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Junior Turkish Mathematical Olympiad

Algebra Difficulty 2.1 Junior Prove it Turkey

Let $x$ , $y$ , $z$ be positive real numbers and $x \le 1$ . Show that
$xy + y + 2z \ge 4\sqrt{xyz}$

Solutions — 2

Solution 1

By applying AM-GM inequality we get
$xy + y + 2z = xy + y + z + z \geq 4\sqrt[4]{xy^2z^2}$
Since $0 < x \leq 1$ we have $x \geq x^2$ and hence
$4\sqrt[4]{xy^2z^2} \geq 4\sqrt[4]{x^2y^2z^2} = 4\sqrt{xyz}$
and we are done.

Solution 2

Since $0 < x \leq 1$ we have $y \geq xy$ and hence
$xy + y + 2z \geq 2xy + 2z \geq 4\sqrt{xyz}$
by AM-GM inequality.

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