Two cubes in unit cube

Question

A cube of side one contains two cubes of sides a and b having non-overlapping interiors. How to prove the inequality $a+b \le 1$? The same question in higher dimensions. It was asked, but not answered in SE.

Answer 1

A better bound, namely $\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ is obtained as follows.

The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respectively. Now, forget these two cubes for a while and maximize the sum $a+b$ of the balls' diameters, assuming that the (variable) balls are contained in the unit cube and their interiors are disjoint. Obviously, the maximum possible value of $a+b$ will be an upper bound for the sum of the edges of the cubes. Easy to see, in the optimal configuration of the balls

(1) the two balls must be tangent to each other,

(2) the balls' centers must lie on the main diagonal of the unit cube,

and

(3) each of the two balls must be tangent to the unit cube's boundary.

A brief explanation for (2): the center of the first ball lies somewhere in a cube of edge length $1-a$, concentric with, and parallel to the unit cube. Likewise, the other ball's center is confined to a cube of edge length $1-b$, concentric with, and parallel to the unit cube. The maximum distance between the balls' centers is reached when the centers lie in the "opposite" corners of their confining cubes.

In this position, by (1), the distance between the balls' centers must be $a+b$. Easy to calculate, $a+b=\frac{2\sqrt n}{\sqrt n+1}$.

Somewhat surprisingly, there is a continuum of optimal configurations, where $\ \frac{\sqrt n-1}{\sqrt n+1}\le a\le1$, $\ b=\frac{2\sqrt n}{\sqrt n+1}-a$.

Neither this, nor the bound given by asatzhh (above) proves the conjectured inequality for cubes, since each of them gives values greater than $1$ for every $n\ge2$, and they both approach $2$ monotonically from below as $n\to\infty$. However, $$\frac{2\sqrt n}{\sqrt n+1}<2^{\frac{n-1}{n}}\ \ {\rm for\ every}\ \ n\ge2.$$ For $n=3,\ $ $\frac{2\sqrt n}{\sqrt n+1}\approx 1.26,\ $ while $\ 2^{\frac{n-1}{n}}\approx 1.58$,

and for $n=9,\ $ $\frac{2\sqrt n}{\sqrt n+1}=1.5,\ $ while $\ 2^{\frac{n-1}{n}}\approx 1.85$.

Answer 2

I have given some comments on Daniel's answer (here) in SE. I think it's OK to repeat it here (with more details).

Using simple calculation of volume we can prove that (assume $n$ is the dimension),

$$a+b\le (2^{n-1}(a^n+b^n))^{\frac{1}{n}}\le2^{\frac{n-1}{n}}.$$

And this is almost the best bound using Daniel's method in SE.

Let's denote $P$ the $n-1$ dimensional plane separating two cube($A$ with side $a$ and $B$ with side $b$), and $v\perp P$ its normal vector. Consider any $k$ dimensional plane $V_k$ which contains $v$. Denote $X\hat\perp V_k$ the projection of $X$ into $V_k$. Note that $P\hat\perp V_k$ separating $A\hat\perp V_k$ and $B\hat\perp V_k$ in $V_k$, so we must have (where $m$ means volume) $$m(V_k\cap I_n)\ge m(A\hat\perp V_k)+m(B\hat\perp V_k).$$

As a corollary of A.Good's conjecture here proved by J.D. Vaaler [A geometric inequality with applications to linear forms, Pacific J. Math. 83(1979), 543–553]. We have $$m(A\hat\perp V_k)\ge a^k,m(B\hat\perp V_k)\ge b^k.$$ which is of course best possible.

And according to K.Ball's deep work in upper bound of $m(V_k\cap I_n)$ [K.Ball, Volumes of sections of cubes and related problems, Lecture Notes in Math. 1376(1989), 251–260.]. We have $$m(V_k\cap I_n)\le\min\{(\frac{n}{k})^\frac{k}{2},2^\frac{n-k}{2}\}$$ where the upper bound is best possible if $k|n$ or $k\ge\frac{n}{2}$.

Combine fact above we reach that $$a+b\le\min_{k}(2^{k-1}(a^k+b^k))^\frac{1}{k}\le\min_k(2^{k-1}\min\{(\frac{n}{k})^\frac{k}{2},2^\frac{n-k}{2}\})^\frac{1}{k}=2^\frac{n-1}{n}.$$

Remark: Indeed we can get a better bound since there must exists some $V_k$ with much larger $m(A\hat\perp V_k),m(B\hat\perp V_k)$ and maybe less $m(V_k\cap I_n)$, but $1$ seems impossible using such a method. But maybe some other constructions or methods is OK.