# Inequality $\sum_{k = 1}^n \log_{x_k} \sqrt {3x_{k + 1} - 2a} \ge \frac {n}{2}$

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.

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

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.