Difference between revisions of "1981 USAMO Problems/Problem 5"
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Show that for any positive real <math>x</math>, <math>[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)</math> | Show that for any positive real <math>x</math>, <math>[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)</math> | ||
− | ==Solution== | + | ==Solution(due to Pavel Zatitskiy)== |
− | + | First of all we write <math>[kx]=kx-\{kx\}</math>. So, we need to prove that | |
− | < | + | <math>\sum_{1}^{n}\left(\frac{\{kx\}}{k}\right)\geq \{nx\}.</math> |
+ | Let's denote <math>a_k=\{kx\}</math>. It is easy to see that <math>a_k+a_m \geq a_{k+m}</math>. We need to prove | ||
+ | <math>\sum_{1}^{n}\left(\frac{a_k}{k}\right)\geq a_n.</math> | ||
+ | |||
+ | We will prove it by induction by <math>n</math>. The base is obvious, so we need to make a step. | ||
+ | |||
+ | Let's take <math>m</math> such that <math>\frac{a_m}{m}</math> is minimal. If <math>m=n</math> then our inequality is obvious. So, <math>m<n</math>. Then, by induction, <math>\sum_{1}^{n-m}\left(\frac{a_k}{k}\right)\geq a_{n-m}</math> and <math>\sum_{n-m+1}^{n}\left(\frac{a_k}{k}\right)\geq m\cdot \frac{a_m}{m}=a_m</math>. Now we can add these two inequalities and get | ||
+ | <cmath>\sum_{1}^{n}\left(\frac{a_k}{k}\right) \geq a_{n-m}+a_m \geq a_n</cmath> | ||
==See Also== | ==See Also== |
Latest revision as of 22:07, 11 May 2021
Problem
Show that for any positive real ,
Solution(due to Pavel Zatitskiy)
First of all we write . So, we need to prove that Let's denote . It is easy to see that . We need to prove
We will prove it by induction by . The base is obvious, so we need to make a step.
Let's take such that is minimal. If then our inequality is obvious. So, . Then, by induction, and . Now we can add these two inequalities and get
See Also
1981 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.