4. Among all tetrahedra with edge lengths 2, 3, 3, 4, 5, 5, what is the maximum volume? Prove your conclusion.
Below, we first prove a lemma: If the three lateral edges of a tetrahedron $O-ABC$ are $OA=a$, $OB=b$, $OC=c$, and the three dihedral angles are $\angle AOB=\alpha$, $\angle BOC=\beta$, $\angle AOC=\gamma$, then the volume $V$ of the tetrahedron $O-ABC$ is 
$$
V=\frac{1}{6} a b c \sqrt{1+2 \cos \alpha \cos \beta \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma}.
$$
As shown in the figure, let $\overrightarrow{OA}=(a, 0,0)$, $\overrightarrow{OB}=(b \cos \alpha, b \sin \alpha, 0)$,
$$
\begin{array}{l}
\overrightarrow{OC}=(x, y, z), \text{ then we have }\left\{\begin{array}{l}
|\overrightarrow{OC}|=c, \\
\overrightarrow{OA} \cdot \overrightarrow{OC}=a c \cos \gamma, \\
\overrightarrow{OB} \cdot \overrightarrow{OC}=b c \cos \beta
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=c^{2}, \\
a x=a c \cos \gamma, \\
b x \cos \alpha+b y \sin \alpha=b c \cos \beta
\end{array} \Rightarrow z^{2}=c^{2}\left[1-\cos ^{2} \gamma-\left(\frac{\cos \beta-\cos \alpha \cos \gamma}{\sin \alpha}\right)^{2}\right]\right. \\
=\frac{c^{2}}{\sin ^{2} \alpha}\left[\sin ^{2} \alpha-\sin ^{2} \alpha \cos ^{2} \gamma-(\cos \beta-\cos \alpha \cos \gamma)^{2}\right] \\
=\frac{c^{2}}{\sin ^{2} \alpha}\left(\sin ^{2} \alpha-\cos ^{2} \gamma-\cos \beta+2 \cos \alpha \cos \beta \cos \gamma\right) \\
=\frac{c^{2}}{\sin ^{2} \alpha}\left(1+2 \cos \alpha \cos \beta \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma\right) \\
\Rightarrow|z|=\frac{c}{\sin ^{2}} \sqrt{1+2 \cos \alpha \cos \beta \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma} \\
\Rightarrow V=\frac{1}{3} S_{\triangle OAB} \cdot|z|=\frac{1}{3} \cdot \frac{1}{2} a b \sin \alpha \cdot|z| \\
=\frac{1}{6} a b c \sqrt{1+2 \cos \alpha \cos \beta \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma} .
\end{array}
$$

Since $2,3,5$ cannot form a triangle, in the triangles sharing the edge of length 2, there are four scenarios: $2,3,3$; $2,3,4$; $2,4,5$; $2,5,5$. Without loss of generality, let $AB=2$, then the combinations of $\triangle ABD$ and $\triangle ABC$ are: $2,3,3$, $2,4,5$; $2,3,3,2,5,5$; $2,3,4,2,5,5$.
For the combination $2,3,3,2,4,5$, the three edges from point $D$ are all $3,3,5$, and the corresponding angles are equal, so they have the same volume. $\cos \angle ADB=\frac{9+9-4}{2 \cdot 3 \cdot 3}=\frac{7}{9}$, $\cos \angle ADC=\frac{9+25-16}{2 \cdot 3 \cdot 5}=\frac{3}{5}$, $\cos \angle BDC=\frac{9+25-25}{2 \cdot 3 \cdot 5}=\frac{3}{10}$. Thus, $V_{1}=\frac{1}{6} \cdot 3 \cdot 3 \cdot 5 \cdot \sqrt{1+2 \cdot \frac{7}{9} \cdot \frac{3}{5} \cdot \frac{3}{10}-\frac{49}{81}-\frac{9}{25}-\frac{9}{100}}=\frac{\sqrt{1823}}{12}$. For the combination $2,3,3,2,5,5$, $\cos \angle ADB=\frac{9+9-4}{2 \cdot 3 \cdot 3}=\frac{7}{9}$, $\cos \angle ADC=\cos \angle BDC=0$, thus $V_{2}=\frac{1}{6} \cdot 3 \cdot 3 \cdot 4 \cdot \sqrt{1-\frac{49}{81}}=\frac{8 \sqrt{2}}{3}$;
For the combination $2,3,4,2,5,5$, $\cos \angle ADB=\frac{9+16-4}{2 \cdot 3 \cdot 4}=\frac{7}{8}$, $\cos \angle ADC=\frac{9+9-25}{2 \cdot 3 \cdot 3}=-\frac{7}{18}$, $\cos \angle BDC=0$.
Thus, $V_{3}=\frac{1}{6} \cdot 3 \cdot 4 \cdot 3 \cdot \sqrt{1-\frac{49}{64}-\frac{49}{18^{2}}}=\frac{\sqrt{431}}{12}$. Clearly, $V_{2}>V_{1}>V_{3}$, in summary, the largest volume is $\frac{8 \sqrt{2}}{3}$.