[proofplan] We prove $|K|^2 \leq |K_{12}| \cdot |K_{13}| \cdot |K_{23}|$ by a compression argument. First, we handle the special case where every horizontal section $K(x_3)$ is a square, reducing the inequality to the pointwise bound $f(x_3) \leq \sup f$. For the general case, we replace each section $K(x_3)$ by a square of the same area, producing a new body $L$ with $|L| = |K|$ and $|L_{12}| \leq |K_{12}|$. For the remaining projections $L_{13}$ and $L_{23}$, the area-to-projection-product bound $|K(x_3)| \leq g(x_3) h(x_3)$ (where $g$ and $h$ are the one-dimensional projections of each section) and the Cauchy--Schwarz inequality together give $|L_{13}| \cdot |L_{23}| \leq |K_{13}| \cdot |K_{23}|$. Chaining through the special case for $L$ yields the result. [/proofplan]

[step:Handle the special case where each section $K(x_3)$ is a square]For each $x_3 \in \mathbb{R}$, define the horizontal section \begin{align*} K(x_3) := \{(x_1, x_2) \in \mathbb{R}^2 : (x_1, x_2, x_3) \in K\} \subseteq \mathbb{R}^2. \end{align*} Suppose that each non-empty section has the form $K(x_3) = (0, f(x_3)) \times (0, f(x_3))$ for a measurable function $f: \mathbb{R} \to [0, \infty)$. Set $M := \sup_{x_3} f(x_3)$. Then: \begin{align*} |K| &= \int_{\mathbb{R}} f(x_3)^2 \, d\mathcal{L}^1(x_3), \\ |K_{12}| &= M^2, \\ |K_{13}| = |K_{23}| &= \int_{\mathbb{R}} f(x_3) \, d\mathcal{L}^1(x_3). \end{align*} The first identity holds because the area of each square section is $f(x_3)^2$, and $|K| = \int |K(x_3)| \, d\mathcal{L}^1(x_3)$ by Fubini's theorem. The second holds because $K_{12} = \bigcup_{x_3} K(x_3) = (0, M) \times (0, M)$, since the squares are nested inside $(0, M)^2$ and the square of side $M$ is achieved in the supremum. The third holds because $K_{13} = \{(x_1, x_3) : 0 < x_1 < f(x_3)\}$, whose Lebesgue measure is $\int f(x_3) \, d\mathcal{L}^1(x_3)$, and $K_{23}$ is identical by the symmetry of the square sections. The desired inequality becomes \begin{align*} \left( \int_{\mathbb{R}} f(x_3)^2 \, d\mathcal{L}^1(x_3) \right)^2 \leq M^2 \left( \int_{\mathbb{R}} f(x_3) \, d\mathcal{L}^1(x_3) \right)^2. \end{align*} Since $0 \leq f(x_3) \leq M$ for all $x_3$, we have $f(x_3)^2 \leq M \cdot f(x_3)$ pointwise. Integrating both sides gives $\int f(x_3)^2 \, d\mathcal{L}^1(x_3) \leq M \int f(x_3) \, d\mathcal{L}^1(x_3)$. Squaring this inequality yields the claim.[/step]

[guided]Why decompose by horizontal sections? The idea is that the three-dimensional body $K$ is a "stack" of two-dimensional slices $K(x_3)$ parameterised by the third coordinate. Its volume is the integral of the areas of these slices (by Fubini's theorem), and its projections can be expressed in terms of the slices' geometry. For the special case, suppose every slice is a square $(0, f(x_3))^2$ of side length $f(x_3) \geq 0$. The key quantities are: - $|K| = \int f(x_3)^2 \, d\mathcal{L}^1(x_3)$, since each slice has area $f(x_3)^2$. - $|K_{12}|$: the projection onto the $(x_1, x_2)$-plane is the union of all slices, which is the largest square $(0, M)^2$ where $M = \sup f$. So $|K_{12}| = M^2$. - $|K_{13}|$: the projection onto the $(x_1, x_3)$-plane records, for each $x_3$, the range of $x_1$ values in $K(x_3)$, which is $(0, f(x_3))$. So $K_{13} = \{(x_1, x_3) : 0 < x_1 < f(x_3)\}$, an epigraph-like region with $|K_{13}| = \int f(x_3) \, d\mathcal{L}^1(x_3)$. By the symmetry of the square sections, $|K_{23}| = |K_{13}|$. The inequality $|K|^2 \leq |K_{12}| \cdot |K_{13}| \cdot |K_{23}|$ becomes $(\int f^2)^2 \leq M^2 (\int f)^2$. Since $f \leq M$ pointwise, we get $f^2 \leq Mf$ pointwise, so $\int f^2 \leq M \int f$. Squaring: $(\int f^2)^2 \leq M^2 (\int f)^2$.[/guided]

