1978 IMO Problems/Problem 5
Revision as of 16:05, 29 January 2021 by Hamstpan38825 (talk | contribs)
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 |