Difference between revisions of "1978 IMO Problems/Problem 5"
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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] | 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- | + | == See Also == {{IMO box|year=1978|num-b=4|num-a=6}} |
Latest revision as of 16:05, 29 January 2021
Problem
Let be an injective function from in itself. Prove that for any we have:
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 with .
From the rearrangement's inequality we know that .
We will denote we permutations of in this form .
So we have .
Let's denote and .
We have . Which comes from .
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 |