Prove: Let $S_{k}=a^{k}+b^{k}+c^{k}+d^{k}(k \in N)$.
(i) $S_{3}=a^{3}+b^{3}+c^{3}-(a+b+c)^{3}$.

Since when $a=\cdots b$ or $b=-c$ or $c=-a$, $0=0$.
Assume $S_{3} \equiv \mu(a+b)(b+c)(c+a)$. When $a=b=c=1$, we get $\mu=-3$.
$$\begin{aligned} \therefore \quad S_{3} & =-3(a+b)(b+c)(c+a) . \\ \text { Also, } S_{2} & =a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \\ & =(a+b)^{2}+(b+c)^{2}+(c+a)^{2}, \end{aligned}$$

By the Arithmetic-Geometric Mean Inequality, we have
$$\begin{array}{l} 27(a+b)^{2}(b+c)^{2}(c+a)^{2} \\ \leqslant\left[(a+b)^{2}+(b+c)^{2}+(c+a)^{2}\right]^{3} . \end{array}$$

Substituting (1) and (2) into (3) and simplifying, we get (i).
(ii) $S_{5}=a^{5}+b^{5}+c^{5}-(a+b+c)^{5}$.

Since when $a=-b$ or $b=-c$ or $c=-a$, $S_{5} \equiv 0$. Hence, we can assume
$$S_{5}=(a+b)(b+c)(c+a)\left[p\left(a^{2}+b^{2}+c^{2}\right)+q(a b\right.$$
$+b c+c a$ )]. (Since $S_{5}$ is a symmetric polynomial of degree 5 in $a, b, c$)
Taking $a=b=c=1$ and $a=b=1, c=0$ respectively, we get
$$p+q=-10,2 p+q=-15 \text {.. }$$

Solving these equations, we get $p=q=-5$.
$$\begin{aligned} \therefore S_{5}= & -\frac{5}{2}(a+b)(b+c)(c+a) \\ & \cdot\left[(a+b)^{2}+(b+c)^{2}+(c+a)^{2}\right] . \end{aligned}$$

Thus, $108 S_{5}^{2}=25\left[27(a+b)^{2}(b+c)^{2}(c+a)^{2}\right][(a+$
$$\left.b)^{2}+(b+c)^{2}+(c+a)^{2}\right]^{2} \text {. }$$

By (3), we get
$$108 S_{5}^{2} \leqslant 25\left[(a+b)^{2}+(b+c)^{2}+(c+a)^{2}\right]^{5} \text {. }$$

Substituting (2) into the equation, we get (ii).