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!

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.
