Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1978 IMO Problems/Problem 5"

(Created page with "yeet")
 
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
−
yeet
+
==Problem==
 +
Let <math>f</math> be an injective function from <math>{1,2,3,\ldots}</math> in itself. Prove that for any <math>n</math> we have: <math>\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.</math>
 +
 
 +
==Solution==
 +
We know that all the unknowns are integers, so the smallest one must greater or equal to 1.
 +
 
 +
Let me denote the permutations of <math>(k_1,k_2,...,k_n)</math> with <math>(y_1,y_2,...,y_n)=y_i (*)</math>.
 +
 
 +
From the rearrangement's inequality we know that <math>\text{Random Sum} \geq \text{Reversed Sum}</math>.
 +
 
 +
We will denote we permutations of <math>y_i</math> in this form <math>y_n \geq ...\geq y_1</math>.
 +
 
 +
So we have <math>\frac{k_1}{1^2}+\frac{k_2}{2^2}+...+\frac{k_n}{n^2} \geq \frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2} \geq 1+\frac{1}{2}+...+\frac{1}{n}</math>.
 +
 
 +
Let's denote <math>\frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2}=T</math> and <math>1+\frac{1}{2}+...+\frac{1}{n}=S</math>.
 +
 
 +
We have <math>T \geq S</math>. Which comes from <math>y_1 \geq1, y_2 \geq2, ...,y_n \geq n</math>.
 +
 
 +
So we are done.
 +
 
 +
The above solution was posted and copyrighted by Davron. The original thread for this problem can be found here: [https://aops.com/community/p509573]
 +
 
 +
== See Also == {{IMO box|year=1978|num-b=4|num-a=6}}

Latest revision as of 16:05, 29 January 2021

Problem

Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$

Solution

We know that all the unknowns are integers, so the smallest one must greater or equal to 1.

Let me denote the permutations of $(k_1,k_2,...,k_n)$ with $(y_1,y_2,...,y_n)=y_i (*)$.

From the rearrangement's inequality we know that $\text{Random Sum} \geq \text{Reversed Sum}$.

We will denote we permutations of $y_i$ in this form $y_n \geq ...\geq y_1$.

So we have $\frac{k_1}{1^2}+\frac{k_2}{2^2}+...+\frac{k_n}{n^2} \geq \frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2} \geq 1+\frac{1}{2}+...+\frac{1}{n}$.

Let's denote $\frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2}=T$ and $1+\frac{1}{2}+...+\frac{1}{n}=S$.

We have $T \geq S$. Which comes from $y_1 \geq1, y_2 \geq2, ...,y_n \geq n$.

So we are done.

The above solution was posted and copyrighted by Davron. The original thread for this problem can be found here: [1]

See Also

1978 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_5&oldid=143971"