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Yet another logarithmic inequality I do not know how to solve.

$$\log_{x_1} \sqrt {3x_2 - 2a} + \log_{x_2} \sqrt {3x_3 - 2a} + ... \log_{x_n} \sqrt {3x_1 - 2a} \ge \frac {n}{2},\\ a > 2, \;x_1, x_2, x_3, ..., x_n \in [a, 2a], 3x_i \gt 2a$$ with the existence conditions.

Since $x_i \ge a \gt 2 \ and \ 3x_i - 2a \ge x_i, $ we have $log_{x_i} (3x_{i + 1} - 2a) \ \ge log_{x_i} x_{i + 1}$, so $\sum_{k = 1} ^ n log_{x_k}(3x_{k + 1} - 2a) \ge \sum_{k = 1} ^ n log_{x_k} x_{k + 1} \ge n\sqrt [n] (\prod_{k = 1} ^ n log_{x_k} x_{k + 1}) = n$, according to AM-GM.

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  • $\begingroup$ I have tried to add the conditions to make it well defined. $\endgroup$
    – andu eu
    Commented Oct 8, 2020 at 10:48
  • $\begingroup$ @andu eu But you did not give the answer on my question. Do you want to solve or to prove this inequality? $\endgroup$ Commented Oct 8, 2020 at 10:49
  • $\begingroup$ I want to prove it. $\endgroup$
    – andu eu
    Commented Oct 8, 2020 at 10:49
  • $\begingroup$ @andu eu If so, fix it please! $\endgroup$ Commented Oct 8, 2020 at 10:50
  • $\begingroup$ I am sorry for the misunderstanding, but I am in the middle of school. $\endgroup$
    – andu eu
    Commented Oct 8, 2020 at 10:51

1 Answer 1

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The original inequality is equivalent to $$ \log_{x_1}(3x_2-2a) + \cdots + \log_{x_n}(3x_{n+1}-2a)\ge n. $$

For each term in the sum, you can consider the inequality $$ \log_{x_i}(3x_{i+1}-2a) \ge 1,$$

which holds in the case $x_i, x_{i+1} \ge a$. So, you conclude that the sum of these terms is $\ge n$ and that the original inequality holds.

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