Library / /1 of 17

, 2012

Algebra Difficulty 4.6 AIME Prove it Canada

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 $x^2 + xy^2 + xyz^2 \ge 4xyz - 4$ .

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
$x^2 + xy^2 + xyz^2 \ge (4x - 4) + x(4y - 4) + xy(4z - 4) = 4xyz - 4.$

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 and solution reproduced as published; topic and difficulty added by this site.