Example 8 Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ satisfy $0<x_{1}<x_{2}$ $<\cdots<x_{n}<x_{n+1}<x_{n+2}$. Try to find the minimum value of $\left.\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{x_{k+1} x_{k+2}}\right)\left(\sum_{i=1}^{n} \frac{x_{l+1}^{2}+x_{l} x_{l+2}}{x^{2}}\right)$ and the real number sequences $x_{k}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that achieve this minimum value, given that $\left(\sum_{k=1}^{n} \frac{x_{k+1} x_{k+2}}{x_{k+1}^{2}+x_{k+1} x_{k+2}}\right)=\left(\sum_{l=1}^{n} \frac{x_{l+1}^{2}+x_{l} x_{l+2}}{x_{l} x_{l+2}}\right)$.
Let $(\bar{I})$ denote $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n$, the expression in the problem can be written as
$$\frac{\left(\sum_{i=1}^{n} t_{i}\right):\left(\sum_{i=1}^{n} t_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)=\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)}$$

We see that by the Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)= \\
\left(\sum_{i=1}^{n} t_{i}-\sum_{i=1}^{n} \frac{t_{i}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)= \\
\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
\left(\sum_{i=1}^{n} \frac{t_{i}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \leqslant \\
\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)- \\
\left(-\sum_{i=1}^{n} \frac{t_{i}}{\sqrt{t_{i}+t_{i+1}}} \cdot \sqrt{t_{i}+t_{i+1}}\right)^{2}= \\
\left(\sum_{i=1}^{n} t_{i}\right)^{2}+\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)-\left(\sum_{i=1}^{n} t_{i}\right)^{2}= \\
\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)
\end{array}$$

Therefore, for the real number sequence $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that meets the conditions, the expression in the problem is no less than 1.
( II ) In the above derivation, the Cauchy-Schwarz inequality is used, and the equality holds if and only if $\frac{\sqrt{t_{i}+t_{i+1}}}{\frac{t_{i}}{\sqrt{t_{i}+t_{i+1}}}}=$
$d(1 \leqslant i \leqslant n)($ constant $)$, which means
$$\frac{t_{i+1}}{t_{i}}=d-1=c,(1 \leqslant i \leqslant n)$$

Let $t_{1}=b$, then $t_{j}=b c^{j-1}, 1 \leqslant j \leqslant n+1$, and accordingly,
$$\frac{x_{j+1}}{x_{j}}=t_{j}=b c^{j-1}, 1 \leqslant j \leqslant n+1$$

Let $x_{1}=a>0$, then
$$x_{k}=t_{k-1} t_{k-2} \cdots t_{1} \cdot a=a b^{k-1} \cdot c^{\frac{(k-1)(k-2)}{2}}, 2 \leqslant k \leqslant n+2$$

Since $x_{2}>x_{1}$, we have
$$b=\frac{x_{2}}{x_{1}}>1$$

Since
$$t_{j}=b c^{j-1}>1,1 \leqslant j \leqslant n+1, c>\sqrt[n]{\frac{1}{b}}\left(\geqslant \sqrt[j-1]{\frac{1}{b}}, 1 \leqslant j \leqslant n+1\right)$$
(III) Conclusion:
( i ) For the real number sequence $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that meets the conditions, the expression in the problem is no less than 1.
( ii ) The sequence that can achieve the minimum value of the expression satisfies $00, b>1, c>\sqrt[n]{\frac{1}{b}}$.