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.

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.