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Algebra Difficulty 4.7 AIME Prove it Greece

If x,y,zx, y, z are positive real numbers, prove that: (3x+y)(3y+z)(3z+x)64xyz(3x + y)(3y + z)(3z + x) \ge 64xyz. When does equality hold?

Solution

We use the inequality of arithmetic and geometric means for four positive terms:
a1+a2+a3+a44a1a2a3a44, \frac{a_1 + a_2 + a_3 + a_4}{4} \ge \sqrt[4]{a_1 a_2 a_3 a_4},
where equality holds when a1=a2=a3=a4a_1 = a_2 = a_3 = a_4.

Applying this to each factor:
(3x+y)4x3y4(3x + y) \ge 4\sqrt[4]{x^3 y}
(3y+z)4y3z4(3y + z) \ge 4\sqrt[4]{y^3 z}
(3z+x)4z3x4(3z + x) \ge 4\sqrt[4]{z^3 x}

Multiplying these inequalities:
(3x+y)(3y+z)(3z+x)4x3y44y3z44z3x4=64x4y4z44=64xyz.(3x + y)(3y + z)(3z + x) \ge 4\sqrt[4]{x^3 y} \cdot 4\sqrt[4]{y^3 z} \cdot 4\sqrt[4]{z^3 x} = 64\sqrt[4]{x^4 y^4 z^4} = 64xyz.

Equality holds when x=y=zx = y = z.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.