When do three closed balls have a nonempty intersection? 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 may be helpful. It also displays that commonly checked candidate intersection points (e.g. center of mass) are not always in the intersection.
 A: A solution is an application of a theorem due to Karl Menger (as an alternative, one can use Schoenberg's theorem).
To simplify matters, I will assume for most of the answer that the points $c_1,...,c_n$ satisfy (*), that is, are in "general position:" they form the vertex set of an $n-1$-dimensional simplex. (I will explain in the end of the answer how to reduce the general case to this one.)
First, let me describe Menger's theorem.
Menger gave a set of necessary and sufficient conditions for  a finite metric space $(X,d)$ to embed isometrically in a Hilbert space $H$; he also gave a characterization of the least dimension of $H$ in terms of the metric $d$; I will denote this number $h(X)$ (I am suppressing the rotation for the metric here and below).
Here is Menger's solution (see my answer here for references).
Given a finite metric space $X=\{x_0, x_1,...,x_n\}$ (I am suppressing the notation for the metric),
Menger uses the following determinant, also known as the Cayley-Menger determinant:
$$
\Delta(X)= \left|\begin{array}{ccccc}
d(x_0,x_0) & d(x_0,x_1) & ... & d(x_0, x_n) & 1\\
d(x_1,x_0) & d(x_1,x_1) & ... & d(x_1, x_n) & 1\\
\vdots     & \vdots     & ... & \vdots      & \vdots\\
d(x_n,x_0) & d(x_n,x_1) & ... & d(x_n, x_n) & 1\\
1          & 1          & ... & 1           & 0
\end{array}\right|. 
$$
The first (and the most important) of Menger's conditions is that $\Delta(X)$ has the sign of $(-1)^{|X|}$ meaning:
$$
\Delta(X) (-1)^{|X|} \ge 0. 
$$
Furthermore, $h(X)= k$ implies $\Delta(X)=0$ (and the converse is true as well as long as $h(Y)=|Y|-1$ for all proper
subsets $Y\subset X$). The rest of Menger's conditions are inductive: For $X$ to embed isometrically in a Hilbert space, all subsets $Y$of $X$ have to be embeddable in Hilbert spaces, i.e. their determinants $\Delta(Y)$ have to have the sign of $(-1)^{|Y|}$ (as above).
Remark. Here is an important observation about the determinant $\Delta(X)$ regarded as a function in the variables $d(x_0,x_1),...,d(x_0,x_n)$:
$\Delta(X)$ is a 2nd degree polynomial in these variables, with the constant terms equal $\pm \Delta(X_0)$, where $X_0= X\setminus \{x_0\}$ (with the restriction of the metric). As a polynomial of $d(x_0,x_i)$  it has the form
$$
A_i d^2(x_0,x_i) + B_i d(x_0,x_i) + C_i, i=1,...,n,
$$
where $A_i= \Delta(X_{0i})\ne 0$ and $X_{0i}\subset X$ is obtained from $X$ by removing the points $x_0, x_i$).
This is where I am using the assumption (*).
I will  use the notation $H$ for an infinite-dimensional Hilbert space, containing all the Euclidean spaces $E^1\subset E^2\subset E^3\subset ...$. I will also use the notation $S(c,r)$ to denote the round sphere in $H$ centered at $c$ and of radius $r$.
Given a subset $C\subset H$, let $span(C)$ denote the affine span of $C$, i.e. the smallest affine subset of $H$ containing $C$.
Let's first solve a slightly different problem than yours:
Given a finite subset $\{c_1,...,c_n\}$ in $H$, what are the  necessary and sufficient conditions on the distances $d_{ij}=||c_i-c_j||$ and radii $r_i\ge 0$, for the intersection
$$
\bigcap_{i=1}^n S(c_i, r_i)
$$
of spheres in $H$ to be nonempty?
Menger's theorem provides an answer to the sphere problem. Namely: Given a tuple
$$
\tau=((c_1,r_1),...,(c_n,r_n)), 
$$
form an abstract pre-metric space $(X,d)=X_\tau$  equal to $\{c_0, c_1,...,c_n\}$ with
$$
d(c_i,c_j)=d_{ij}, d(c_0, c_k)=r_k, k=1,...,n.
$$
(The adjective pre-metric refers to the fact that $d$ might violate triangle inequalities when applied to triples $c_0, c_i, c_j$.)  Then the following are equivalent:

*

*$X_\tau$ embeds isometrically in $H$.


