Let x and y be two positive real numbers such that x + y = 1. Then the minimum value of `1/x+1/y` is-

Options

A. 2
B. `5/2`
C. 3
D. 4

Answer

To find the minimum value of \( \frac{1}{x} + \frac{1}{y} \) given that \( x + y = 1 \) and both \( x \) and \( y \) are positive real numbers, we can use the concept of the Arithmetic Mean-Harmonic Mean (AM-HM) inequality.

### Step-by-step Solution:

1. **Given Condition**: We know that \( x + y = 1 \).

2. **Expressing the Function**: We want to minimize the expression \( \frac{1}{x} + \frac{1}{y} \).

3. **Using AM-HM Inequality**: According to the AM-HM inequality, for any two positive numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \frac{2}{\frac{1}{a} + \frac{1}{b}} \]
Here, let \( a = x \) and \( b = y \). Therefore, we have:
\[ \frac{x + y}{2} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}} \]

4. **Substituting the Given Condition**: Since \( x + y = 1 \), we can substitute:
\[ \frac{1}{2} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}} \]

5. **Cross Multiplying**: Cross-multiplying gives us:
\[ \frac{1}{2} \left( \frac{1}{x} + \frac{1}{y} \right) \geq 2 \]

6. **Rearranging the Inequality**: Rearranging the inequality leads to:
\[ \frac{1}{x} + \frac{1}{y} \geq 4 \]

7. **Conclusion**: Thus, the minimum value of \( \frac{1}{x} + \frac{1}{y} \) is \( 4 \).

8. **Equality Condition**: The equality holds when \( x = y \). Since \( x + y = 1 \), this occurs when \( x = y = \frac{1}{2} \).

### Final Answer:
The minimum value of \( \frac{1}{x} + \frac{1}{y} \) is \( 4 \).

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