[step:Compress the general body to one with square sections of equal area]For a general bounded open body $K \subseteq \mathbb{R}^3$, define the compressed body $L \subseteq \mathbb{R}^3$ by replacing each horizontal section with a square of the same area: \begin{align*} L(x_3) := \left(0, \sqrt{|K(x_3)|}\right) \times \left(0, \sqrt{|K(x_3)|}\right) \subseteq \mathbb{R}^2, \end{align*} where $|K(x_3)|$ denotes the two-dimensional Lebesgue measure of $K(x_3)$. Then $|L(x_3)| = |K(x_3)|$ for every $x_3$, so by Fubini's theorem, \begin{align*} |L| = \int_{\mathbb{R}} |L(x_3)| \, d\mathcal{L}^1(x_3) = \int_{\mathbb{R}} |K(x_3)| \, d\mathcal{L}^1(x_3) = |K|. \end{align*} The body $L$ has square sections with side length $f(x_3) := \sqrt{|K(x_3)|}$, so the special case applies to $L$.[/step]

[guided]The compression idea is to replace each complicated two-dimensional section with a geometrically simpler shape -- a square -- of the same area. The square is the natural choice because it treats the $x_1$- and $x_2$-directions symmetrically, which will ensure $|L_{13}| = |L_{23}|$ and simplify the final estimate. The critical property is that compression preserves volume: $|L| = |K|$. This is immediate because we have only rearranged mass within each horizontal slice, not moved mass between slices.[/guided]

[step:Bound $|L_{12}|$ by $|K_{12}|$]The projection $L_{12}$ is the union of all square sections: \begin{align*} L_{12} = \bigcup_{x_3} L(x_3) = \left(0, \sup_{x_3} \sqrt{|K(x_3)|}\right)^2. \end{align*} Its measure is $|L_{12}| = \sup_{x_3} |K(x_3)|$. Meanwhile, $K_{12} = \bigcup_{x_3} K(x_3)$, so \begin{align*} |L_{12}| = \sup_{x_3} |K(x_3)| \leq \left| \bigcup_{x_3} K(x_3) \right| = |K_{12}|, \end{align*} where the inequality holds because the measure of a union is at least the supremum of the measures of its members.[/step]

[guided]Why does $\sup_{x_3} |K(x_3)| \leq |K_{12}|$? Each section $K(x_3)$ is a subset of the projection $K_{12} = \bigcup_{x_3} K(x_3)$. Since $K(x_3) \subseteq K_{12}$ for every $x_3$, monotonicity of Lebesgue measure gives $|K(x_3)| \leq |K_{12}|$. Taking the supremum over $x_3$ yields $\sup_{x_3} |K(x_3)| \leq |K_{12}|$. This step uses a genuine inequality: replacing general sections by squares of the same area can only decrease the projection onto the $(x_1, x_2)$-plane, because the sections may have had diverse shapes whose union covered a large region, while the squares are all centred at the origin and nested.[/guided]

[step:Bound $|L_{13}| \cdot |L_{23}|$ by $|K_{13}| \cdot |K_{23}|$ via Cauchy--Schwarz]For each $x_3$, define the one-dimensional projections of the section $K(x_3)$: \begin{align*} g(x_3) &:= |K(x_3)_1| = \mathcal{L}^1\bigl(\{x_1 \in \mathbb{R} : \exists\, x_2, \, (x_1, x_2) \in K(x_3)\}\bigr), \\ h(x_3) &:= |K(x_3)_2| = \mathcal{L}^1\bigl(\{x_2 \in \mathbb{R} : \exists\, x_1, \, (x_1, x_2) \in K(x_3)\}\bigr). \end{align*} By the elementary projection bound in $\mathbb{R}^2$, the area of any planar set is at most the product of its one-dimensional projections: $|K(x_3)| \leq g(x_3) \cdot h(x_3)$. The square section $L(x_3)$ has side length $\sqrt{|K(x_3)|}$, so \begin{align*} \sqrt{|K(x_3)|} \leq g(x_3)^{1/2} \cdot h(x_3)^{1/2}. \end{align*} Since $L_{13} = \{(x_1, x_3) : 0 < x_1 < \sqrt{|K(x_3)|}\}$, its measure is $|L_{13}| = \int_{\mathbb{R}} \sqrt{|K(x_3)|} \, d\mathcal{L}^1(x_3)$. By the same reasoning, $|L_{23}| = |L_{13}|$. Therefore \begin{align*} |L_{13}| \cdot |L_{23}| = \left( \int_{\mathbb{R}} \sqrt{|K(x_3)|} \, d\mathcal{L}^1(x_3) \right)^2 \leq \left( \int_{\mathbb{R}} g(x_3)^{1/2} \cdot h(x_3)^{1/2} \, d\mathcal{L}^1(x_3) \right)^2. \end{align*} We apply the Cauchy--Schwarz inequality to the functions $g^{1/2}$ and $h^{1/2}$ (both non-negative and measurable): \begin{align*} \left( \int_{\mathbb{R}} g(x_3)^{1/2} \cdot h(x_3)^{1/2} \, d\mathcal{L}^1(x_3) \right)^2 \leq \int_{\mathbb{R}} g(x_3) \, d\mathcal{L}^1(x_3) \cdot \int_{\mathbb{R}} h(x_3) \, d\mathcal{L}^1(x_3). \end{align*} Now $\int g(x_3) \, d\mathcal{L}^1(x_3) = |K_{13}|$ because $K_{13} = \{(x_1, x_3) : x_1 \in K(x_3)_1\}$ and applying Fubini's theorem gives $|K_{13}| = \int g(x_3) \, d\mathcal{L}^1(x_3)$. Similarly, $\int h(x_3) \, d\mathcal{L}^1(x_3) = |K_{23}|$. Combining: \begin{align*} |L_{13}| \cdot |L_{23}| \leq |K_{13}| \cdot |K_{23}|. \end{align*}[/step]

