MathDB
Floor and ceiling inequality with 2020

Source: Mexican Quarantine Mathematical Olympiad P1

April 25, 2020
ceiling functionfloor functioninequalities

Problem Statement

Let a,ba, b and cc be real numbers such that
⌈a⌉+⌈b⌉+⌈c⌉+⌊a+b⌋+⌊b+c⌋+⌊c+a⌋=2020\lceil a \rceil + \lceil b \rceil + \lceil c \rceil + \lfloor a + b \rfloor + \lfloor b + c \rfloor + \lfloor c + a \rfloor = 2020
Prove that
⌊a⌋+⌊b⌋+⌊c⌋+⌈a+b+c⌉≥1346\lfloor a \rfloor + \lfloor b \rfloor + \lfloor c \rfloor + \lceil a + b + c \rceil \ge 1346
Note: ⌊x⌋\lfloor x \rfloor is the greatest integer less than or equal to xx, and ⌈x⌉\lceil x \rceil is the smallest integer greater than or equal to xx. That is, ⌊x⌋\lfloor x \rfloor is the unique integer satisfying ⌊x⌋≤x<⌊x⌋+1\lfloor x \rfloor \le x < \lfloor x \rfloor + 1, and ⌈x⌉\lceil x \rceil is the unique integer satisfying ⌈x⌉−1<x≤⌈x⌉\lceil x \rceil - 1 < x \le \lceil x \rceil.
Proposed by Ariel García
Back to ProblemsView on AoPS