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 solve the problem, we need 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 \(x, y, z\) are positive real numbers. We will denote this minimum value as \(K^2\) and find \(|K|\).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
The expression can be rewritten as:
\[ (1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \left(1 + \frac{y + z}{yz}\right)\left(1 + \frac{z + x}{zx}\right)\left(1 + \frac{x + y}{xy}\right) \]

2. **Apply AM-HM Inequality**:
By the Arithmetic Mean-Harmonic Mean (AM-HM) inequality, we have:
\[ \frac{1/x + 1/y + 1/z}{3} \geq \frac{3}{x + y + z} \]
Since \(x + y + z = 1\), this simplifies to:
\[ \frac{1/x + 1/y + 1/z}{3} \geq 3 \implies 1/x + 1/y + 1/z \geq 9 \]

3. **Add 3 to Each Term**:
Now, we add 1 to each of the fractions:
\[ (1 + \frac{1}{x}) + (1 + \frac{1}{y}) + (1 + \frac{1}{z}) = 3 + \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 3 + 9 = 12 \]

4. **Use AM-GM Inequality**:
By the AM-GM inequality:
\[ \frac{(1 + \frac{1}{x}) + (1 + \frac{1}{y}) + (1 + \frac{1}{z})}{3} \geq \sqrt[3]{(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z})} \]
Thus, we have:
\[ 4 \geq \sqrt[3]{(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z})} \]

5. **Cubing Both Sides**:
Cubing both sides gives:
\[ 64 \geq (1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) \]

6. **Finding Minimum Value**:
The minimum value of \((1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z})\) is 64. Therefore, we have:
\[ K^2 = 64 \implies K = 8 \]

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