I was wondering if the reverse Loomis-Whitney inequality holds for general functions:


Let $n\geq 2$. Let $(X\_i,\mu\_i)$, $1\leq i\leq n$ be measure spaces. Write $x=(x\_1,\dots,x\_n)$ and for each $1\leq i\leq n$, write $$
 \pi\_i(x)=x\_i'=(x\_1,\dots,x\_{i-1},x\_{i+1},\dots,x\_n)\in X'\_i:=X\_1\times \cdots X\_{i-1}\times X\_{i+1}\times\cdots X\_n.
 $$ 
 Let $f\_i$ be nonnegative measurable functions defined on $X'\_i$. Then do we have the following inequality:
 $$
 \gamma\_n\prod\_{i=1}^n \left\| f\_i\right\|\_{L^{n-1}(X'\_i)}\leq \int\_{X\_1}\cdots \int\_{X\_n}\prod\_{i=1}^n f\_i(x\_i')dx.
 $$
 where $0<\gamma\_n<1$ is an absolute constant?


It is known that if $f\_i=1\_{\pi\_i(K)}$ for some compact set $K$, then the reverse inequality holds. (See, for example, <https://arxiv.org/pdf/1607.07891.pdf>)



Reverse Loomis-Whitney Inequality for functions
This conjecture is trivially true for $n=2$, but false for any $n\ge3$. Indeed, take any $n\ge3$ and for all $i$ let $X\_i=\mathbb R$ and $\mu\_i=\lambda$, the Lebesgue measure. Next, let
$$f\_i(x'\_i):=\prod\_{j:\ j\ne i}1\_{i<x\_j<i+1}. 
$$
Then 
$$\prod\_i f\_i(x'\_i)=\prod\_i\prod\_{j:\ j\ne i}1\_{i<x\_j<i+1}
=\prod\_j\prod\_{i:\ i\ne j}1\_{i<x\_j<i+1}=0 
$$
for all $x$, but 
$$\|f\_i\|\_1=\int f\_i=\prod\_{j:\ j\ne i}\int\_i^{i+1}dx\_j=1
$$
for all $i$. 
So, the right side of your inequality is $0$, whereas its left side is $\gamma\_n>0$. 


