98. Let positive numbers $a, b, c$ satisfy $a+b+c \leqslant \frac{3}{2}$, find the minimum value of $S=abc+\frac{1}{abc}$. (2008 Moldova Mathematical Olympiad Problem)
98. By the AM-GM inequality, we have $\frac{3}{2} \geqslant a+b+c \geqslant 3 \sqrt[3]{a b c}$, thus $a b c \leqslant \frac{1}{8}$. By the AM-GM inequality, we get
$$\begin{aligned}
S= & a b c+\frac{1}{64 a b c}+\frac{63}{64 a b c} \geqslant 2 \sqrt{a b c \cdot \frac{1}{64 a b c}}+\frac{63}{64 a b c} \geqslant \\
& \frac{1}{4}+\frac{63}{8}=\frac{65}{8}
\end{aligned}$$