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!

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 $a=b=c=\frac23$.
