Balti Tee 2019 · Ülesanne 1

From AM-GM inequality we have $\sqrt{xyz} \le \frac{xy+z}{2}$. Hence it suffices to prove that

$$\frac{x^3 - y^3 + z^3 + 1}{3} \ge (x - y)(xy + z),$$

which is equivalent to

$$\frac{(x - y)^3 + z^3 + 1}{3} \ge z(x - y).$$

This inequality follows directly from AM-GM inequality applied to the numbers $(x-y)^3$, $z^3$ and $1$.

Notice that

$$\frac{x^3 - y^3 + z^3 + 1}{6} = \frac{(x - y)^3 + xy(x - y) + xy(x - y) + xy(x - y) + z^3 + 1}{6}.$$

Applying AM-GM inequality this yields

$$\frac{x^3 - y^3 + z^3 + 1}{6} \ge \sqrt[6]{(x-y)^3 \cdot (xy(x-y))^3} \cdot z^3 \cdot 1 = (x-y)\sqrt{xyz}.$$