2023 IMO Problems/Problem 4

Problem

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

Solution

We first solve for $a_1.$ $a_1 = \sqrt{(x_1)(\frac{1}{x_1})} = \sqrt{1} = 1.$ Now we solve for $a_{n+1}$ in terms of $a_n$ and $x.$ $a_{n+1}^2\\ = (\sum^{n+1}_{k=1}x_k)(\sum^{n+1}_{k=1}\frac1{x_k}) \\= (x_{n+1}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\sum^{n}_{k=1}\frac1{x_k}) \\= 1 + \frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k})\\=1+\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k}+a_n^2$ By AM-GM, $a_{n+1}^2 \ge 1+a_n^2 + 2 \sqrt{(\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k)(x_{n+1} \sum^{n}_{k=1} \frac1{x_k})} = 1 + a_n^2 + 2a_n = (a_n+1)^2 \$ (Error compiling LaTeX. Unknown error_msg)

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2023 IMO Problems/Problem 4

Problem

Solution