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.

  • $\begingroup$ I would have thought that all those terms were positive. If you had all $x_i=\frac12$ then the sum of the logarithms would be $2017\times 2=4034$ $\endgroup$ – Henry Jan 30 '18 at 8:14
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Jan 31 '18 at 22:36

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 $\sum_{i=1}^{2017}2= 2 \cdot 2017=4034$.

  • $\begingroup$ Does that imply the minimum value is 2017? $\endgroup$ – Sonal_sqrt Jan 30 '18 at 8:28
  • $\begingroup$ so its $2\times2017$ $\endgroup$ – Sonal_sqrt Jan 30 '18 at 8:35
  • $\begingroup$ Yes of course, I was misleading the equality case! It is 2*2017! $\endgroup$ – user Jan 30 '18 at 8:58

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