Baltic Way 2019

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 $1$.

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Alternative solution.

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