*$X_\tau$ is a metric space which satisfies the conditions in Menger's theorem, i.e.
(a) $d(c_i, c_k)\le  d(c_i, c_j) + d(c_j, c_k)$ for all triples $i, j, k\in \{0,...,n\}$ such that the product
$$
ijk=0.
$$
(b) For all subsets $Y\subset X_\tau$ containing $c_0$, $\Delta(Y) (-1)^{|Y|}\ge 0$
3.
$$
\bigcap_{i=1}^n S(c_i, r_i)\ne \emptyset. 
$$
Moreover,
$$
span(\{c_1,...,c_n\})\cap \bigcap_{i=1}^n S(c_i, r_i)\ne \emptyset 
$$
if and only if, additionally, $\Delta(X)=0$.
Note also that, for each finite-dimensional Euclidean subspace $A$ containing $\{c_1,...,c_n\}$, the intersection
$$
A\cap \bigcap_{i=1}^n S(c_i, r_i)
$$
is either empty, or is a single point, equal to the intersection of the above spheres in the Hilbert space $H$, as well as in $span(\{c_1,...,c_n\})$, or is a round sphere of dimension $\dim(A) - n$.
Now, let's turn to the original problem of intersection of closed balls in Euclidean spaces. It is easy to see that, if
$$
\bigcap_{i=1}^n S(c_i, r_i)\ne \emptyset, 
$$
then
$$
span(\{c_1,...,c_n\})\cap \bigcap_{i=1}^n B(c_i, R_i)\ne \emptyset, 
$$
for any $n$-tuples of real numbers $R_i\ge r_i$.
Definition. A collection of round balls ${\mathcal G}= \{B(c_1,r_1),..., B(c_n,r_n)\}$ in a (possibly infinite-dimensional) Euclidean space $E^\alpha$ will be called redundant if there is a proper subset $I\subset [n]=\{1,...,n\}$ such that
$$
\bigcap_{i\in [n]} B(c_i,r_i)= \bigcap_{i\in I} B(c_i,r_i). 
$$
The collection  of balls will be called irredundant otherwise. The same terminology applies to the tuple of centers and radii:
$$
\tau=((c_1,r_1),...,(c_n,r_n)). 
$$
It is easy to see that a tuple is redundant if and only if it is redundant in $span(\{c_1,...,c_n\})$.
If one knows that a tuple  $\tau$ is redundant, then one can describe necessary and sufficient conditions for nonemptyness of the intersection of a collection balls using a smaller subcollection, hence, give and inductive description this way.
As an example: For $n=3$, a tuple is redundant if and only if the 4-point pre-metric space $(X,d)$ as above
violates triangle inequalities, i.e. is not a metric space.
Lemma. A tuple $\tau=((c_1,r_1),...,(c_n,r_n))$ is redundant if and only if
$$
\bigcap_{i=1}^n S(c_i, r_i)= \emptyset, 
$$
where the intersection is taken in $H$.
The proof of this lemma is a straightforward induction on $n$ and I omit it.
This lemma allows one to give a numerical criterion for redundancy:
Corollary. Suppose that $n\ge 3$. Unless the intersection of balls
$$
\bigcap_{i\in [n]} B(c_i,r_i)
$$
is empty, the tuple $\tau=((c_1,r_1),...,(c_n,r_n))$ is irredundant if and only if:
(a) For each proper subset $I\subset [n]$, the corresponding tuple
$$
