2007 IMO Problems/Problem 1
Real numbers are given. For each () define
(a) Prove that, for any real numbers ,
(b) Show that there are real numbers such that equality holds in (*)
Since , all can be expressed as , where .
Thus, can be expressed as for some and ,
Assume for contradiction that , then for all ,
Then, is a non-decreasing function, which means, , and , which means, .
Then, and contradiction.
If , is the maximum of a set of non-negative number, which must be at least .
Case 2) (We can ignore because of lemma)
Using the fact that can be expressed as for some and , .
Assume for contradiction that .
Then, , .
Thus, and .
Subtracting the two inequality, we will obtain:
--- contradiction ().
A set of where the equality in (*) holds is:
Since is a non-decreasing function, is non-decreasing.
Let , .
Thus, ( because is the max of a set including )
Since and ,
This is written by Mo Lam--- who is a horrible proof writer, so please fix the proof for me. Thank you. O, also the formatting.
--> Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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