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 $0 \leq x, y, z \leq 1, (1-x), (1-y), (1-z) \geq 0$
Using A.M - G.M,
$(1-x) + (1-y) = z \geq 2 \sqrt{(1-x)(1-y)} \ $ (as $ \ x + y + z = 2$)
Similarly for $x, y$.
Multiplying them, we have
$xyz \geq 8 (1-x)(1-y)(1-z)$
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
We let $a=1-x,b=1-y,c=1-z$ so that we have $a+b+c=1$ and $0 < a,b,c < 1$.
Our new inequality now becomes $$8abc \leq (1-a)(1-b)(1-c)\\ \iff 8 \leq \left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right) \\ \iff 8 \leq \left(\dfrac{b+c}{a}\right)\left(\dfrac{c+a}{b}\right)\left(\dfrac{a+b}{c}\right) $$ which means $(a+b)(b+c)(c+a)\geq 8abc$ which is true by A.M-G.M inequality.
Equality occurs at $x=y=z=\dfrac{2}{3}$.
As for your approach, it is also correct. If you are substituting the value of $z$, making it a two variable inequality, you can form a quadratic equation taking everything to one side and set it's discriminant less than or equal to $0$.