[guided]The functions $g$ and $h$ measure how "wide" each section $K(x_3)$ is when projected onto the $x_1$- and $x_2$-axes, respectively. The planar projection bound $|K(x_3)| \leq g(x_3) h(x_3)$ is the two-dimensional analogue of the partition inequality: the coordinates $\{1\}$ and $\{2\}$ partition $\{1, 2\}$, so the area of a planar set is at most the product of its one-dimensional projections. The side length of the compressed square is $\sqrt{|K(x_3)|} \leq \sqrt{g(x_3) h(x_3)} = g(x_3)^{1/2} h(x_3)^{1/2}$. Integrating this over $x_3$ gives an upper bound on $|L_{13}|$ (and $|L_{23}|$). The Cauchy--Schwarz inequality is applied to the $L^2$ inner product of the functions $x_3 \mapsto g(x_3)^{1/2}$ and $x_3 \mapsto h(x_3)^{1/2}$: \begin{align*} \left( \int g^{1/2} h^{1/2} \, d\mathcal{L}^1 \right)^2 \leq \left( \int g \, d\mathcal{L}^1 \right) \left( \int h \, d\mathcal{L}^1 \right). \end{align*} The right-hand side equals $|K_{13}| \cdot |K_{23}|$ by Fubini's theorem: $|K_{13}|$ is the measure of $\{(x_1, x_3) : (x_1, x_2, x_3) \in K \text{ for some } x_2\}$, which equals $\int g(x_3) \, d\mathcal{L}^1(x_3)$ since $g(x_3)$ is the length of the $x_1$-section at height $x_3$. Similarly for $|K_{23}|$.[/guided]

[step:Chain the estimates to conclude the inequality for $K$]Applying the special-case result (from the first step) to the body $L$, which has square sections: \begin{align*} |K|^2 = |L|^2 \leq |L_{12}| \cdot |L_{13}| \cdot |L_{23}|. \end{align*} Using the bounds $|L_{12}| \leq |K_{12}|$ (from the third step) and $|L_{13}| \cdot |L_{23}| \leq |K_{13}| \cdot |K_{23}|$ (from the fourth step): \begin{align*} |K|^2 \leq |L_{12}| \cdot |L_{13}| \cdot |L_{23}| \leq |K_{12}| \cdot |K_{13}| \cdot |K_{23}|. \end{align*}[/step]

[guided]The argument follows the compression paradigm: replace a complicated object ($K$) with a simpler one ($L$) that is easier to analyse, verify that the replacement does not increase the quantities of interest (the projection measures), prove the inequality for the simpler object, and transfer the bound back. Concretely, the three ingredients are: 1. $|L| = |K|$ (compression preserves volume), 2. $|L_{12}| \leq |K_{12}|$ and $|L_{13}| \cdot |L_{23}| \leq |K_{13}| \cdot |K_{23}|$ (compression does not increase projections), 3. $|L|^2 \leq |L_{12}| \cdot |L_{13}| \cdot |L_{23}|$ (the inequality holds for square-section bodies). Chaining: $|K|^2 = |L|^2 \leq |L_{12}| \cdot |L_{13}| \cdot |L_{23}| \leq |K_{12}| \cdot |K_{13}| \cdot |K_{23}|$.[/guided]