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Sun Life Financial Canadian Mathematical Olympiad

Algebra Difficulty 4.8 AIME Prove it Canada

Problem:
Let $x$ , $y$ and $z$ be positive real numbers. Show that $x^{2} + x y^{2} + x y z^{2} \geq 4 x y z - 4$ .

Solution

Solution:
Note that
$x^{2} \geq 4x - 4, \quad y^{2} \geq 4y - 4, \quad \text{and} \quad z^{2} \geq 4z - 4$
and therefore
$x^{2} + x y^{2} + x y z^{2} \geq (4x - 4) + x(4y - 4) + x y (4z - 4) = 4 x y z - 4.$

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