2007 IMO Shortlist Problems/A1
Problem
(New Zealand) You are given a sequence of numbers. For each () define
and let
(a) Prove that for arbitrary real numbers ,
(b) Show that there exists a sequence of real numbers such that we have equality in (a).
Solution
Since , all can be expressed as , where . Thus, can be expressed as for some and , Lemma) Assume for contradiction that , then for all , Then, is a non-decreasing function, which means, , and , which means, . Then, and contradiction.
a) Case 1) 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, , . , and Thus, and . Subtracting the two inequality, we will obtain: --- contradiction (). Thus,
(b) 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 , .\ (Taken from https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_1)