambda \|x' - x\| < \lambda s = r
$$
so $y \in \mathrm B(x,r)$. A symmetric argument shows $y \in \mathrm B(x',r')$, so the balls are not disjoint.
$y$ is a convex combination of $x$ and $x'$.
A: Here is how you can establish the result in a normed space without using connectedness directly
(this follows the usual proof that an interval is connected).
Let $p(t) = x+t(x'-x)$, note that $p(0) \in B(x,r), p(1) \in B(x',r')$
and $B(x,r), B(x',r')$ are disjoint.
Let $t^* = \sup \{ t \in [0,1] | p(t) \in B(x,r) \}$. Note that $t^* >0$
since $B(x,r)$ is open and $t^* < 1$ since $B(x',r')$ is open. Also
note that $\|p(t_1)-p(t_2)\| = \|(t_1-t_2)(x'-x)\| = |t_1-t_2| \|x'-x\|$.
Note that $t^* \notin B(x,r)$ since $B(x,r)$ is open (otherwise this would
contradict the definition of $t^*$) and $t^* \notin B(x',r')$ since
$B(x',r')$ is open and the two balls are disjoint.
Hence $p(t^*) \notin B(x,r) \cup B(x',r')$.
Note that $p(1)-p(0) = p(1)-p(t^*) + p(t^*)-p(0)$ and
$\|p(1)-p(0) \| = \|p(1)-p(t^*)| + \|p(t^*)-p(0)\|$ (that is, we have equality) and hence
$\|x'-x\| = \|x'-p(t^*)\| + \|p(t^*)-x\| \ge r' + r$.
