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Algebra Difficulty 2.1 Junior Prove it Turkey

Let xx, yy, zz be positive real numbers and x≤1x \le 1. Show that
xy+y+2z≥4xyz 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≥4xy2z24 xy + y + 2z = xy + y + z + z \geq 4\sqrt[4]{xy^2z^2}
Since 0<x≤10 < x \leq 1 we have x≥x2x \geq x^2 and hence
4xy2z24≥4x2y2z24=4xyz 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≤10 < x \leq 1 we have y≥xyy \geq xy and hence
xy+y+2z≥2xy+2z≥4xyz 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.