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Algebra Difficulty 4.8 AIME Prove it Canada

Problem:
Let xx, yy and zz be positive real numbers. Show that x2+xy2+xyz2≥4xyz−4x^{2} + x y^{2} + x y z^{2} \geq 4 x y z - 4.

Solution

Solution:
Note that
x2≥4x−4,y2≥4y−4,andz2≥4z−4 x^{2} \geq 4x - 4, \quad y^{2} \geq 4y - 4, \quad \text{and} \quad z^{2} \geq 4z - 4
and therefore
x2+xy2+xyz2≥(4x−4)+x(4y−4)+xy(4z−4)=4xyz−4. 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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