Four. (20 points) In the right triangle $\triangle ABC$, $\angle ACB = 90^{\circ}$, $BC = a$, $CA = b$ ($a > 0$, $b > 0$), and point $P$ is on side $AB$. The right triangle $\triangle ABC$ is folded along the line $PC$ to form a tetrahedron $PABC$. Find the maximum possible volume of this tetrahedron.
Let the volume of tetrahedron $PABC$ be $V$.
As shown in Figure 7, construct $PD \perp AC$ at point $D$, and $BE \perp PC$ at point $E$. Then,
$$
\begin{array}{l}
V \leqslant \frac{1}{3}\left(\frac{1}{2} PD \cdot AC\right) \cdot BE \\
=\frac{b}{6} PD \cdot a \sin \theta=\frac{ab}{6} PD \sin \theta \\
=\frac{ab}{6} PD \sin \left(90^{\circ}-\varphi\right)=\frac{ab}{6} PD \cos \varphi .
\end{array}
$$
The equality holds if and only if the plane $PBC \perp$ plane $PCA$.
In $\triangle PAC$, by $AD + DC = b$, we get
$PD \cot \alpha + PD \cot \varphi = b$, where $\cot \alpha = \frac{b}{a}$.
$$
\begin{array}{l}
\text{Then } PD = \frac{b}{\frac{b}{a} + \cot \varphi}, \\
V \leqslant \frac{ab}{6} \cdot \frac{b \cos \varphi}{\frac{b}{a} + \cot \varphi} = \frac{a^2 b^2}{6} \cdot \frac{\cos \varphi}{b + a \cot \varphi} \\
= \frac{a^2 b^2}{6} \sqrt{\frac{\cos^2 \varphi}{(b + a \cot \varphi)^2}} \\
= \frac{a^2 b^2}{6} \cdot \frac{1}{\sqrt{\left(1 + \tan^2 \varphi\right)(b + a \cot \varphi)^2}}. \\
\text{Let } x = \tan \varphi (x > 0), \\
f(x) = \left(1 + x^2\right)\left(b + \frac{a}{x}\right)^2 \\
= b^2 x^2 + 2abx + a^2 + b^2 + \frac{2ab}{x} + \frac{a^2}{x^2} \\
= \left(b^2 x^2 + \frac{ab}{x} + \frac{ab}{x}\right) + \left(abx + abx + \frac{a^2}{x^2}\right) + a^2 + b^2 \\
\geqslant 3 \sqrt[3]{a^2 b^4} + 3 \sqrt[3]{a^4 b^2} + a^2 + b^2 \\
= \left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^3. \\
\end{array}
$$

The equality holds if and only if $x = \sqrt[3]{\frac{a}{b}}$.
Thus, $V \leqslant \frac{a^2 b^2}{6} \cdot \frac{1}{\sqrt{f(x)}} \leqslant \frac{a^2 b^2}{6} \cdot \frac{1}{\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}}$
$$
= \frac{1}{6} \cdot \frac{a^2 b^2}{\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}}.
$$

The equality holds if and only if the plane $PCB \perp$ plane $PCA$, and $\tan \varphi = \left(\frac{a}{b}\right)^{\frac{1}{3}}$.
Therefore, the maximum volume of the tetrahedron is
$$
\frac{1}{6} \cdot \frac{a^2 b^2}{\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}}.
$$