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 $0&lt;a,b,c&lt;1$ 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!  $\begin{split}L.H.S.-R.H.S.&amp;=8(1-a)(1-b)(1-c)-abc\\&amp;=8-8(a+b+c) +8(ab+bc+ca)-9abc\\&amp;=-(a^3+b^3+c^3)+(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)-3abc\\&amp;=\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}a^3\Bigg)+\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}abc\Bigg)\end{split}$ 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!
How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$? Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$  My effort:   Since $a+b+c =1$, we can write $$  (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1  +(ab+bc+ca) - abc =  (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?
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