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? 

@Saaqib Mahmood, I'm expanding on @5xum's answer here. Using $AM\geq HM$,
$$$$${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\over{3}$$\geq$${3}\over{a+b+c}$
$$$$So, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$
