9-46 Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$, find
$$
S=(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4}
$$
the maximum value.
[Solution]
$$
\begin{aligned}
S \leqslant & (a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4} \\
& +(c+d)^{4}+(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b- \\
& c)^{4}+(b-d)^{4}+(c-d)^{4} \\
= & 6\left(a^{4}+b^{4}+c^{4}+d^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}+2 a^{2} d^{2}+2 b^{2} c^{2}\right. \\
& \left.+2 b^{2} d^{2}+2 c^{2} d^{2}\right) \\
= & 6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2} \leqslant 6 .
\end{aligned}
$$

If $a=b=c=d=\frac{1}{2}$, then $S=6$. Therefore, the maximum value of $S$ is 6.