Difference between revisions of "2023 IMO Problems/Problem 4"
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==Problem== | ==Problem== | ||
Let <math>x_1, x_2, \cdots , x_{2023}</math> be pairwise different positive real numbers such that | Let <math>x_1, x_2, \cdots , x_{2023}</math> be pairwise different positive real numbers such that | ||
− | <cmath>a_n = \sqrt{(x_1+x_2+···+x_n)(\frac1{x_1} + \frac1{x_2} +···+\frac1{x_n})}</cmath> | + | <cmath>a_n = \sqrt{(x_1+x_2+ \text{···} +x_n)(\frac1{x_1} + \frac1{x_2} + \text{···} +\frac1{x_n})}</cmath> |
is an integer for every <math>n = 1,2,\cdots,2023</math>. Prove that <math>a_{2023} \ge 3034</math>. | is an integer for every <math>n = 1,2,\cdots,2023</math>. Prove that <math>a_{2023} \ge 3034</math>. | ||
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==Solution== | ==Solution== | ||
− | We | + | We solve for <math>a_{n+2}</math> in terms of <math>a_n</math> and <math>x.</math> |
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<math>a_{n+2}^2 \\ | <math>a_{n+2}^2 \\ | ||
= (\sum^{n+2}_{k=1}x_k)(\sum^{n+2}_{k=1}\frac1{x_k}) \\ | = (\sum^{n+2}_{k=1}x_k)(\sum^{n+2}_{k=1}\frac1{x_k}) \\ | ||
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Hence, <math>a_{n+2} \ge a_{n} + 2,</math> but equality is achieved only when <math>\frac{x_{n+1}}{x_{n+1}},\frac{x_{n+1}}{x_{n+2}},\frac{x_{n+2}}{x_{n+1}}, </math> and <math>\frac{x_{n+2}}{x_{n+2}}</math> are equal. They can never be equal because there are no two equal <math>x_k.</math>So <math>a_{2023} \ge a_1 + 3\times \frac{2023-1}{2} = 1 + 3033 = 3034</math> | Hence, <math>a_{n+2} \ge a_{n} + 2,</math> but equality is achieved only when <math>\frac{x_{n+1}}{x_{n+1}},\frac{x_{n+1}}{x_{n+2}},\frac{x_{n+2}}{x_{n+1}}, </math> and <math>\frac{x_{n+2}}{x_{n+2}}</math> are equal. They can never be equal because there are no two equal <math>x_k.</math>So <math>a_{2023} \ge a_1 + 3\times \frac{2023-1}{2} = 1 + 3033 = 3034</math> | ||
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+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2023|num-b=3|num-a=5}} |
Latest revision as of 21:29, 17 January 2024
Contents
Problem
Let be pairwise different positive real numbers such that is an integer for every . Prove that .
Video Solution
https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]
Solution
We solve for in terms of and
Again, by AM-GM, the above equation becomes
Hence, but equality is achieved only when and are equal. They can never be equal because there are no two equal So
See Also
2023 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |