If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-x-y$ and replace:
I get LHS = $x^2y+xy^2+2(x+y)-x^2-y^2-3xy-1$ but don't know how to proceed.
Thank you for your help.

as $1-x=\frac{1}{2}(y+z-x)$ and similarly for others we have to prove  $$(x+y-z)(y+z-x)(x+z-y)\le xyz$$ which is schur's inequality
