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!

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:
$$\prod_{cyc}(x+y)\geq\prod_{cyc}2\sqrt{xy}=8xyz.$$
