Two cubes in unit cube

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.

Here's a proof of the (much weaker) fact that $a+b\leq\sqrt{n}$. Maybe someone can find a way to elaborate upon it to get a solution of the problem.

Let $n$ be the dimension we are considering. Let $C_n\subset\mathbb{R}^n$ be the cube of side $1$. Let $a$ be the side of the first cube (we will call it $A$), $b$ the side of the second cube ($B$). By the classical theorems of convex separation, there is an $(n-1)$-plane $P_{n-1}$ separating the two cubes (since they are disjoint). By translating everything if necessary, assume that $0$ is contained in $P_{n-1}$. Let $L$ denote the line through $0$ that is orthogonal to $P_{n-1}$ and denote by $\pi:\mathbb{R}^n\rightarrow L\cong\mathbb{R}$ be the projection on $L$. It is easy to prove that the projection of a cube of side $x$ on a line is an interval of maximal length $x\sqrt{n}$. Let $\ell$ be the function associating to an interval in $\mathbb{R}$ its length. Then: $$\ell(\pi(C_n))=:z\leq\sqrt{n}$$ $$\ell(\pi(A))=:x\geq a$$ $$\ell(\pi(B))=:y\geq b$$ Thus: $$a+b\leq x+y\leq z\leq \sqrt{n}$$ where the second inequality is given by the fact that $P_{n-1}$ separates $A$ and $B$.

A possible elaboration of the above would be treating higher dimensional planes instead of the $1$-plane (= straight line) $L$. For example we could obtain an $(n-1)$-plane $Q$ by taking some vector $v\in P_{n-1}$ and defining $Q$ to be its orthogonal complement. Then the projections of $A$ and $B$ on $Q$ would be disjoint, since the $(n-2)$-plane $P_{n-1}\cap Q$ would separate them in $Q$. Maybe some work in this sense could give us better estimates or some way to prove the initial statement by induction.