Track / Stage 4 / 78 of 340 #338 of 1964

Problem 338

AMC 12 late, AIME early

Algebra Difficulty 4.6 Prove it Kanada · Canada · 2012

Let $x$ , $y$ and $z$ be positive real numbers. Show that $x^2 + xy^2 + xyz^2 \ge 4xyz - 4$ .

Soit $x$ , $y$ et $z$ trois nombres réels positifs. Démontrez que $x^2 + xy^2 + xyz^2 \ge 4xyz - 4$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Note that
$x^2 \ge 4x - 4, \quad y^2 \ge 4y - 4, \quad \text{and} \quad z^2 \ge 4z - 4,$
and therefore
$x^2 + xy^2 + xyz^2 \ge (4x - 4) + x(4y - 4) + xy(4z - 4) = 4xyz - 4.$

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