Difference between revisions of "2023 IMO Problems/Problem 4"
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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> <math>a_{n+1} = \ | + | Now we solve for <math>a_{n+1}</math> in terms of <math>a_n</math> and <math>x.</math> <math>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</math> |
| + | By AM-GM, <math>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 \\ </math> | ||
Revision as of 00:01, 14 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 
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)
