<s>[INST] 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? 
 [/INST] You can rewrite $(1-a)(1-b)(1-c)$ as $abc(\frac1a +\frac1b + \frac1c -1)$, meaning that your inequality is equivalent to proving that $$\frac 1a + \frac 1b + \frac 1c \geq 9,$$
an inequality that is simpler to show.
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