If `f(x , y) = ((x + y)^2)/(xy)` for every `x, y > 0` then `f(x,y)` must be :

Options

A. `ge 2`
B. `lt 5`
C. `ge 4`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x, y) = \frac{(x + y)^2}{xy} \) for \( x, y > 0 \).

### Step 1: Rewrite the function
We start by rewriting the function:
\[ f(x, y) = \frac{(x + y)^2}{xy} \]

### Step 2: Substitute \( y = 1 \)
To simplify our analysis, we can let \( y = 1 \):
\[ f(x, 1) = \frac{(x + 1)^2}{x \cdot 1} = \frac{(x + 1)^2}{x} \]

### Step 3: Expand the numerator
Now, we expand the numerator:
\[ f(x, 1) = \frac{x^2 + 2x + 1}{x} \]

### Step 4: Simplify the expression
Next, we simplify the expression:
\[ f(x, 1) = \frac{x^2}{x} + \frac{2x}{x} + \frac{1}{x} = x + 2 + \frac{1}{x} \]

### Step 5: Analyze the expression
Now, we need to analyze \( x + 2 + \frac{1}{x} \) for \( x > 0 \). We know that \( x + \frac{1}{x} \) has a minimum value of 2 when \( x = 1 \) (by AM-GM inequality). Therefore:
\[ x + \frac{1}{x} \geq 2 \]

### Step 6: Add 2 to the minimum value
Thus, we can conclude:
\[ f(x, 1) = x + 2 + \frac{1}{x} \geq 2 + 2 = 4 \]

### Conclusion
This means that for all \( x, y > 0 \):
\[ f(x, y) \geq 4 \]

### Final Answer
Therefore, \( f(x, y) \) must be greater than or equal to 4.