Let $A \subset \mathbb Z^3$ / $|A| < \infty$. Prove that: $|A| \le \sqrt{|A_x| |A_y| |A_z|}$

Question

Here is the problem statement word by word:

$1)$ Prove that if $a_{ij}$, $b_{jk}$ and $c_{ki}$ are non-negative reals with $1 \le i,j,k \le n$, then:

$$\sum_{i,j,k = 1}^n \sqrt{a_{ij} \times b_{jk} \times c_{ki}} \le \sqrt{ \sum_{i,j = 1}^n a_{ij} \times \sum_{j,k=1}^n b_{jk} \times \sum_{j,i=1}^nc_{ki} }$$

$2)$ Let $A \subset \mathbb Z^3$ be finite. Let $A_x$ denote the projection of $A$ on the $zy$ plane, and define $A_y$ and $A_z$ in a similar manner. Prove that:

$|A| \le \sqrt{|A_x| |A_y| |A_z|}$

The first part is easy to prove by applying the Cauchy-Schwarz inequality twice. Obviously, $2)$ should be a result of $1)$, but I'm not used to deal with $\mathbb Z^n$, and I don't have any idea about what is meant by the projection of $A$ (i.e., neither algebraically nor geometrically).

I'm unable to express the cardinality of $A$ as a triple sum, and I don't know how to deal with the $A_x$, $A_y$ and $A_z$ because I do not understand what is meant by them to start with.

Please provide some explanation about the nature of each of these sets and how to understand projections in $\mathbb Z^3$ geometrically.

Thank you.

Answer 1

Without loss of generality, suppose the finite set $A$ is contained in ${1,\dots,n}^3$ (one can translate it if necessary). Define $a_{ij}$ to be $1$ if $(i,j) \in A_x$ and $0$ otherwise; define $b_{jk}$ to be $1$ if $(k,j) \in A_y$ and $0$ otherwise; define $c_{ki}$ to be $1$ if $(k,i) \in A_z$ and $0$ otherwise. Then the right-hand side of the inequality in 1) is exactly $\sqrt{\#A_x \cdot \#A_y \cdot \#A_z}$. On the other hand, every point $(k,i,j)\in A$ contributes $1$ to the left-hand side, and these contributions are distinct; hence the left-hand side is at least $\#A$.

Answer 2

The projection $A_x$ is the set of all points in $\mathbb Z^2$ you get by stripping away the $x$ coordinate from points in $A$. Geometrically, one can visualize this as setting the $x$-coordinate to zero, which means projecting each point onto the shadow it casts on the $yz$-plane. In the process some pairs of points may overlap, in which case they only count as a single point in the projection (so that $|A_x| \le |A|$ and the inequality may be strict).

Once you understand what $A_x$ looks like, I suggest you try to define $a_{ij}$ so that the corresponding sum on the right-hand side matches $|A_x|$. If you do this in the most natural way, then the left-hand side won't equate exactly to $|A|$ (which explains why it is hard to express as a triple sum), but it will differ from $|A|$ in the right direction: as long as you can show the sum is $\ge |A|$, you will still get the desired inequality.

Answer 3

This is the Loomis-Whitney inequality. Related: Why is Volume^2 at most product of the 3 projections?

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For (2) let $a_{ij} = 1$ if $ (i,j) \in A_x$ and $0$ if $(i,j) \not \in A_x$. Similarly let $b,c$ correspond to the $y,z$ projections.

Then $a_{ij}b_{jk}c_{ki} = 1$ iff if $(i,j,k) \in A$. So with this choice of $a, b, c$ we have (1) one implies (2).

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(1) follows from Cauchy-Schwarz so you are done.

$$ \sum_{i,j,k = 1}^n \sqrt{a_{ij} \times b_{jk} \times c_{ki}} \leq \sqrt{ \sum_{i,j = 1}^n a_{ij} } \times\sum_{i,j,k = 1}^n \sqrt{b_{jk} \times c_{ki}} \le \sqrt{ \sum_{i,j = 1}^n a_{ij} \times \sum_{j,k=1}^n b_{jk} \times \sum_{j,i=1}^nc_{ki} } $$