To solve the problem (assuming the common context: finding the minimum value of ((\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)) where (x,y,z>0) and (x+y+z=1)):

Step 1: Rewrite the expression

Since (x+y+z=1), we have:
[ \frac{1}{x}-1 = \frac{1-x}{x} = \frac{y+z}{x} ]
Similarly:
[ \frac{1}{y}-1 = \frac{x+z}{y}, \quad \frac{1}{z}-1 = \frac{x+y}{z} ]
The product becomes:
[ \left(\frac{y+z}{x}\right)\left(\frac{x+z}{y}\right)\left(\frac{x+y}{z}\right) = \frac{(y+z)(x+z)(x+y)}{xyz} ]

Step 2: Apply AM ≥ GM

For positive (a,b), (a+b \geq 2\sqrt{ab}):
[ y+z \geq 2\sqrt{yz}, \quad x+z \geq 2\sqrt{xz}, \quad x+y \geq 2\sqrt{xy} ]
Multiplying these:
[ (y+z)(x+z)(x+y) \geq 2\sqrt{yz} \cdot 2\sqrt{xz} \cdot 2\sqrt{xy} = 8xyz ]

Step 3: Equality condition

Equality holds when (x=y=z=\frac{1}{3}). Substituting:
[ \left(3-1\right)^3 = 2^3 = 8 ]

Answer: (\boxed{8})

作者声明:本文包含人工智能生成内容。