Greatest open ball in the unit $n$-cell 
Let $n\geq2$ be an integer, let $C=[-1,1]^n$ and let $A$ be the set of all real numbers $r$ such that $r$ is the radius of some open ball $V$ such that $V$ is contained in $C$ and $V$ is disjoint to the open ball whose center is $\bf 0$ and radius is $1.$ Find $\sup A.$

I think the problem of finding $y=\sup A$ (without proof) isn't very difficult, because we can look at the case when $n=2$ and then "generalize" (graphically and using the pythagorean theorem and to generalize, we use the pythagorean theorem in $n$ dimensions). I found that $y=(\sqrt n-1)/(\sqrt n +1)$ 
I intituively see that a point $x$ (there are exactly $2^n$ such points) at which we can center an open ball $V$ of radius $y$ such that $V\subseteq C$ and $V\cap B(\mathbf{0},1)=\varnothing$ belongs to the line connecting the vector $\bf 0$ and the vector $\bf 1$, all of whose coordinates are $1.$  Thus $x=t\mathbf{1}$ for some real $0<t<1$ and also $|x|>1$ (because $x\notin B(\mathbf{0},1)$ and also $x$ must be an interior point of the complement of this ball). But then I don't know what to do. I would like to be as rigorous as possible.
Thank you for any help.
 A: Consider a point $p ∈ [-1, 1]^n \setminus B(\mathbf{0}, 1)$. Observe that $B(p, r) ⊆ [-1, 1]^n \setminus B(\mathbf{0}, 1)$ if and only if $r$ is not greater than the distance from $p$ to any of the faces of the cube and also not greater than the distance from $p$ to $B(\mathbf{0}, 1)$. But we can calculate those!
For $n = 2$ and $p = (x, y)$ the distances from the faces are $1 - x$, $1 - y$, $x - (-1) = 1 + x$, and $y - (-1) = 1 + y$. The distance from the ball is $\sqrt{x^2 + y^2} - 1$.
To find the supremum it is enough to maximize the minimum of the distances. By the symmetry of the problem, we may suppose $0 ≤ x ≤ y ≤ 1$, and hence consider only minimum of $1 - y$ and $\sqrt{x^2 + y^2} - 1$. We consider two cases, depending on which of the quantities is lesser. This is decided by condition $1 - y ≤ \sqrt{x^2 + y^2} - 1$, which is in our situation equivalent to $2 \sqrt{1 - y} ≤ x$.
If $2 \sqrt{1 - y} ≤ x$, then $r = 1 - y$, which we are maximizing. Equivalently, we are minimizing $y$ under condition $0 ≤ 2 \sqrt{1 - y} ≤ x ≤ y ≤ 1$. Note that as $y$ decreases from $1$ to $0$, the quantity $2 \sqrt{1 - y}$ increaces from $0$ to $2$. Hence, when miniminzing $y$ under the condition we obtain $x = y = 2\sqrt{1 - y}$.
If $2 \sqrt{1 - y} ≥ x$, then $r = \sqrt{x^2 + y^2} - 1$, which we are maximizing. $r$ inscreases as $x$ increases, so we put $x = \min(y,\, 2\sqrt{1 - y})$ and we maximize $\sqrt{\min(y,\, 2\sqrt{1 - y})^2 + y^2} - 1$ for $0 ≤ y ≤ 1$. We again consider two cases: if $y ≤ 2 \sqrt{1 - y}$, then we are maximizing $\sqrt{y^2 + y^2} - 1 = \sqrt{2} y - 1$ or equivalently $y$ under conditions $0 ≤ y ≤ 2\sqrt{1 - y},\, 1$. By the same observation on behavior of the two quantities we obtain again $y = 2 \sqrt{1 - y}$. Otherwise if $y ≥ 2 \sqrt{1 - y}$, we maximize $\sqrt{4 (1 - y) + y^2} - 1 = 1 - y$ and therefore minimize $y$ under $0 ≤ 2 \sqrt{1 - y} ≤ y ≤ 1$, which we have already done.
In all cases we get $x = y = 2 \sqrt{1 - y}$ with solution $x = y = 2 \sqrt{2} - 2$ and $r = 1 - y = \sqrt{x^2 + y^2} - 1 = 3 - 2 \sqrt{2} = (\sqrt{2} - 1)/(\sqrt{2} + 1)$, which is the desired result.

The computation can be hopefully generalized to arbitrary $n$. We can again suppose $0 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 1$ and maximize $\min(1 - x_n,\, \sqrt{∑_{i = 1}^n x_i^2} - 1)$.

Also note that $(\sqrt{n} - 1)/(\sqrt{n} + 1) \to 1$ as $n \to ∞$. Therefore, at higher dimensions the smaller balls have almost the same radius as $B(\mathbf{0}, 1)$, and so are intersecting. They are thouching exactly when $n = 9$. More precisely, every two of the $2^9$ balls with centers sharing all but one coordinate are touching.
I gave this problem to a colleague of mine and he found the result for $n = 2$ a different way: he considered a copy of the square rotated by $45$ degrees and then inscribed the small circle we a looking for to one of the resulting triangles. Then by using simple trigonometry we get $r = \tan^2(π/8)$. As a byproduct we obtain $\tan^2(π/8) = (\sqrt{2} - 1)/(\sqrt{2} + 1)$.
