Given positive real numbers $a_1, a_2, \dots, a_n$. Let \[ m = \min \left( a_1 + \frac{1}{a_2}, a_2 + \frac{1}{a_3}, \dots, a_{n - 1} + \frac{1}{a_n} , a_n + \frac{1}{a_1} \right). \]Prove the inequality \[ \sqrt[n]{a_1 a_2 \dots a_n} + \frac{1}{\sqrt[n]{a_1 a_2 \dots a_n}} \ge m. \]
Problem
Source: Tuymaada 2020 Senior, Problem 2
Tags: Inequality, inequalities, n-variable inequality
06.10.2020 05:08
Note. Thanks, @below. I have fixed the argument.
07.10.2020 01:51
@IndoMath: The solution I came up with is exactly the same the first one. However, I am not convinced with the second solution: it appears a contradiction argument is used to establish the claim. But the negation of the assertion of claim is the following: for each $i$, either $G<a_i$ or $G>a_{i+1}$ (as opposed to $a_i \le G$ for all $i$ or $a_i\ge G$ for all $i$). It is not clear to me how one would conclude from that.
12.11.2020 12:44
The $\textbf{direct method}$ above is the official solution (while in the actual test I came up with the same solution as @IndoMathXdZ) $\rule{25cm}{0.5pt}$ Note: To elaborate a bit on the motivation; I decided it would be a good idea to multiply $\textbf{ALL}$ equations because the geometric mean $\sqrt[n]{a_1a_2 \ldots a_n}$ really invited me to bet on the belief that "geometric mean is really much, much weaker than the polarized sum of reciprocals. Moreover, the minimal condition sounds as if it mimics the idea of IMO $\#5$ where the "inequality" is best homogenised due to order/size reasons to simplify "absurd minimality".
18.12.2021 06:53
Denote by $M$ the GM. If there is any $i$ for which $a_i \leq M, a_{i+1} \geq M$ then we win, as\[m \leq a_i + \frac{1}{a_{i+1}} \leq M + \frac{1}{M}\]is guaranteed. We will show that such an $i$ must exist (for simplicity let indices cycle modulo $n$). First, pick a $k$ for which $a_k\leq M$ and an $\ell > k$ for which $a_{\ell} \geq M$. The subsequence $a_i$ as $i$ goes from $k$ to $\ell$ goes from being upper bounded by $M$ to being lower bounded by $M$, so there must be two consecutive terms that go from $\leq M$ to $\geq M$, as desired. If not, then the sequence stays $\leq M$ forever, impossible unless all terms are equal, which is a special case that clearly works. Now we win.