Given an integer $n > 3$, let real numbers $x_{1}, x_{2}, \cdots, x_{n}$, $x_{n+1}, x_{n+2}$ satisfy the condition
$$
0 < x_{1} < x_{2} < \cdots < x_{n} < x_{n+1} < x_{n+2}.
$$

Find the minimum value of
$$
\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left(\sum_{k=1}^{n} \frac{x_{k+1} x_{k+2}}{x_{k+1}^{2} + x_{k} x_{k+2}}\right)\left(\sum_{l=1}^{n} \frac{x_{l+1}^{2} + x_{l} x_{l+2}}{x_{l} x_{l+1}}\right)}
$$

and determine all sets of real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that achieve this minimum value.
(Zhang Zhusheng provided)
Solution: (I) Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n+1$. The expression in the problem can be written as
$$
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} i_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i=1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)} \text {. }
$$

We observe that
$$
\begin{aligned}
( & \left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= & \left(\sum_{i=1}^{n} t_{i}-\sum_{i=1}^{n} \frac{t_{i,}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
& -\left(\sum_{i=1}^{n} \frac{t_{i}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
\leqslant & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
& \quad-\left(\sum_{i=1}^{n} \frac{t_{i}}{\sqrt{t_{i}+t_{i+1}}} \sqrt{t_{i}+t_{i+1}}\right)^{2} \\
= & \left(\sum_{i=1}^{n} t_{i}\right)^{2}+\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)-\left(\sum_{i=1}^{n} t_{i}\right)^{2} \\
= & \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right) .
\end{aligned}
$$

Therefore, for the real number sequence $00$, we have
$$
x_{k}=t_{k-1} t_{k-2} \cdots t_{1} a=a b^{k-1} c^{\frac{(k-1)(k-2)}{2}}, 2 \leqslant k \leqslant n
$$
+2 .
$$
\begin{array}{l}
\because x_{2}>x_{1}, \therefore b=\frac{x_{2}}{x_{1}}>1 . \\
\text{Also, since } t_{j}=b c^{j-1}>1,1 \leqslant j \leqslant n+1, \\
\therefore c>\sqrt[n]{\frac{1}{b}}\left(\geqslant \sqrt[j]{\frac{1}{b}}, 1 \leqslant j \leqslant n+1\right) .
\end{array}
$$
(III) Conclusion:
(i) For the real number sequence $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}$, $x_{n+2}$, the minimum value of the expression in the problem is 1.
(ii) The conditions for the expression to achieve its minimum value are $00, b>1, c>\sqrt[\pi]{\frac{1}{b}}$.