Find the min value of $$\left\vert \log_{x_1} \left ( x_2 - \frac {1}{4}\right) +\log_{x_2} \left ( x_3 - \frac {1}{4}\right) +\log_{x_3} \left ( x_4 - \frac {1}{4}\right) +\cdot \cdot \cdot \cdot \cdot +\log_{x_{2016}} \left ( x_{2017} - \frac {1}{4}\right) +\log_{x_{2017}} \left ( x_{1} - \frac {1}{4}\right) \right\vert$$
Where $x_1,x_2,x_3,......, x_{2017}\in \left(\frac {1}{4}, 1\right)$
I am not getting any idea over where and how to start approaching this problem.
Note that
$$\log_{x_i} \left ( x_j - \frac {1}{4}\right)=\frac {\log \left ( x_j - \frac {1}{4}\right)}{\log x_i}$$
thus by Rearrangement inequality
$$\left\vert \log_{x_1} \left ( x_2 - \frac {1}{4}\right) +\log_{x_2} \left ( x_3 - \frac {1}{4}\right) +\log_{x_3} \left ( x_4 - \frac {1}{4}\right) +\cdot \cdot \cdot \cdot \cdot +\log_{x_{2016}} \left ( x_{2017} - \frac {1}{4}\right) +\log_{x_{2017}} \left ( x_{1} - \frac {1}{4}\right) \right\vert=$$
$$=\left\vert\frac {\log \left ( x_2 - \frac {1}{4}\right)}{\log x_1}+\frac {\log \left ( x_3 - \frac {1}{4}\right)}{\log x_2}+...+\frac {\log \left ( x_1 - \frac {1}{4}\right)}{\log x_{2017}}\right\vert \ge \sum_{i=1}^{2017} \frac {\log \left ( x_i - \frac {1}{4}\right)}{\log x_i}$$
and equality holds when $\forall i,j\quad x_i=x_j$.
Thus we get the minimum when $\frac {\log \left ( x_i - \frac {1}{4}\right)}{\log x_i}$ is minimum that is 2 for $x_i=\frac12$.
Thus the minimum for the given sum is $\sum_{i=1}^{2017}2= 2 \cdot 2017=4034$.