15. (2001 China National Team Selection Test for IMO) Given an integer $n$ greater than 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 $\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\left(\sum_{k=1}^{n} \frac{x_{k+1} x_{k+2}}{x_{k+1}^{2}+x_{k} x_{k+2}}\right)\left(\sum_{i=1}^{n} \frac{x_{l+1}^{2}+x_{l} x_{l+2}}{x_{l} x_{l+1}}\right)}$
and determine all real number sequences $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ that achieve this minimum value.
15. (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
\[
\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right)}.
\]

We observe that
\[
\begin{array}{l}
\left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) = \left(\sum_{i=1}^{n} t_{i} - \sum_{i=1}^{n} \frac{t_{i}^{2} + t_{i+1}}{}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
= \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) - \left(\sum_{i=1}^{n} \frac{t_{i}^{2}}{t_{i}+t_{i+1}}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) \\
\leqslant \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right) - \left(\sum_{i=1}^{n} \frac{t_{i}}{\sqrt{t_{i}+t_{i+1}}} \sqrt{t_{i}+t_{i+1}}\right)^{2} \\
= \left(\sum_{i=1}^{n} t_{i}\right)^{2} + \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right) - \left(\sum_{i=1}^{n} t_{i}\right)^{2} = \left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right).
\end{array}
\]

Therefore, for the real number sequence \( 00 \), we have
\[ x_{k} = t_{k \cdots 1} t_{k-2} \cdots t_{1}, a = a b^{k-1} c^{\frac{(k-1)(k-2)}{2}}, 2 \leqslant k \leqslant n+2. \]
Since \( x_{2} > x_{1} \), then \( b = \frac{x_{2}}{x_{1}} > 1 \).
Also, since \( t_{j} = b c^{j \cdot 1} > 1, 1 \leqslant j \leqslant n+1 \),
then \( c > \sqrt[n]{\frac{1}{b}} \left(\geqslant \sqrt[j]{\frac{1}{b}}, 1 \leqslant j \leqslant n+1\right) \).
(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 minimum value of the expression in the problem is 1.
(ii) The conditions for the expression to achieve its minimum value are \( 00, b > 1, c > \sqrt[n]{\frac{1}{b}} \).