If `x,y,z` are positive real number, such that `x+y+z=1`, if the minimum value of `(1+1/x)(1+1/y)(1+1/z)` is `K^(2)`, then `|K|` is ……….

Explanation

To find the minimum value of the expression \((1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z})\) given that \(x + y + z = 1\) and \(x, y, z > 0\), we will use the AM-GM inequality.

### Step-by-step Solution:

1. **Rewriting the Expression**:
We start by rewriting the expression:
\[ (1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \left(1 + \frac{y + z}{yz}\right)\left(1 + \frac{x + z}{xz}\right)\left(1 + \frac{x + y}{xy}\right) \]

2. **Using AM-GM Inequality**:
According to the AM-GM inequality, for any positive real numbers \(a\), \(b\), and \(c\):
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
We apply this to the terms \(\frac{1}{x}\), \(\frac{1}{y}\), and \(\frac{1}{z}\):
\[ \frac{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{3} \geq \sqrt[3]{\frac{1}{xyz}} \]

3. **Finding the Minimum of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \)**:
Since \(x + y + z = 1\), we can express \(xyz\) in terms of \(x, y, z\). By AM-HM inequality:
\[ \frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]
Therefore:
\[ 1 \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 9 \]

4. **Calculating the Minimum Value**:
Now we can substitute back into our expression:
\[ (1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = (1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) \geq (1 + 3)(1 + 3)(1 + 3) = 4 \cdot 4 \cdot 4 = 64 \]

5. **Finding \(K\)**:
Given that the minimum value is \(K^2\), we have:
\[ K^2 = 64 \implies K = 8 \]

6. **Final Answer**:
Thus, the value of \(|K|\) is:
\[ |K| = 8 \]