AM-GM for three numbers

For $x,y,z\ge0$ we have the identity
$$x^3+y^3+z^3-3xyz=\tfrac12(x+y+z)\left[(x-y)^2+(y-z)^2+(z-x)^2\right]\ge0.$$

Apply it to $x=a^{1/3}$, $y=b^{1/3}$, $z=c^{1/3}$. Then $xyz=(abc)^{1/3}=1$ and $x^3+y^3+z^3=a+b+c$, so $a+b+c\ge3$.

Equality holds exactly when $x=y=z$, that is $a=b=c$, and with $abc=1$ this gives $a=b=c=1$.