# Inequality between volume and its projections

Let $A \subset \mathbb{R}^3$ be connected and let's define $A_1, A_2, A_3 \subset \mathbb{R}^2$ as projections of $A$ onto three perpendicular (to each other) planes. Show that: $$|A| \le \sqrt{|A_1| |A_2| |A_3|}\;,$$ where $|\cdot|$ is volume when applied to $A$ and area whenas $A_{1,2,3}$.

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It's a fine question, and he may just be posting it because he saw it and thought it was nice. Not everyone posting is posting for help. – Potato Jun 13 '12 at 10:19
It is obvious, it is correct for parallelepiped. And $|A|$ is estimated by volumes of parallepipeds. – Nikita Evseev Jun 13 '12 at 10:42
There's a solution below, but I would like to see something elementary, if anyone has one! You should also probably include the hypothesis A is measurable. – Potato Jun 13 '12 at 11:36
@Potato sure it should be measurable. – qoqosz Jun 13 '12 at 21:43
@nikita2 even obvious facts need a proof in mathematics :) – qoqosz Jun 13 '12 at 21:44

This may be overkill, but this inequality is a simple consequence of the inequality on the first page of this document. It says that, for any three sigma-finite measures spaces $(X,\mu)$, $(Y,\nu)$ and $(Z,\xi)$ and any three square-integrable functions $f:X \times Y \to \mathbb{R}$, $g:Y \times Z \to \mathbb{R}$ and $h:Z \times X \to \mathbb{R}$,

$$\int_{X \times Y \times Z} f(x,y) g(y,z) h(z,x) d\mu (x) d\nu (y) d\xi (z) \leq \sqrt{\int_{X \times Y} f^2(x,y) d\mu (x) d\nu (y) \int_{Y \times Z} g^2(y,z) d\nu (y) d\xi (z) \int_{Z \times X} h^2(z,x) d\xi (z) d\mu (x)}.$$

We can assume without loss of generality that the three perpendicular planes are the canonical planes $(x,y,0)$, $(0,y,z)$ and $(x,0,z)$. Then, taking $f(x_0,y_0)=1$ if there exists a $z$ suchat that $(x_0,y_0,z)$ belongs to $A$ (in other words, $f$ is conjugated to $1_{A_1}$ via the trivial morphism between $\mathbb{R}^2$ and the first canonical plane), an the same for $g$ and $h$, one gets the result.

A few remarks:

• Connectedness is not necessary.

• The article I linked works with probability measures. However, since the inequality is homogenous in the measures, one can extend it for free to finite measures (which solves the case of bounded $A$), and with some work the sigma-finite case is obvious - at least for positive functions.

• There are simpler proofs, if only because we work with $\{0,1\}$-valued functions, but the trick is nice to know anyway.

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