Consider a real Hilbert space $\mathcal{X}$. For $(c,\rho)\in\mathcal{X}\times
\mathbb{R}$, I denote the closed ball $B(c;\rho) = \{x \in \mathcal{X}\, |\, \|x -c\|\leq\rho\}$.
I am curious if y'all know of a generalization of the following equivalence:

\begin{equation}
 \label{2sets} B(c\_1;\rho\_1) \cap B(c\_2;\rho\_2)
\neq \varnothing \quad \Leftrightarrow \quad \|c\_1 - c\_2\| \leq
\rho\_1 + \rho\_2. \tag{\*}
\end{equation}

My question:

> 
> Let $n\in\mathbb{N}$, let $(c\_i)\_{1\leq i\leq n}\in\mathcal{X}^n$, and let $(\rho\_i)\_{1\leq i\leq n}\in[0,+\infty[^n$. Is there an equivalent statement for
> $$\bigcap\_{1\leq i\leq n} B(c\_i;\rho\_i) \neq\varnothing,$$
> in the same spirit of (\*)? i.e. do you know of an equivalence which has **(1) no quantifiers** and **(2) phrasing via finitely-many inequalities involving only $(c\_i)\_{1\leq i\leq n}$ and $(\rho\_i)\_{1\leq i \leq n}$**? I have not found a result, even for $n=3$ and $\mathcal{X}=\mathbb{R}^N$.
> 
> 
> 

Here's a proof of (\*), where $B\_i$ denotes $B(c\_i;\rho\_i)$:

$(\implies)$: Let $x \in B\_1 \cap B\_2$. From the triangle inequality, $\|c\_1 - c\_2\| \leq \|c\_1 - x\| + \|c\_2 - x\| \leq \rho\_1 + \rho\_2$.

$(\impliedby)$: Construct
$x =\left(1 - \frac{\rho\_1}{\rho\_1+\rho\_2}\right)c\_1 + \frac{\rho\_1}{\rho\_1+\rho\_2}c\_2.$
Then $\|c\_2 - x\| = (1 - \frac{\rho\_1}{\rho\_1+\rho\_2})\|c\_1-c\_2\| \leq \rho\_2$, so $x \in B\_2$. Similarly, $\|c\_1 - x\| = \frac{\rho\_1}{\rho\_1+\rho\_2} \|c\_2 - c\_1\| \leq \rho\_1$, so $x \in B\_1 \cap B\_2 \neq \varnothing$ $\square$

Comments with partial results are appreciated!

EDIT: [This graph](https://www.desmos.com/calculator/wdvxlxde5t) may be helpful. It also displays that commonly checked candidate intersection points (e.g. center of mass) are not always in the intersection.