2023 IMO Problems/Problem 4: Difference between revisions
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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> | ||
Now we solve for <math>a_{n+1}</math> in terms of <math>a_n</math> and <math>x.</math> | 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 
be pairwise different positive real numbers such that
![\[a_n = \sqrt{(x_1+x_2+···+x_n)(\frac1{x_1} + \frac1{x_2} +···+\frac1{x_n})}\]](http://latex.artofproblemsolving.com/9/e/d/9edc277a4c56ac586454738c4b2238a040695b47.png)
is an integer for every 
. Prove that 
.
Solution
We first solve for 
Now we solve for 
in terms of 
and 
