If `x+y+z=1`, then the least value of `(1)/(x)+(1)/(y)+(1)/(z)`, is

Options

A. 3
B. 9
C. 27
D. 1

Explanation

To find the least value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) given that \( x + y + z = 1 \), we can use the concept of inequalities, specifically the relationship between the arithmetic mean and harmonic mean.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to minimize the expression \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) under the constraint \( x + y + z = 1 \).

2. **Using the Arithmetic Mean-Harmonic Mean Inequality (AM-HM Inequality)**: According to the AM-HM inequality, for any positive numbers \( a, b, c \):
\[ \text{AM} \geq \text{HM} \]
where
\[ \text{AM} = \frac{a + b + c}{3} \quad \text{and} \quad \text{HM} = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \]

3. **Applying AM-HM to Our Expression**: Let \( a = x, b = y, c = z \). Then we have:
\[ \frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]

4. **Substituting the Constraint**: Since \( x + y + z = 1 \):
\[ \frac{1}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]

5. **Rearranging the Inequality**: Cross-multiplying gives:
\[ \frac{1}{3} \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3 \]
Thus,
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 9 \]

6. **Conclusion**: The least value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) is \( 9 \).

### Final Answer:
The least value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) is \( 9 \).

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