Let $y_0,y_1,x_1,\dots,x_N$ be distinct points in $\mathbb{R}^n$. Clearly there exists a piecewise linear curve $\gamma:[0,1]\rightarrow \mathbb{R}^n$ joining $y_0$ to $y_1$ for which $$ \min_{t \in [0,1]}\,\|\gamma(t) - x_i\|>\delta \quad \text{where} \quad \delta < \min \{ \min_{i\neq j}\frac{\|x_i-x_j\|}{2},\min_{1≤j≤N\\i=0,1} \|y_i-x_j\|\} \,. $$
What is the infimum of the length of such a curves?
On
HINT:
Might as well consider the relaxed condition $\min\|\gamma(t) - x_i\|\ge \delta$, since the infimum of the lengths will be the same. The advantage is that now we do have optimal paths, since their image lies in a closed set.
The problem is to find the shortest path between two points that avoids a family of (open) disks. Intuitively, we understand what these path look like in some simple cases. They must consist of some pieces of tangents to the circles and some arcs. For instance, in the case of one disk, the shortest path consist of a tangent from $y_0$ to the disk, then an ark, then completed with a tangent from $y_1$ to the disk.
In any homotopy class of paths from $y_0$ to $y_1$ that avoid the (open) disks there is a unique shortest path. This is like tightening a rope around some trees.
This is an elaboration on my comments.
I do not think there are anything like a nice "closed" formula for the infimum $L$ of length such paths. However, one can estimate $L$ as follows. I will assume that ${\mathbb R}^n$ is equipped with the Euclidean metric $d=||x-y||$, making it a Euclidean space $E^n$.
Let $p, q\in E^n$ be points in $$ U=E^n - \bigcup_{i=1}^N B(x_i, \delta). $$ Connect $p$ to $q$ by the Euclidean line segment $pq$ (of length $d(p,q)$, of course). This path need not be contained in $U$, so we modify it as follows. Whenever $pq$ crosses a ball $B(x_i,\delta)$ along a segment $p_iq_i$, we replace this subsegment of $pq$ with the shortest arc $\widehat{p_iq_i}$ of a great circle on the boundary sphere of $B(x_i,\delta)$. Let $c_{pq}$ denote the new path from $p$ to $q$. It lies outside of the union of open balls $B(x_i,\delta)$.
The length of the arc $\widehat{p_iq_i}$ equals $2\delta\alpha_i$, where $\alpha_i$ is the angle $\frac{1}{2}\angle p_i x_i q_i$. Denoting $D_i$ the length of the chord $p_iq_i$, and doing some trigonometry observe that $$ \frac{2\delta\alpha}{D_i}= \frac{2\alpha}{\sin \alpha_i}=f(\alpha_i). $$
The function $f(\alpha)= \frac{2\alpha}{\sin \alpha}$ has no critical points on the open interval $(0,\pi/2)$, so its maximum is realized when $p_iq_i$ is the diameter, i.e.
$$ \frac{2\delta\alpha_i}{D_i} \le \frac{\pi}{2}. $$ Therefore, replacing the chords $p_iq_i$ with the arcs $\widehat{p_i q_i}$ throughout $pq$ can increased the overall length at most by the factor $\pi/2$. I am not claiming that the path $c_{pq}$ I constructed is the shortest path from $p$ to $q$ which avoids all the open balls $B(x_i,\delta)$: In many cases it is not.
To conclude: The minimal length $L$ of a path from $p$ to $q$ in the complement to the open balls $B(x_i,\delta)$ satisfies $$ d(p,q)\le L\le \frac{\pi}{2} d(p,q) \le 1.6 d(p,q). $$
To be off by a linear factor is not too bad for this type of problems.
Incidentally, finding a path from $p$ to $q$ in a more general setting (when obstacles overlap and their location and shapes are a priori unknown) is an interesting problem in autonomous robot navigation. (Think of a GPS-equipped Roomba robot trying to get from one room to another.) There is a general "ellipse method" for solving this in the plane due to Lumelsky. From what I remember, his solution is off by a uniform multiplicative factor from an optimal solution. The situation in dimension 3 (and higher) is far from clear. However, I am not an expert in this staff.