If `x gt 0, y gt 0` and `xy=5`, then the minimum value of `x+y` is …………… .

Answer

To find the minimum value of \(x + y\) given that \(xy = 5\) and \(x > 0\), \(y > 0\), we can follow these steps:

### Step 1: Express \(y\) in terms of \(x\)
Given the equation \(xy = 5\), we can express \(y\) as:
\[ y = \frac{5}{x} \]

### Step 2: Write the function \(f(x)\)
We want to minimize the function \(f(x) = x + y\). Substituting \(y\) from the previous step, we have:
\[ f(x) = x + \frac{5}{x} \]

### Step 3: Differentiate \(f(x)\)
To find the minimum value, we need to differentiate \(f(x)\) with respect to \(x\):
\[ f'(x) = 1 - \frac{5}{x^2} \]

### Step 4: Set the derivative to zero
To find critical points, we set the derivative equal to zero:
\[ 1 - \frac{5}{x^2} = 0 \]
This simplifies to:
\[ \frac{5}{x^2} = 1 \implies x^2 = 5 \implies x = \sqrt{5} \]
(Note: We only consider the positive root since \(x > 0\).)

### Step 5: Verify if it's a minimum
To confirm that this critical point is a minimum, we can use the second derivative test. We differentiate \(f'(x)\):
\[ f''(x) = \frac{10}{x^3} \]
Since \(f''(x) > 0\) for \(x > 0\), this indicates that \(f(x)\) is concave up at \(x = \sqrt{5}\), confirming a local minimum.

### Step 6: Find the corresponding \(y\) value
Now, we substitute \(x = \sqrt{5}\) back into the equation for \(y\):
\[ y = \frac{5}{\sqrt{5}} = \sqrt{5} \]

### Step 7: Calculate the minimum value of \(x + y\)
Now we can find the minimum value of \(x + y\):
\[ x + y = \sqrt{5} + \sqrt{5} = 2\sqrt{5} \]

### Final Answer
Thus, the minimum value of \(x + y\) is:
\[ \boxed{2\sqrt{5}} \]

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