How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:

If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$

I tried many times, and finally I used Muirhead, but it failed!

$\begin{split}L.H.S.-R.H.S.&=8(1-a)(1-b)(1-c)-abc\\&=8-8(a+b+c) +8(ab+bc+ca)-9abc\\&=-(a^3+b^3+c^3)+(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)-3abc\\&=\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}a^3\Bigg)+\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}abc\Bigg)\end{split}$
But as $(3,0,0)$ majorizes $(2,1,0)$ but $(2,1,0)$ majorizes $(1,1,1)$, so it fails.

Could someone help? Any help is appreciated!
 A: By AM-GM you have
$$(1 - a)^{1\over 2}(1 - b)^{1 \over 2} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$(1 - b)^{1\over 2}(1 - c)^{1 \over 2} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$(1 - c)^{1\over 2}(1 - a)^{1 \over 2} \leq {1 - c + 1 - a \over 2} = {b \over 2}$$
Multiplying these three inequalities together gives the desired inequality.
A: We need to prove that
$$\prod_{cyc}(a+b-c)\leq abc$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is true by the following reasoning.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$
$$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0.$$
A: Also, we can use Muirhead here.
Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Thus, $$x=a+b+c-2a=2(1-a)>0.$$
Similarly, $y>0$ and $z>0$ and we need to prove that
$$8xyz\leq(x+y)(x+z)(y+z)$$ or
$$\sum_{cyc}(x^2y+x^2z-2xyz)\geq0,$$ which is true by Muirhead.
Also, we can use AM-GM:
$$\prod_{cyc}(x+y)\geq\prod_{cyc}2\sqrt{xy}=8xyz.$$
A: Denote: $1-a=x,1-b=y,1-c=z$. Then:
$$x+y+z=1 \stackrel{AM-GM}{\Rightarrow} \color{red}1=x+y+z\color{red}{\ge 3\sqrt[3]{xyz}}$$
Hence:
$$8(1-a)(1-b)(1-c)\le abc \Rightarrow 8xyz\le (1-x)(1-y)(1-z)\iff \\
8xyz\le 1-(x+y+z)+(xy+yz+zx)-xyz \iff \\
9xyz\le xy+yz+zx\iff \\
9xyz\le 3\sqrt[3]{(xyz)^2} \stackrel{AM-GM}{\le} xy+yz+zx\iff \\
\color{red}{3\sqrt[3]{xyz}\le 1}.$$
equality occurs for $x=y=z=\frac13$, consequently, for $a=b=c=\frac23$.
