# Applying a contraction to balls' centers increases the size of the balls' intersection?

The following statement seems clearly true, but I'm having a hard time proving it:

Fix $\alpha\in[0,1]$. Let $\mu$ be Lebesgue measure. For $r\ge 0$, let $B(c,r)\equiv[c-r,c+r]$. Fix $r_1,\ldots,r_n\in[0,\infty)$, and $c_1,\ldots,c_n\in\mathbb R$. Then, $\mu\left(\bigcap_{i=1}^{n}B(\alpha c_{i},r_{i})\right)\ge\mu\left(\bigcap_{i=1}^{n}B(c_{i},r_{i})\right)$

In words, the statement just says that contracting the 1-dimensional balls' centers towards $0$, by a constant proportion, while leaving the radii unchanged, increases the amount of overlap. This seems straightforward enough: if the centers were contracted all the way to the origin, then the intersection would just be the smallest ball. There can't be more overlap than that.

Can anyone suggest a simple proof, or point me to a relevant theorem?

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Let $B(c,r)=\bigcap_{i=1}^n B(c_i,r_i)$. We have $|c_i-c|\le r_i-r$ for all $i$. Hence, $|\alpha c_i-\alpha c|\le r_i-r$, which implies $B(\alpha c,r)\subseteq \bigcap_{i=1}^n B(\alpha c_i,r_i)$.
Explanation: a ball $B$ is contained in ball $B'$ if and only if the distance between centers is at most radius(B')-radius(B).
Here it helps that in 1 dimension the intersection of balls is also a ball. I wonder what a proof would be like in higher dimensions. As far as I know, for general 1-Lipschitz maps $f\colon \mathbb R^n\to\mathbb R^n$ the inequality $\mu\left(\bigcap_{i=1}^{n}B(f(c_{i}),r_{i})\right)\ge\mu\left(\bigcap_{i=1}^{n}‌​B(c_{i},r_{i})\right)$ remains conjectural. – user31373 Jun 21 '12 at 21:12