All-Soviet-Union Mathematical Olympiad

Solution:

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

By AM/GM, $x^2 + x^2 + y^2 + z^2 \geq 4x\sqrt{yz}$. Adding the similar results gives
$$x^2 + y^2 + z^2 \geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}.$$

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