To solve the problem (assuming the common constraint (a+b+c=1) for positive reals (a,b,c) and finding the minimum of ((\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1))):

Step 1: Rewrite the expression

For (a+b+c=1):
[ \frac{1}{a}-1 = \frac{1-a}{a} = \frac{b+c}{a} ]
Similarly:
[ \frac{1}{b}-1=\frac{a+c}{b}, \quad \frac{1}{c}-1=\frac{a+b}{c} ]
The product becomes:
[ \frac{(b+c)(a+c)(a+b)}{abc} ]

Step 2: Apply AM ≥ GM

By AM ≥ GM:
[ b+c \geq 2\sqrt{bc}, \quad a+c \geq 2\sqrt{ac}, \quad a+b \geq 2\sqrt{ab} ]
Multiply these inequalities:
[ (b+c)(a+c)(a+b) \geq 8\sqrt{bc \cdot ac \cdot ab} = 8abc ]

Step 3: Simplify

[ \frac{(b+c)(a+c)(a+b)}{abc} \geq 8 ]
Equality holds when (a=b=c=\frac{1}{3}).

Answer: (\boxed{8})

作者声明:本文包含人工智能生成内容。