Maths Olympiad Prep

Library / /19 of 71

Algebra Difficulty 4.9 AIME Prove it United States

Problem:
Suppose that xx, yy, and zz are non-negative real numbers such that x+y+z=1x + y + z = 1. What is the maximum possible value of x+y2+z3x + y^{2} + z^{3}?

Solution

Solution:
Since 0y,z10 \leq y, z \leq 1, we have y2yy^{2} \leq y and z3zz^{3} \leq z. Therefore x+y2+z3x+y+z=1x + y^{2} + z^{3} \leq x + y + z = 1. We can get x+y2+z3=1x + y^{2} + z^{3} = 1 by setting (x,y,z)=(1,0,0)(x, y, z) = (1, 0, 0).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.