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

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==Problem==
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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>
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==Solution==
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We know that all the unknowns are integers, so the smallest one must greater or equal to 1.
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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>.
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From the rearrangement's inequality we know that <math>\text{Random Sum} \geq \text{Reversed Sum}</math>.
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We will denote we permutations of <math>y_i</math> in this form <math>y_n \geq ...\geq y_1</math>.
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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>.
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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>.
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We have <math>T \geq S</math>. Which comes from <math>y_1 \geq1, y_2 \geq2, ...,y_n \geq n</math>.
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So we are done.
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The above solution was posted and copyrighted by Davron. The original thread for this problem can be found here: [https://aops.com/community/p509573]
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== See Also == {{IMO box|year=1978|num-a=4|num-a=6}}

Revision as of 17:05, 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
[[1978 IMO Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions