What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$? The primary question was:
What is the minimum value for $(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
$\color{red}{\text{But sorry guys! I messed it up! my question is:}}$ (Edited)

What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?

I feel like this needs to deal with AM-GM inequality but I don't know how to apply that onto this case. It is easy but like I still don't know how to do it! Can someone help me with it?

Given $$ \frac{(1-a)(1-b)(1-c)}{abc}$$
Now Using $a+b+c=1\;,$ We get
$$\frac{(b+c)(c+a)(a+b)}{abc} = \left[\left(\frac{b}{a}+\frac{c}{a}\right)\cdot \left(\frac{c}{b}+\frac{a}{b}\right)\cdot \left(\frac{b}{c}+\frac{a}{c}\right) \right]$$
Now Using $\bf{A.M\geq G.M},$ We get $$\left(\frac{b}{a}+\frac{c}{a}\right)\cdot \left(\frac{c}{b}+\frac{a}{b}\right)\cdot \left(\frac{b}{c}+\frac{a}{c}\right)\geq 2\sqrt{\frac{bc}{a^2}}\cdot 2\sqrt{\frac{ca}{b^2}} \cdot 2\sqrt{\frac{ab}{c^2}} = 8$$
and equality hold when $\displaystyle a=b=c = \frac{1}{3}$
