1.5.7 $* \star$ Let $x_{1}, \cdots, x_{n}$ be real numbers in the interval $\left(\frac{1}{4}, 1\right)$. Find the minimum value of the sum $\sum_{k=1}^{n} \log _{x_{4}}\left(x_{k+1}-\right.$ $\left.\frac{1}{4}\right)$, where $x_{n+1}=x_{1}$.
For any real number $x$, we have $x-\frac{1}{4} \leqslant x^{2}\left(\right.$ i.e., $\left.\left(x-\frac{1}{2}\right)^{2} \geqslant 0\right)$, combined with $\frac{1}{4} < x_{k} < 1$, and the properties of logarithmic functions, we can deduce that $\sum_{k=1}^{n} \log _{x_{k}}\left(x_{k+1}-\frac{1}{4}\right) \geqslant \sum_{k=1}^{n} \log _{x_{k}} x_{k+1}^{2} = 2 \sum_{k=1}^{n} \log _{x_{k}} \cdot x_{k+1} \geqslant 2 n \sqrt[n]{\prod_{k=1}^{n} \log _{x_{k}} x_{k+1}} = 2 n$. (The last equality uses the property $\log _{x} y \log _{y} z = \log _{x} z$.) Furthermore, when $x_{1} = \cdots = x_{n} = \frac{1}{2}$, we have $\sum_{k=1}^{n} \log _{x_{k}}\left(x_{k+1}-\frac{1}{4}\right) = 2 n$. Therefore, the minimum value sought is $2 n$.