Problem 1338

I. solution. Let the orthogonal projections of the tetrahedron's vertices onto the intersecting plane be $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$, and let the distances of the vertices from the intersecting plane be denoted by $d_{A}, d_{B}, d_{C}, d_{D}$, respectively. (If any of these distances is 0, the statement is obvious.) Then, due to the equality of the corresponding angles in the right triangles $A A^{\prime} K$ and $B B^{\prime} K$, we have

$$\frac{A K}{B K}=\frac{d_{A}}{d_{B}}$$

and similarly, we obtain

$$\frac{B L}{C L}=\frac{d_{B}}{d_{C}}, \quad \frac{C M}{D M}=\frac{d_{C}}{d_{D}} \quad \text { and } \quad \frac{D N}{A N}=\frac{d_{D}}{d_{A}}$$

Thus, $A K \cdot B L \cdot C M \cdot D N=B K \cdot C L \cdot D M \cdot A N$.

Multiply both sides by the fraction

$$\frac{A K \cdot B L \cdot C M \cdot D N}{(A B \cdot B C \cdot C D \cdot A D)^{2}}$$

and then transform the right side using the fact that $B K=A B-A K, C L=B C-B L, D M=C D-C M$ and $A N=A D-D N$, and finally apply the inequality between the arithmetic and geometric means:

$$\begin{aligned} & \left(\frac{A K \cdot B L \cdot C M \cdot D N}{A B \cdot B C \cdot C D \cdot A D}\right)^{2}= \\ = & \frac{A K(A B-A K)}{A B^{2}} \cdot \frac{B L(B C-B L)}{B C^{2}} \cdot \frac{C M(C D-C M)}{C D^{2}} \cdot \frac{D N(A D-D N)}{A D^{2}} \leq \\ \leq & \left(\frac{A K+(A B-A K)}{2}\right)^{2} \cdot \frac{1}{A B^{2}} \cdot\left(\frac{B L+(B C-B L)}{2}\right)^{2} \cdot \frac{1}{B C^{2}} \\ & \cdot\left(\frac{C M+(C D-C M)}{2}\right)^{2} \cdot \frac{1}{C D^{2}} \cdot\left(\frac{D N+(A D-D N)}{2}\right)^{2} \cdot \frac{1}{A D^{2}}=\left(\frac{1}{2^{2}}\right)^{4} \end{aligned}$$

Taking the square root of both sides of the inequality, we obtain the statement to be proved.

II. solution. Let us take a spatial Cartesian coordinate system such that the equation of the intersecting plane is $Z=0$. Project the points mentioned in the problem orthogonally onto the $z$-axis, and let the projection of any point $T$ be $T^{\prime}$. The orthogonal projection preserves the ratio of segments, so it is sufficient to show that

$$\frac{A^{\prime} K^{\prime}}{A^{\prime} B^{\prime}} \cdot \frac{B^{\prime} L^{\prime}}{B^{\prime} C^{\prime}} \cdot \frac{C^{\prime} M^{\prime}}{C^{\prime} D^{\prime}} \cdot \frac{D^{\prime} N^{\prime}}{A^{\prime} D^{\prime}} \leq \frac{1}{16}$$

If the intersecting plane passes through any vertex of the tetrahedron, the inequality is clearly satisfied, because one of the fractions on the left side is 0. If the plane does not pass through any vertex, then the vertices $A$ and $C$ are on one side of the plane, and the vertices $B$ and $D$ are on the other side. Since the intersecting plane is perpendicular to the $z$-axis, the same is true for the points $A^{\prime}, C^{\prime}$ and $B^{\prime}, D^{\prime}$. By symmetry, we can assume that the third coordinates of $A^{\prime}$ and $C^{\prime}$ are positive, while the third coordinates of $B^{\prime}$ and $D^{\prime}$ are negative. Thus, $A^{\prime}(0,0, a), B^{\prime}(0,0,-b), C^{\prime}(0,0, c)$ and $D^{\prime}(0,0,-d)$ with suitable positive numbers $a, b, c, d$. Since $K^{\prime} \equiv L^{\prime} \equiv M^{\prime} \equiv N^{\prime}=(0,0,0)$

$$\frac{A^{\prime} K^{\prime}}{A^{\prime} B^{\prime}}=\frac{a}{a+b}, \quad \frac{B^{\prime} L^{\prime}}{B^{\prime} C^{\prime}}=\frac{b}{b+c}, \quad \frac{C^{\prime} M^{\prime}}{C^{\prime} D^{\prime}}=\frac{c}{c+d} \quad \text { and } \quad \frac{D^{\prime} N^{\prime}}{A^{\prime} D^{\prime}}=\frac{d}{d+a}$$

Thus, we need to prove that for any positive numbers $a, b, c, d$, the inequality

$$\frac{a}{a+b} \cdot \frac{b}{b+c} \cdot \frac{c}{c+d} \cdot \frac{d}{d+a} \leq \frac{1}{16}$$

holds. This, however, easily follows from the inequalities obtained by rearranging the arithmetic and geometric means:

$$\frac{\sqrt{a b}}{a+b} \leq \frac{1}{2}, \quad \frac{\sqrt{b c}}{b+c} \leq \frac{1}{2}, \quad \frac{\sqrt{c d}}{c+d} \leq \frac{1}{2} \quad \text { and } \quad \frac{\sqrt{d a}}{d+a} \leq \frac{1}{2}$$

by multiplying them together.

It also follows from the proof that equality holds precisely when the intersecting plane passes through the midpoints of four edges of the tetrahedron.