Can anyone prove this statement? It seems true, but I'm finding it tricky to give a concise proof.


Fix $\alpha\in[0,1]$. Let $\mu$ be Lebesgue measure. Define $B(c,r)\equiv[c-r,c+r]$, where $[\cdot, \cdot]$ denotes an interval. For $i=1,\ldots,n$, fix $r\_1,\ldots,r\_n\in[0,\infty)$, and $c\_1,\ldots,c\_n\in\mathbb R$. Let $\bar c, \bar r, \tilde c$, and $\tilde r$ satisfy $B(\bar c, \bar r)=\bigcap\_{i=1}^{n}B(c\_{i},r\_{i})$ and $B(\tilde c, \tilde r)=\bigcap\_{i=1}^{n}B(c\_{i},\alpha r\_{i})$.
Then, $\tilde r \le \alpha r$.



Length of intersection of intervals
The problem is reduced to the case $n=2$. Indeed, let $\cap B(c\_i,r\_i)=(a,b)$.
Changing the numeration we can assume $a=c\_1-r\_1$. Then $b=c\_k+r\_k$ for some $k$.
Now it is clear that if $\cap B(c\_i\alpha r\_i)=(a',b')$ then $a'\geq c\_1-\alpha r\_1$ and $b'\leq c\_k+\alpha r\_k$. So it is enough to prove that 
$$c\_k-c\_1+\alpha r\_k+\alpha r\_1\leq \alpha(c\_k-c\_1+r\_k+r\_1),$$
that is $c\_1\geq c\_k$.
But this last inequality is easy, just draw a picture of intersection of two intervals.


