5、In Rt $\triangle A B C$, $C=\frac{\pi}{2}, B C=a, C A=b$, point $P$ is on side $A B$, along line $P C$ fold
$R t \triangle A B C$ into tetrahedron $P A B C$, the maximum volume of this tetrahedron is $\qquad$ (expressed in terms of $\boldsymbol{a}, \boldsymbol{b}$)
5. $\frac{a^{2} b^{2}}{6\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}}$

Analysis: As shown in the right figure, $P D \perp A C, B E \perp P C$, let $|P C|=x, \angle P C A=\theta$, then we know $|B E|=a \cdot \cos \theta$, and we have
$\frac{|P D|}{a}=\frac{b-|C D|}{b}$, i.e., $\frac{x \cdot \sin \theta}{a}=\frac{b-x \cdot \cos \theta}{b}$, thus,
$$
x=\frac{a b}{a \cos \theta+b \sin \theta} \text { . }
$$

Clearly, when $B E \perp$ plane $P C D$, the volume of the tetrahedron is maximized. At this time, the volume of the tetrahedron $B-P C D$ is
$$
V_{B-P C D}=\frac{1}{3} S_{\mathrm{VPCA}} \cdot|B E|=\frac{a^{2} b^{2} \sin \theta \cos \theta}{6(a \cos \theta+b \sin \theta)}=\frac{a^{2} b^{2}}{6} \cdot \frac{1}{\frac{a}{\sin \theta}+\frac{b}{\cos \theta}} .
$$

Consider the minimum value of $S=\frac{a}{\sin \theta}+\frac{b}{\cos \theta}$, by the AM-GM inequality
$$
S=\frac{a}{2 \sin \theta}+\frac{a}{2 \sin \theta}+k \sin ^{2} \theta+\frac{b}{2 \cos \theta}+\frac{b}{2 \cos \theta}+k \cos ^{2} \theta-k \geq 3\left(\frac{a^{2} k}{4}\right)^{\frac{1}{3}}+3\left(\frac{b^{2} k}{4}\right)^{\frac{1}{3}}-k
$$
, equality holds when $\frac{a}{2 \sin \theta}=k \sin ^{2} \theta$, and $\frac{b}{2 \cos \theta}=k \cos ^{2} \theta$. At this time, $\sin \theta=\frac{a^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}, \cos \theta=\frac{b^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}, k=\frac{\sqrt{a^{2}+3 a^{\frac{4}{3}} b^{\frac{2}{3}}+3 a^{\frac{2}{3}} b^{\frac{4}{3}}+b^{2}}}{2}$, substituting in we get
$$
V_{B-P C D}=\frac{a^{2} b^{2}}{6} \cdot \frac{1}{\frac{a}{\sin \theta}+\frac{b}{\cos \theta}} \leq \frac{a^{2} b^{2}}{6} \cdot \frac{1}{\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}}=\frac{a^{2} b^{2}}{6\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}}
$$