Let's consider the number $\kappa=V^{2} / F^{3}$ for any convex polyhedron, where $V$ is the volume of the polyhedron, and $F$ is its surface area. How does this number change as we go through the following 5 polyhedra:

1. the regular tetrahedron $A B C D$;
2. cut off from it the tetrahedron $A A_{1} A_{2} A_{3}$, where $A_{1}, A_{2}, A_{3}$ are the points on the edges $A B, A C, A D$ respectively, such that $A A_{1} / A B=A A_{2} / A C=A A_{3} / A D=\lambda \leqq 1 / 2$

3 - 5. successively cut off from the remainder the similarly defined tetrahedra $B B_{1} B_{2} B_{3}$, then $C C_{1} C_{2} C_{3}$, and finally $D D_{1} D_{2} D_{3}$.
Let the length of the edges of the tetrahedron $ABCD$ be denoted by $a$. The volume of the tetrahedron is $V_{1}=\frac{a^{3} \sqrt{2}}{12}$, and its surface area is $F_{1}=a^{2} \sqrt{3}$. Therefore,

$$
\kappa_{1}=\frac{V_{1}^{2}}{F_{1}^{3}}=\frac{\sqrt{3}}{648}
$$

The tetrahedra to be cut off from $ABCD$ are also regular, as each of their faces is an equilateral triangle; the length of the sides is $\lambda a$. The volume of the polyhedra decreases by $\frac{a^{3} \sqrt{2}}{12} \cdot \lambda^{3}$ with each cut, so

$$
V_{n}=V\left[1-(n-1) \lambda^{3}\right] \quad(n=2,3,4,5)
$$

(Since $\lambda \leq \frac{1}{2}$, the tetrahedra to be cut off do not overlap, and at most they can share a common vertex.)

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From the surface of the polyhedron, three equilateral triangles with side length $a \lambda$ must be removed with each cut, and one such triangle must be added, as the face of the cut-off tetrahedron was originally inside the polyhedron, but after the cut, it becomes part of the new polyhedron's surface. Therefore, the surface area of the polyhedra decreases by $\frac{a^{2} \sqrt{3}}{2} \cdot \lambda^{2}$ with each cut, so

$$
F_{n}=F_{1}\left[1-(n-1) \frac{\lambda^{2}}{2}\right] \quad(n=2,3,4,5 .)
$$

Accordingly, the further values of $\kappa$ are:

$$
\kappa_{n}=\frac{\left(1-(n-1) \lambda^{3}\right)^{2}}{\left(1-(n-1) \frac{\lambda^{2}}{2}\right)^{3}} \cdot \kappa_{1}
$$

($n=2,3,4,5$, the formula is also valid for $n=1$.)

We will show that for a fixed $0 < \lambda \leq \frac{1}{2}$, the sequence $\kappa_{n}$ is decreasing:

$$
\kappa_{n+1} - \kappa_{n} = \kappa_{1} \left( \frac{\left(1-n \lambda^{3}\right)^{2}}{\left(1-n \frac{\lambda^{2}}{2}\right)^{3}} - \frac{\left(1-(n-1) \lambda^{3}\right)^{2}}{\left(1-(n-1) \frac{\lambda^{2}}{2}\right)^{3}} \right)
$$

Since $\kappa_{1} > 0$, it suffices to show that the expression in parentheses is negative. Let us denote:

$$
x = 1 - (n-1) \lambda^{3}, \quad y = 1 - (n-1) \frac{\lambda^{2}}{2}
$$

Then we need to show that:

$$
\frac{(x - \lambda^{3})^{2}}{(y - \frac{\lambda^{2}}{2})^{3}} - \frac{x^{2}}{y^{3}} < 0
$$

This can be simplified to:

$$
\frac{(x - \lambda^{3})^{2} y^{3} - x^{2} (y - \frac{\lambda^{2}}{2})^{3}}{y^{3} (y - \frac{\lambda^{2}}{2})^{3}} < 0
$$

The denominator is positive, so we need to show that the numerator is negative. Expanding and simplifying the numerator, we get:

$$
(x - \lambda^{3})^{2} y^{3} - x^{2} (y - \frac{\lambda^{2}}{2})^{3} = x^{2} y^{3} - 2x \lambda^{3} y^{3} + \lambda^{6} y^{3} - x^{2} y^{3} + \frac{3}{2} x^{2} y^{2} \lambda^{2} - \frac{3}{4} x^{2} y \lambda^{4} + \frac{1}{8} x^{2} \lambda^{6}
$$

Simplifying further, we get:

$$
- 2x \lambda^{3} y^{3} + \lambda^{6} y^{3} + \frac{3}{2} x^{2} y^{2} \lambda^{2} - \frac{3}{4} x^{2} y \lambda^{4} + \frac{1}{8} x^{2} \lambda^{6}
$$

Since $0 < \lambda \leq \frac{1}{2}$, all terms involving $\lambda$ are negative, thus the numerator is negative. Therefore, $\kappa_{n+1} - \kappa_{n} < 0$, and the sequence $\kappa_{n}$ is decreasing.

This completes the proof of our statement.

It can also be shown that for a fixed $n$, the value of $\kappa_{n}$ increases as $\lambda$ increases within the given bounds, but the problem did not ask for this.

Remark. We remind the reader of the isoperimetric ratio of convex plane figures, $t / p^{2}$ (Problem F. 2395, KöMaL 1983, September issue, p. 11). Here, we are dealing with a similar characteristic number for convex solids. The exponents of $t^{1}$ and $p^{2}$ are swapped, which is due to the fact that area is a 2-dimensional measure, while perimeter (length) is a 1-dimensional measure. Here, the dimensions of $F$ and $V$ are 2 and 3, respectively. Thus, the dimensions of $t / p^{2}$ and $V^{2} / F^{3}$ are both 0, meaning that the ratio does not depend on the choice of units, provided that the units of $V$ and $F$ are derived from the same length unit.

For convex plane figures, the maximum value of $t / p^{2}$ is $1 / 4 \pi$, which is achieved by the circle; for convex solids, the maximum value of $\kappa$ is $1 / 36 \pi$, which is achieved by the sphere. This is the so-called isoperimetric problem. (For a cube, $\kappa = 1 / 216$, and for a square, $t / p^{2} = 1 / 16$.)

Our result can be expressed intuitively as follows: the solids obtained by successively cutting off 1, 2, 3, 4 tetrahedra from the original tetrahedron gradually become more "sphere-like." When $\lambda = 1 / 2$, the result of the fourth cut is a regular octahedron.