# 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)