\tau_I=((c_{i_1},r_{i_1}),...,(c_{i_k},r_{i_k})), I= (i_1,...,i_k)$$
is irredundant (in particular, $(X,d)$ is a metric space).
(b) $\Delta(X) (-1)^{n+1}\ge 0$.
Note that this corollary does not directly solve the problem of nonemptyness of the intersection of balls.
At last, here is an answer to the problem of nonemptyness of intersection of round balls $B(c_i,r_i)$ in $span(\{c_1,...,c_n\})$ (which we still assume to have dimension $n-1$). The solution is inductive in $n$. For $n=2$ the answer is in the form of "triangle a inequality"
$$
B(c_1,r_1)\cap B(c_2,r_2)\ne \emptyset
$$
if and only if $r_1+r_2\ge d_{12}=||c_1-c_2||$.
Assume the problem is solved for all $m<n$. In particular, we have a test for redundancy for sets of $m$ balls, $m<n$, i.e. in addition to the numerical criterion we can also tell if the intersection of $m$ balls is nonempty. Now, given a tuple $\tau=((c_1,r_1),...,(c_n,r_n))$, either:
(i) There exists a proper subtuple $\tau_I$ which is redundant (and which is something we can test), hence, $\tau$ itself is redundant
and, thus, the problem of nonemptyness of $\tau$ is reduced to a smaller set of balls.
(ii) Suppose that all proper subtuples $\tau_I$ are irredundant; in particular,
the subtuple $\sigma=((c_1,r_1),...,(c_{n-1},r_{n-1}))$ is irredundant.
Solve the equation $\Delta(X_\tau)=0$ for the unknown $y=d(c_0, c_n)$; this equation has the form
$$
A_n y^2 + B_n y + C_n=0,
$$
with $A_n\ne 0$, where $A_n, B_n, C_n$ are functions of the tuple $\sigma$. This quadratic equation has two (possibly equal) solutions
$$
y_\pm= -\frac{B_n}{2A_n} \pm 
\sqrt{
\left(\frac{B_n}{2A_n}\right)^2 - C_i}. 
$$
Both solutions are real and nonnegative. Geometrically speaking, they correspond to the following:
Consider the intersection of spheres in $span(\{c_1,...,c_n\})$:
$$
\bigcap_{i=1}^{n-1} S(c_i, r_i)=S^0.
$$
This intersection is nonempty (by the irredundancy assumption!)
and is either a singleton (contained in $span(\{c_1,...,c_{n-1}\})$) or it a 2-point set $s_-, s_+\}$,
one of its points $s_-$ is closer to $c_n$ than the other. Then
$$
y_\pm= ||c_n - s_{\pm}||. 
$$
The case when $S^0$ is a singleton happens precisely when $y_+=y_-$. Then
\begin{equation}
\bigcap_{i=1}^n B(c_i, r_i)\ne \emptyset, 
\end{equation}
if and only if $r_n\ge y_-$, i.e. either $\Delta(X_\tau) (-1)^{n}\ge 0$ or the tuple $\tau$ is redundant because
$B(c_n,r_n)$ strictly contains the intersection
\begin{equation}
\bigcap_{i=1}^{n-1} B(c_i, r_i).  
\end{equation}
As an example, here is this solution implemented in the case of intersection of three balls in the Euclidean plane,
$\tau=((c_1,r_1),...,(c_3,r_3))$.  I will use the notation
$$
d_{ij}= ||c_i-c_j||
$$

