Problem 1322

1. We start with the given inequality to prove:
$$x^4 + y^4 + z^2 \ge xyz \sqrt{8}$$
for all positive reals $x, y, z$.

2. We will use the well-known inequality for positive reals $a, b, c, d$:
$$a + b + c + d \ge 4 \sqrt[4]{abcd}$$
This is a form of the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

3. To apply this inequality, we rewrite $x^4 + y^4 + z^2$ by introducing two additional terms:
$$x^4 + y^4 + z^2 = x^4 + y^4 + \frac{z^2}{2} + \frac{z^2}{2}$$

4. Now, we apply the AM-GM inequality to the terms $x^4, y^4, \frac{z^2}{2}, \frac{z^2}{2}$:
$$x^4 + y^4 + \frac{z^2}{2} + \frac{z^2}{2} \ge 4 \sqrt[4]{x^4 \cdot y^4 \cdot \frac{z^2}{2} \cdot \frac{z^2}{2}}$$

5. Simplify the expression inside the fourth root:
$$x^4 \cdot y^4 \cdot \frac{z^2}{2} \cdot \frac{z^2}{2} = x^4 y^4 \left(\frac{z^2}{2}\right)^2 = x^4 y^4 \frac{z^4}{4}$$

6. Therefore, we have:
$$4 \sqrt[4]{x^4 y^4 \frac{z^4}{4}} = 4 \sqrt[4]{\frac{x^4 y^4 z^4}{4}}$$

7. Simplify the fourth root:
$$4 \sqrt[4]{\frac{x^4 y^4 z^4}{4}} = 4 \cdot \frac{\sqrt[4]{x^4 y^4 z^4}}{\sqrt[4]{4}} = 4 \cdot \frac{xyz}{\sqrt[4]{4}}$$

8. Since $\sqrt[4]{4} = \sqrt{2}$, we get:
$$4 \cdot \frac{xyz}{\sqrt{2}} = 4 \cdot \frac{xyz}{\sqrt{2}} = 4 \cdot \frac{xyz}{\sqrt{2}} = xyz \sqrt{8}$$

9. Therefore, we have shown that:
$$x^4 + y^4 + z^2 \ge xyz \sqrt{8}$$

$\blacksquare$