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

In all tetrahedrons with edge lengths of $2,3,3,4,5,5$, what is the maximum volume? Prove your conclusion.
$\frac{8 \sqrt{2}}{3}$
Four, [Analysis and Solution] According to the property that the difference between two sides of a triangle is less than the third side, according to the given data, for a triangle with one side being 2, the other two sides can only be:
(1) 3,3;
(2) 5,5;
(3) 4,5;
(4) 3,4
Thus, in the tetrahedron given in the problem, the other two sides of the two side faces sharing the common side 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, $A C=B C=3, A D=B D=5$,
Since $3^{2}+4^{2}=5^{2}$,
Therefore, $C D \perp A C, C D \perp B C$,
Thus, $C D$ is perpendicular to the plane $A B C$.
By symmetry, there is only one such tetrahedron. Its volume is:
$$
V_{1}=\frac{1}{3} C D \cdot S_{\triangle A B C}=\frac{4}{3} \times \frac{1}{2} \times 2 \times \sqrt{3^{2}-1}=\frac{8 \sqrt{2}}{3} \text {. }
$$
(2) There are two such tetrahedrons, as shown in the figure below. It is easy to know that their volumes are equal, denoted as $V_{2}$,

Since $2^{2}+4^{2}<5^{2}$, hence $\angle A B D$ is an obtuse angle, i.e., the edge $B D$ intersects the plane $A B C$ obliquely. Let the height from $D$ to the base $A B C$ be $h_{2}$, then $h_{2}<B D=4$,
Therefore, $V_{2}=\frac{1}{3} h_{2} \cdot S_{\triangle A B C}<\frac{4}{3} S_{\triangle A B C}=V_{1}=\frac{8 \sqrt{2}}{3}$.
(3) There are also two such tetrahedrons, as shown in the figure below, their volumes are also equal, denoted as $V_{3}$.
Since $2^{2}+4^{2}<5^{2}$. Hence $\angle B A C$ is an acute angle, i.e., the edge $A B$ intersects the plane $A C D$ obliquely. Let the height from $B$ to the base $A C D$ be $h_{3}$, then $h_{3}<A B=2$.
Therefore, $V_{3}=\frac{1}{3} h_{3} \cdot S_{\triangle A C D}<\frac{2}{3} S_{\triangle A C D}=\frac{2}{3} \times \frac{1}{2} \times 5 \sqrt{3^{2}-\left(\frac{5}{2}\right)^{2}}=\frac{5}{6} \sqrt{11}$.
Since $V_{3}^{2}<\frac{275}{36}, V_{1}^{2}<\frac{128}{9}$.
Therefore, $V_{3}<V_{1}$.
Thus, the largest volume is $V_{1}=\frac{8 \sqrt{2}}{3}$.