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Algebra Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

Prove that for positive real numbers x,y,zx, y, z,
xyz(x+y+z+x2+y2+z2)(x2+y2+z2)(xy+yz+zx)≤3+39. \frac{xyz(x + y + z + \sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \le \frac{3 + \sqrt{3}}{9}.

Solution

By the QM-AM inequality, we have
x+y+z≤3(x2+y2+z2). x + y + z \le \sqrt{3(x^2 + y^2 + z^2)}.
This implies
xyz(x+y+z+x2+y2+z2)(x2+y2+z2)(xy+yz+zx)≤(3+1)xyzx2+y2+z2(x2+y2+z2)(xy+yz+zx)=(3+1)xyzx2+y2+z2⋅(xy+yz+zx). \frac{xyz(x + y + z + \sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \le \frac{(\sqrt{3} + 1)xyz\sqrt{x^2 + y^2 + z^2}}{(x^2 + y^2 + z^2)(xy + yz + zx)} = \frac{(\sqrt{3} + 1)xyz}{\sqrt{x^2 + y^2 + z^2} \cdot (xy + yz + zx)}.
Next, by the AM-GM inequality, we have
x2+y2+z2≥3x2y2z23,xy+yz+zx≥3x2y2z23. x^2 + y^2 + z^2 \ge 3\sqrt[3]{x^2y^2z^2}, \quad xy + yz + zx \ge 3\sqrt[3]{x^2y^2z^2}.
It follows that
xyz(x+y+z+x2+y2+z2)(x2+y2+z2)(xy+yz+zx)≤(3+1)xyzxyz3⋅3x2y2z2=3+39. \frac{xyz(x + y + z + \sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \le \frac{(\sqrt{3} + 1)xyz}{\sqrt[3]{xyz} \cdot 3\sqrt{x^2y^2z^2}} = \frac{3 + \sqrt{3}}{9}.

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