1975 IMO Problems/Problem 1
Problem
Let be real numbers such that Prove that, if is any permutation of then
Solution
We can rewrite the summation as Since , the above inequality is equivalent to We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of . Obviously, if or , the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of s that are not ordered in the form such that is at a maximum out of all possible permutations and is greater than the sum . This necessarily means that in the sum there exists two terms and such that and . Notice that which means if we make the terms and instead of the original and , we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality is proved, which is equivalent to what we wanted to prove. ~Imajinary
See Also
1975 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |