If a and b are two positive quantities whose sum is `lamda` then the minimum value of `sqrt((1+1/a)(1+1/b))` is

Options

A. `lamda-1/(lamda)`
B. `lamda-2/(lamda)`
C. `1+1/(lamda)`
D. `1+2/(lamda)`

Answer

To find the minimum value of \( \sqrt{(1 + \frac{1}{a})(1 + \frac{1}{b})} \) given that \( a + b = \lambda \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ x = \sqrt{(1 + \frac{1}{a})(1 + \frac{1}{b})} \]
Expanding this, we get:
\[ x = \sqrt{1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{ab}} \]

### Step 2: Substitute \( a + b = \lambda \)
Since \( a + b = \lambda \), we can express \( \frac{1}{a} + \frac{1}{b} \) as:
\[ \frac{1}{a} + \frac{1}{b} = \frac{b + a}{ab} = \frac{\lambda}{ab} \]
Thus, we can rewrite \( x \) as:
\[ x = \sqrt{1 + \frac{\lambda}{ab} + \frac{1}{ab}} \]
This simplifies to:
\[ x = \sqrt{1 + \frac{\lambda + 1}{ab}} \]

### Step 3: Maximize \( ab \)
To minimize \( x \), we need to maximize \( ab \). Using the AM-GM inequality:
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
This implies:
\[ \frac{\lambda}{2} \geq \sqrt{ab} \implies ab \leq \left(\frac{\lambda}{2}\right)^2 = \frac{\lambda^2}{4} \]

### Step 4: Substitute maximum \( ab \) into \( x \)
Substituting \( ab = \frac{\lambda^2}{4} \) into the expression for \( x \):
\[ x = \sqrt{1 + \frac{\lambda + 1}{\frac{\lambda^2}{4}}} \]
This simplifies to:
\[ x = \sqrt{1 + \frac{4(\lambda + 1)}{\lambda^2}} = \sqrt{1 + \frac{4\lambda + 4}{\lambda^2}} \]

### Step 5: Simplify the expression
Combining the terms under the square root:
\[ x = \sqrt{\frac{\lambda^2 + 4\lambda + 4}{\lambda^2}} = \sqrt{\frac{(\lambda + 2)^2}{\lambda^2}} = \frac{\lambda + 2}{\lambda} \]

### Step 6: Final result
Thus, the minimum value of \( x \) is:
\[ x_{\text{min}} = 1 + \frac{2}{\lambda} \]

### Conclusion
The minimum value of \( \sqrt{(1 + \frac{1}{a})(1 + \frac{1}{b})} \) given \( a + b = \lambda \) is:
\[ \boxed{1 + \frac{2}{\lambda}} \]