$$
\begin{array}{c}
\text { Example 2. Let } \sum_{k=1}^{11} x_{k}=100, x_{k} \in N(k=1,2, \\
\cdots, 11) \text {, find the } F=\prod_{1 \leqslant i<j \leqslant 11}\left(x_{1}+x_{j}\right)!\text { }
\end{array}
$$

minimum value.
To find the minimum value of $F$ of type $N(100,0)$.
First, prove the reverse inequality in (4): Note
$$
\begin{array}{c}
F\left(a, b, c_{3}, \cdots, c_{11}\right) \\
=(a+b)!\cdot \prod_{3 \leqslant j \leqslant 11}\left(a+c_{j}\right)!\left(b+c_{j}\right)! \\
\cdot \prod_{3 \leqslant i < j \leqslant 11}\left(a+c_{i}+c_{j}\right)!\left(b+c_{i}+c_{j}\right)!.
\end{array}
$$
Since $a-1-c_{j} < a+c_{j}$, we have
$$
\begin{array}{l}
\left(a-1-c_{j}\right)!\left(b+1+c_{j}\right)! \\
=\frac{\left(a+c_{j}\right)!}{\left(a+c_{j}\right)} \cdot\left(b+1+c_{j}\right) \cdot\left(b+c_{j}\right)! \\
>\left(a+c_{j}\right)!\left(b+c_{j}\right)!.
\end{array}
$$

It is clear that $F\left(a-1, b+1, c_{3}, \cdots, c_{11}\right)$
$$
>F\left(a, b, c_{3}, \cdots, c_{11}\right).
$$

Since $100=9 \times 11+1$, we have $q=9, r=1$, hence the minimum value is
$$
\begin{aligned}
& F(9,9, \ldots, 9,10) \\
= & \prod_{1 \leqslant i < j \leqslant 11} \left(9 + c_i + c_j\right)! \left(9 + c_i + c_j\right)! \cdot 10! \cdot 10!.
\end{aligned}
$$

Given $\varphi(x) > 0$, and for $x \geqslant 2$ we have
$$
\varphi(x-1) \cdot \varphi(x+1)<[\varphi(x)]^{2}, \quad \text { (8) }
$$

Then $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\prod_{k=1}^{n} \varphi\left(x_{k}\right)$
is the maximum value of type $N(S, d)$
$$
\begin{array}{l}
\pi_{\max }=\prod_{k=1}^{n-x} \varphi(q+(k-1) d) \\
- \prod_{n=n-r+1}^{n} \varphi(q+(h-1) d+1). \\
\end{array}
$$

Let $f(x)=\ln \varphi(x)$, and by Theorem 2 it is easy to prove.
Note 3. Let $d=0$ to remove condition (2), then the conclusions of Theorem 2 and 3 are $\sigma_{\max }=(n-r) f(q)+r f(q+1)$, $\pi_{\max }=[\varphi(q)]^{n-r}[\varphi(q+1)]^{r}$.