AoPS - Problem

Source: Tuymaada 2020 Senior, Problem 2

Tags: Inequality, inequalities, n-variable inequality

IndoMathXdZ

06.10.2020 04:48

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 $\sqrt[n]{a_1 a_2 \dots a_n} + \frac{1}{\sqrt[n]{a_1 a_2 \dots a_n}} \ge m.$

IndoMathXdZ

06.10.2020 05:08

First Solution - Contradiction method, mine First, notice that if

$a_1 = a_2 = \dots = a_n$ , then the inequality obviously holds as it is an equality. Now, we assume that not all of

$a_i$ s are the same. For the sake of contradiction, we assume that

$\min \left( a_1 + \frac{1}{a_2} , a_2 + \frac{1}{a_3}, \dots , a_n + \frac{1}{a_1} \right) > \sqrt[n]{a_1 a_2 \dots a_n} + \frac{1}{\sqrt[n]{a_1 a_2 \dots a_n}}$ For simplicity, let

$G = \sqrt[n]{a_1 a_2 \dots a_n}$ . Now, notice that

$\min_i a_i < \sqrt[n]{a_1 a_2 \dots a_n} = G$ Therefore, we must have

$a_i + \frac{1}{a_{i + 1}} > G + \frac{1}{G}$ by our assumption, which gives us

$a_{i + 1} < G$ as well. Continue doing this cyclically, we get

$a_i < G$ for all

$i$ . Therefore,

$\prod_{i = 1}^{n} a_i < \prod_{i = 1}^n G = G^n = \prod_{i = 1}^n a_i$ which is a contradiction Second solution - Direct approach, Stephen Sanjaya Let

$G = \sqrt[n]{a_1 a_2 \dots a_n}$ . Claim. There must exists index

$i$ such that

$a_i \le G$ and

$a_{i + 1} \ge G$ . Proof. Take

$\min_i a_i$ and

$\max_j a_j$ . This must satisfy

$\min_i a_i \le G \le \max_j a_j$ . Now, let

$a_k = \min_j a_j$ . Then increase the index one by one, there must exists a smallest index

$\ell$ such that

$a_{\ell} \ge G$ . Then

$\ell - 1$ satisfy the criteria. Now, take such index

$i$ , then we have

$a_i + \frac{1}{a_{i + 1}} \le G + \frac{1}{G}$ We are done. Remark.In the contest, I didn't notice the solution immediately. It's quite tempting to use smoothing to fix RHS and increase the LHS, or even induction. However, as far as I know, none of these approaches work and only the approach above works.

Note. Thanks, @below. I have fixed the argument.

grupyorum

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.

alexiaslexia

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".

(To put it in perspective one can only find

$\textit{the idea}$ in that IMO

$\#5$ if and only if one decided to isolate the minimum from any other term, while in this problem the first solution is at hand once a person

$\textbf{unifies}$ the strangely loose minimal condition)

jj_ca888

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.