# 2023 IMO Problems/Problem 4

## Problem

Let $x_1, x_2, \cdots , x_{2023}$ be pairwise different positive real numbers such that $$a_n = \sqrt{(x_1+x_2+ \text{···} +x_n)(\frac1{x_1} + \frac1{x_2} + \text{···} +\frac1{x_n})}$$ is an integer for every $n = 1,2,\cdots,2023$. Prove that $a_{2023} \ge 3034$.

## Video Solution

https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]

## Solution

We solve for $a_{n+2}$ in terms of $a_n$ and $x.$ $a_{n+2}^2 \\ = (\sum^{n+2}_{k=1}x_k)(\sum^{n+2}_{k=1}\frac1{x_k}) \\ = (x_{n+1}+x_{n+2}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}+\sum^{n}_{k=1}\frac1{x_k}) \\ = \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} + \frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k}) \\ = \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} + \frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + a_n^2 \\ \text{}$

Again, by AM-GM, the above equation becomes $a_{n+2}^2 \ge 4 \sqrt[4]{(\frac{x_{n+1}}{x_{n+1}})(\frac{x_{n+1}}{x_{n+2}})(\frac{x_{n+2}}{x_{n+1}})(\frac{x_{n+2}}{x_{n+2}})} + 4\sqrt[4]{ (\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k)(x_{n+1}\sum^{n}_{k=1}\frac1{x_k})(\frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k)(x_{n+2}\sum^{n}_{k=1}\frac1{x_k}) } + a_n^2 = a_n^2+4a_n+4 = (a_n+2)^2 \\ \text{}$

Hence, $a_{n+2} \ge a_{n} + 2,$ but equality is achieved only when $\frac{x_{n+1}}{x_{n+1}},\frac{x_{n+1}}{x_{n+2}},\frac{x_{n+2}}{x_{n+1}},$ and $\frac{x_{n+2}}{x_{n+2}}$ are equal. They can never be equal because there are no two equal $x_k.$So $a_{2023} \ge a_1 + 3\times \frac{2023-1}{2} = 1 + 3033 = 3034$