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Difference between revisions of "2023 IMO Problems/Problem 4"

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==Solution==
 
==Solution==
 
We first solve for <math>a_1.</math> <math>a_1 = \sqrt{(x_1)(\frac{1}{x_1})} = \sqrt{1} = 1.</math>
 
We first solve for <math>a_1.</math> <math>a_1 = \sqrt{(x_1)(\frac{1}{x_1})} = \sqrt{1} = 1.</math>
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Now we solve for <math>a_{n+1}</math> in terms of <math>a_n</math> and <math>x.</math>
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Now we solve for <math>a_{n+1}</math> in terms of <math>a_n</math> and <math>x.</math> <math>a_{n+1} = \sqrt{\sum^{n+1}_{k=1}} = \sqrt{}</math>

Revision as of 23:51, 13 July 2023

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 $x.$ $a_{n+1} = \sqrt{\sum^{n+1}_{k=1}} = \sqrt{}$

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