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Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:

Show that (x+y+z)2/3≥xyz+yzx+zxy\left(x + y + z\right)^2 / 3 \geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy} for all non-negative reals xx, yy, zz.

Solution

Solution:

By AM/GM, xy+yz≥2xyzxy + yz \geq 2x\sqrt{yz}. Adding the similar results gives
2(xy+yz+zx)≥2(xyz+yzx+zxy). 2(xy + yz + zx) \geq 2\left(x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}\right).

By AM/GM, x2+x2+y2+z2≥4xyzx^2 + x^2 + y^2 + z^2 \geq 4x\sqrt{yz}. Adding the similar results gives
x2+y2+z2≥xyz+yzx+zxy. x^2 + y^2 + z^2 \geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}.

Adding the first result gives
(x+y+z)23≥xyz+yzx+zxy. \frac{(x + y + z)^2}{3} \geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}.

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