I would like to prove that for any family of balls $\{B(c_i,r_i)\}_i \subset \mathbb{R}^d$ such that $\{c_1, \dots, c_n\} \subset \bigcap_i B(c_i,r_i) $ and $\forall i, r_i \geq 1$, there exists a ball of radius $1-\frac{\theta_d}{2}$ included in the intersection $\bigcap_i B(c_i,r_i)$ where $\theta_d/2 = \frac{1}{2}\sqrt{\frac{2d}{d+1}}$ denotes the ratio between the diameter and the radius of the smallest enclosing ball of a regular simplex.

Intuitively, it seems that the way of making the smallest intersection is to assign all points $c_i$ to the vertices of a regular simplex of diameter $1$ and all $r_i$ to $1$. By doing so, one can check that the ball of radius $1-\frac{\theta_d}{2}$ centered at $x$ the barycenter of $\{c_1,\dots,c_n\}$ is included in the intersection of balls $ \bigcap_i B(c_i,r_i)$. Indeed, in the case of a simplex, the radius of the biggest ball centered at $x$ and included in the intersection of balls is $1-\text{Radius}(\sigma) = 1 - \frac{\theta_d}{2}$ (hence the constant is tight in this case).

I am having difficulties to prove that this case is indeed the worst case. I was just able to prove that the result holds when all balls have the same radius. Does this result seems familiar to someone? I would really appreciate any comment, idea or reference.

ps : The topology tag is here for several reason. One of them is that the biggest radius of the ball included in the intersection corresponds to the weak feature size of the complement of the intersection. Another one is that this result is linked to a collapsibily result.