<s>[INST] 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?
 [/INST] Since $a+b+c=1$, Re-write all $1$s to $a+b+c$, then
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\left(\frac{a+b+c}{a}-1\right)\left(\frac{a+b+c}{b}-1\right)\left(\frac{a+b+c}{c}-1\right)=\left(\frac{b+c}{a}\right)\left(\frac{a+c}{b}\right)\left(\frac{a+b}{c}\right)$$
Using AM-GM inequality, 
$$b+c\geq2\sqrt{bc}$$
$$a+c\geq2\sqrt{ac}$$
$$a+b\geq2\sqrt{ab}$$
For these equations, equality holds when $a=b=c=1/3$.
So we get
$$\left(\frac{b+c}{a}\right)\left(\frac{a+c}{b}\right)\left(\frac{a+b}{c}\right)\geq\frac{2\sqrt{bc}}{a}\frac{2\sqrt{ac}}{b}\frac{2\sqrt{ab}}{c}=8\frac{abc}{abc}=8$$
Thus the minimum is $8$ when $a=b=c=\frac{1}{3}$.
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