Simple inequality over positive reals: $2(x+y+z) \geq 3xyz + xy+yz+zx$ for $xyz=1$

with your Substitution we get $$a(a-b)(b-c)+b(a-c)(b-c)+c(a-c)(a-b)\geq 0$$ this is true for $$a\geq b\geq c$$

It is false. Consider $x=\frac1{20},$ $y=4,$ and $z=5.$

If $x=y=4$ and $z=1/16$, then $xyz=1$, and $$2(x+y+z)-(3xyz+xy+yz+zx)=-27/8<0.$$