Super cool Algebra from the IMO!!

 (This article was written 15-1-2025)

Problem (23IMO4). Let \(x_1,x_2,\dots,x_{2023}\) be pairwise different positive real numbers such that \[a_n = \sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\] is an integer for every \(n = 1,2,\dots,2023\). Prove that \(a_{2023} \geqslant 3034\).


Let's try to get small bounds on the \(a_i\).


Well, \(a_1 = 1\) is obvious.  

Similarly, we have \[a_2 = \sqrt{(x_1 + x_2) \left(\frac{1}{x_1} + \frac{1}{x_2} \right)} \ge 2,\] where the inequality is by Cauchy-Schwarz. But the sequence has distinct terms (conveniently enough), so in fact \(a_2 \ge 3\).


Is there any general inductive bound we can have on \(a_{n+1}\) given \(a_n\)? Let's see: \[a_{n+1} = \sqrt{(x_1+x_2+\dots+x_{n+1})\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_{n+1}}\right)}.\] But we can already see a hidden \(a_n\): \[a_{n+1} = \sqrt{\left((x_1+x_2+\dots+x_n) + x_{n+1}\right)\left(\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right) + \frac{1}{x_{n+1}}\right)}\] \[= \sqrt{a_n^2 + \frac{(x_1 + \dots + x_n)}{x_{n+1}} + x_{n+1} \left(\frac{1}{x_1} + \dots + \frac{1}{x_n}\right) + 1}.\]


Now we can use AM-GM: \[\ge \sqrt{a_n^2 + 2(x_1 + \dots + x_n)\left(\frac{1}{x_1} + \dots + \frac{1}{x_n}\right) + 1} = \sqrt{a_n^2 + 2a_n + 1} = a_n + 1.\]


This is a bit of a bummer — if we follow this bound we'll only reach \(2024\), not the required \(3034\).  

So the question becomes — why can't equality hold everywhere? (Note the similarity to the proof that \(a_2 \ge 3\).)


Claim 1. \(a_{n+1} \ge a_n + 1\), with equality if and only if \[x_{n+1} = \frac{x_1 + \dots + x_n}{a_n}.\]


Proof. We have \[a_{n+1} = \sqrt{a_n^2 + 1 + x_{n+1} \left(\frac{1}{x_1} + \dots + \frac{1}{x_n} \right) + \frac{x_1 + \dots + x_n}{x_{n+1}}} \ge \sqrt{a_n^2 + 1 + 2a_n} = a_n + 1,\] by AM-GM. Equality holds if and only if equality holds in AM-GM, i.e. \[x_{n+1}^2 = \frac{x_1 + \dots + x_n}{\frac{1}{x_1} + \dots + \frac{1}{x_n}} = \frac{x_1 + \dots + x_n}{\frac{a_n}{x_1 + \dots + x_n}} \iff x_{n+1} = \frac{x_1 + \dots + x_n}{a_n}.\] \(\square\)


Claim 2. We cannot simultaneously have \(a_{n+2} = a_{n+1} + 1\) and \(a_{n+1} = a_n + 1\).


Proof. For equality to hold in both cases, we would need \[x_{n+2} = \frac{x_1 + \dots + x_n + x_{n+1}}{a_n + 1}.\] But since \[x_{n+1} = \frac{x_1 + \dots + x_n}{a_n},\] we substitute to get \[x_{n+2} = \frac{x_1 + \dots + x_n + \frac{x_1 + \dots + x_n}{a_n}}{a_n + 1} = \frac{x_1 + \dots + x_n}{a_n} = x_{n+1},\] a contradiction to the terms being distinct. \(\square\)


Due to Claim 2, we must have \(a_{n+2} \ge a_n + 3\). Therefore, \[a_{2023} \ge a_{2021} + 3 \ge a_{2019} + 6 \ge a_{2017} + 9 \ge \cdots \ge a_1 + 3 \times 1011 = 3034,\] as desired.\(\square\)


Motivational Note.  

While most inequalities rely on studying the equality case, this is a refreshing twist — instead of proving when equality holds, we show it cannot hold for two consecutive steps. This is a form of reverse sharpness: the bounds are tight only in cases that are actually forbidden by the problem conditions (distinct \(x_i\)).  

The integrality of the \(a_i\) forces jumps of at least 1, and violating equality at every step means those jumps quickly accumulate into a much higher bound than initially expected.

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