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?
 A: 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 $a=b=c=\frac{1}{3}$.
A: 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 $\displaystyle a=b=c = \frac{1}{3}$
A: Hint: As you know, since $a+b+c=1$, we have
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right) =  \left(\frac1a +\frac1b +\frac1c\right)-1$$
So, you need to find the minimum of $\ \ \ \frac1a +\frac1b +\frac1c\ \ \ $ when $a+b+c=1$. 
We know that the minimum of $\ \ \ \frac1a +\frac1b +\frac1c\ \ \ $ is obtained when $a =b =c=\frac13$. So, the answer of your question is $\ \ \ \color{red}{9-1=8}$
A: Let $f(a,b,c)=(1/a-1)(1/b-1)(1/c-1).$ Fix $a$ while letting $b,c$ vary with $b+c$ fixed at $1-a.$ Then $dc/db=-1,$ and   $df/db<0$ for $b<c,$ while $df/db>0$ for $b>c.$  So $f$ is not minimum when $b\ne c.$ Similarly $f$ is not minimum when $c\ne a,$ nor when  $a\ne b.$ So a minimum is only possible at $a=b=c=1/3.$ What remains is to prove that a minimum is achieved.