*

*Test proper subtuples for emptyness: If for some $1\le i<j\le 3$
$$
r_i+r_j< d_{ij},
$$
then $B(c_i,r_i)\cap B(c_j,r_j)=\emptyset$ and we are done. Suppose, therefore that all these intersections are nonempty.


*Test proper subtuples for redundancy: If for some $1\le i\ne j\le 3$
$$
r_i> r_j+ d_{ij}
$$
then we can eliminate the ball $B(c_i,r_i)$ from the collection $B(c_k,r_k), k=1,2,3$ without changing the intersection, and, hence, nonemptyness of the triple intersection is guaranteed by the triangle inequality
$$
r_k+r_j\ge d_{jk}, i\notin \{j,k\}, j\ne k. 
$$


*Suppose, lastly that $X_\tau$ is a metric space and each proper subtuple $\sigma$ in $\tau$
is irredundant.
Then
\begin{equation}
\bigcap_{i=1}^3 B(c_i, r_i)\ne \emptyset, 
\end{equation}
if and only if $r_3\ge y_-$, where $y_-$ is the smaller root of the polynomial
$$
A_3 y^2 + B_3 y + C_3. 
$$
The coefficients $A_3, B_3, C_3$ are computed as follows:
$$
A_3= 2 d_{12},
$$
$$
B_3= -2r_1(d_{12}+d_{23}-d_{13}) - 2r_2(d_{31}+d_{12} -d_{23}),
$$
$$
C_3= \Delta(X_0)= 
\left|\begin{array}{cccc}
0 & d_{12} & d_{13} & 1\\
d_{21} & 0 & d_{23} & 1\\
d_{31}       & d_{32}     & 0       & 1\\
1          & 1          & 1           & 0
\end{array}\right|. 
$$
Lastly, let me explain the solution for $n$ points in $H$,  which are not in general position, i.e. their affine
span has dimension $\le n$. Again, I will take solution for $<n$ points for granted. Then, according to Haley's theorem,
$$
\bigcap_{i\in [n]} B(c_i, r_i)\ne \emptyset
$$
if and only if for each proper subset $I\subset [n]$,
$$
\bigcap_{i\in I} B(c_i, r_i)\ne \emptyset. 
$$
The intersection problem for $< n$ balls is solved by the inductive assumption. A bit more concretely,
inductively applying Haley's theorem, we reduce the problem to the intersection problem of balls centered at configurations of points in general position in some affine subspaces of $H$.
A: For $\mathbb{R}^d$, there is a useful result on convex sets. Helly's theorem is: Let $X_1, ..., X_n$ be a finite collection of convex subsets of $\mathbb{R}^d$, with $n>d+1$. If the intersection of every $d+1$ of these sets is nonempty, then the whole collection has a nonempty intersection, so that $\cap_{j=1}^n X_j \neq \varnothing$.
If $n$ is small, there are ${n}\choose{d+1}$ subsets to check. If that's a reasonable number for your application, that might not be so bad. Maybe by doing induction on Helly's theorem you can get a useful procedure for checking joint subsets to economize on the process.
Or, note the contrapositive is, "if $\cap_{j=1}^n X_j = \varnothing$, then there exists a set of $d+1$ sets with an empty intersection." So one approach might be to take your $n$ sets and instead direct your efforts to finding a $|d+1|$ element subset with an empty intersection. Perhaps normalize the balls so that one is translated to the origin with radius 1, and find the ball furthest away in the Hausdorff distance, and continue like this, building up a $|d+1|$ element set by adding a new ball each time that is as far as possible from the existing set in the Hausdorff metric. Continue this way until you generate a disjoint set, or if it fails, pick a new ball and start again, until you've run through all your basis balls. The nice part about this approach is that you can use your existing test to do pairwise checks very quickly.
I am not sure the extent to which this generalizes to Hilbert space. Flipping through the proof, it uses induction and basic convexity arguments. Since there's no reference to compactness, it might generalize to function spaces as long as the intersection is finite (if countable, I can imagine there being problems around the existence of a convergent subsequence; then again, every Hilbert space is reflexive, so maybe there's a convergent subsequence in the weak topology).  I guess it depends what you actually want to do.
