Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Romania

Let $x$ , $y$ , $z$ be positive real numbers satisfying $2x^2 + 3y^2 + 6z^2 + 12(x + y + z) = 108$ . Find the maximum value of $x^3y^2z$ .

Solution

Guessing that the maximum is obtained when $x = 3$ , $y = 2$ , $z = 1$ , we multiply the given equality by $6$ , and apply AM-GM:

$6 \cdot 108 = 4x^2 + 4x^2 + 4x^2 + 9y^2 + 9y^2 + 36z^2 + 12x + 12x + 12x + 12x + 12x + 12x + 18y + 18y + 18y + 18y + 36z + 36z \ge 18 \sqrt{2^{28} \cdot 3^{24} \cdot x^{12} \cdot y^8 \cdot z^4},$

from which it follows immediately that $108$ is the desired maximum and this value is obtained when all the numbers are equal, i.e. when $x = 3$ , $y = 2$ , $z = 1$ .

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