If `x,y in R^(+)` such that x + y = 8, then find the minimum value of `(1 + (1)/(x)) (1 + (1)/(y))`

Explanation

To find the minimum value of the expression \( (1 + \frac{1}{x})(1 + \frac{1}{y}) \) given that \( x + y = 8 \) and \( x, y \in \mathbb{R}^{+} \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ (1 + \frac{1}{x})(1 + \frac{1}{y}) \]
Expanding this, we have:
\[ = 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \]

### Step 2: Substitute \( y \)
Since we know \( x + y = 8 \), we can express \( y \) in terms of \( x \):
\[ y = 8 - x \]
Now, substituting \( y \) into the expression gives:
\[ = 1 + \frac{1}{x} + \frac{1}{8 - x} + \frac{1}{x(8 - x)} \]

### Step 3: Combine Terms
Combining the terms, we get:
\[ = 1 + \frac{8 - x + x}{x(8 - x)} + \frac{1}{x(8 - x)} = 1 + \frac{8}{x(8 - x)} \]

### Step 4: Simplify Further
Thus, we can rewrite our expression as:
\[ = 1 + \frac{8}{x(8 - x)} \]

### Step 5: Find the Maximum of \( xy \)
To minimize \( \frac{8}{xy} \), we need to maximize \( xy \). Using the AM-GM inequality:
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
Substituting \( x + y = 8 \):
\[ \frac{8}{2} \geq \sqrt{xy} \implies 4 \geq \sqrt{xy} \implies 16 \geq xy \]
The maximum value of \( xy \) occurs when \( x = y = 4 \).

### Step 6: Substitute Back to Find Minimum Value
Substituting \( xy = 16 \) back into our expression:
\[ = 1 + \frac{8}{16} = 1 + \frac{1}{2} = \frac{3}{2} \]

### Step 7: Final Expression
Thus, the minimum value of \( (1 + \frac{1}{x})(1 + \frac{1}{y}) \) is:
\[ \frac{3}{2} \]

### Conclusion
The minimum value of \( (1 + \frac{1}{x})(1 + \frac{1}{y}) \) given \( x + y = 8 \) is \( \frac{3}{2} \).

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