Four, among all tetrahedra with edge lengths of $2,3,3,4,5,5$, what is the maximum volume? Prove your conclusion.

In all tetrahedra with edge lengths of $2,3,3,4,5,5$, what is the maximum volume? Prove your conclusion.
Four, Solution: The maximum volume is $\frac{8 \sqrt{2}}{3}$.
According to the property that the difference of the lengths of any two sides of a triangle is less than the length of the third side, given the data in the problem, for all triangles with one side of 2, the other two sides can only be:
(1) 3,3;
(2) 5,5;
(3) 4,5;
(4) 3,4.

Thus, in the tetrahedron described in the problem, the other two sides of the two side faces sharing the common edge of 2 can only have the following three scenarios:
(1): (1) and (2);
(2): (1) and (3);
(3): (2) and (4).
We will discuss these three scenarios separately:
(1) As shown in the right figure, $A C=B C=3$,

Since
$$
\begin{array}{l}
A D=B D=5, \\
3^{2}+4^{2}=5^{2}, \\
C D \perp A C, C D \perp B C,
\end{array}
$$

Thus, $C D$ is perpendicular to the plane $A B C$.
By symmetry, there is only one such tetrahedron. Its volume is:
$$
\begin{aligned}
V_{1} & =\frac{1}{3} C D \cdot \triangle A B C \\
& =\frac{4}{3} \cdot \frac{1}{2} \cdot 2 \cdot \sqrt{3^{2}-1}=\frac{8 \sqrt{2}}{3}
\end{aligned}
$$
(Where $\triangle A B C$ represents the area of $\triangle A B C$, the same applies below).
(2) There are two such tetrahedra, as shown in the figure below. It is easy to see that their volumes are equal, denoted as $V_{2}$,
Since $2^{2}+4^{2}<5^{2}$, $\angle B A C$ is an acute angle, i.e., the edge $A B$ intersects the plane $A C D$ externally. Let the height from $B$ to the base $A C D$ be $h_{3}$, then $h_{3}<A B=2$.
Thus, $V_{3}=\frac{1}{3} h_{3} \triangle A C D<\frac{2}{3} \triangle A C D$
$$
=\frac{2}{3} \cdot \frac{1}{2} \cdot 5 \cdot \sqrt{3^{2}-\left(\frac{5}{2}\right)^{2}}=\frac{5}{6} \sqrt{11} \text {. }
$$

Since $V_{3}^{2}<\frac{275}{36}$,
$$
\begin{array}{c}
V_{1}^{2}=\frac{128}{9} . \\
V_{3}<V_{1} .
\end{array}
$$

Therefore,
So, the maximum volume is $\quad V_{1}=\frac{8 \sqrt{2}}{3}$.