AM-GM for three numbers

Identity. We use the factorization
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
The second factor equals $\frac12[(x-y)^2+(y-z)^2+(z-x)^2]\ge0$, so
$$x^3+y^3+z^3\ge3xyz,$$
with equality iff $x=y=z$.

Substitution. Set $x=\sqrt[3]{a}$, $y=\sqrt[3]{b}$, $z=\sqrt[3]{c}$. Then $x^3=a$ etc., so
$$a+b+c\ge3\sqrt[3]{abc},$$
i.e. $\sqrt[3]{abc}\le\frac{a+b+c}{3}$.

Equality case. Equality holds iff $x=y=z$, i.e. $a=b=c$.