Difference between revisions of "1981 USAMO Problems/Problem 5"

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(Solution)
 
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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==
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==Solution(due to Pavel Zatitskiy)==
We know that <math>x\geq\lfloor x \rfloor</math>. Also, <math>nx\geq\lfloor nx \rfloor</math>, so <math>x\geq\frac{\lfloor nx \rfloor}{n}</math>. Thus, each of the terms in the sum is <math>x\geq</math>, so the total sum is <math>nx\geq\sum_{1}^{n}\left(\frac{[kx]}{k}\right)</math>.
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First of all we write <math>[kx]=kx-\{kx\}</math>. So, we need to prove that
<math>\blacksquare</math>
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<math>\sum_{1}^{n}\left(\frac{\{kx\}}{k}\right)\geq \{nx\}.</math>
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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
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<math>\sum_{1}^{n}\left(\frac{a_k}{k}\right)\geq a_n.</math>
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We will prove it by induction by <math>n</math>. The base is obvious, so we need to make a step.
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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
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<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 $x$, $[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)$

Solution(due to Pavel Zatitskiy)

First of all we write $[kx]=kx-\{kx\}$. So, we need to prove that $\sum_{1}^{n}\left(\frac{\{kx\}}{k}\right)\geq \{nx\}.$ Let's denote $a_k=\{kx\}$. It is easy to see that $a_k+a_m \geq a_{k+m}$. We need to prove $\sum_{1}^{n}\left(\frac{a_k}{k}\right)\geq a_n.$

We will prove it by induction by $n$. The base is obvious, so we need to make a step.

Let's take $m$ such that $\frac{a_m}{m}$ is minimal. If $m=n$ then our inequality is obvious. So, $m<n$. Then, by induction, $\sum_{1}^{n-m}\left(\frac{a_k}{k}\right)\geq a_{n-m}$ and $\sum_{n-m+1}^{n}\left(\frac{a_k}{k}\right)\geq m\cdot \frac{a_m}{m}=a_m$. Now we can add these two inequalities and get \[\sum_{1}^{n}\left(\frac{a_k}{k}\right) \geq a_{n-m}+a_m \geq a_n\]

See Also

1981 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Question
1 2 3 4 5
All USAMO Problems and Solutions

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