MathDB
solve 8xyz <= 1 and 8xyz = 1 if x^2 + y^2 + z^2 + 2xyz = 1$

Source: 2019 Irish Mathematical Olympiad paper 2 p9

October 5, 2020
algebrainequalities

Problem Statement

Suppose x,y,zx, y, z are real numbers such that x2+y2+z2+2xyz=1x^2 + y^2 + z^2 + 2xyz = 1. Prove that 8xyz≤18xyz \le 1, with equality if and only if (x,y,z)(x, y,z) is one of the following: (12,12,12),(−12,−12,12),(−12,12,−12),(12,−12,−12)\left( \frac12, \frac12, \frac12 \right) , \left( -\frac12, -\frac12, \frac12 \right), \left(- \frac12, \frac12, -\frac12 \right), \left( \frac12,- \frac12, - \frac12 \right)
Back to ProblemsView on AoPS