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 |