Okay, so I need to find the minimum value of the function (x + 5)(x + 2)/(x + 1) where x is greater than 0. Hmm, let's start by understanding the problem. The function is a rational function, and since x is positive, the denominator x + 1 is always positive, so the function is well-defined. 

First, maybe I should simplify the expression. Let me expand the numerator: (x + 5)(x + 2) = x² + 7x + 10. So the function becomes (x² + 7x + 10)/(x + 1). Maybe I can perform polynomial division here to simplify it further. Let's divide x² + 7x + 10 by x + 1.

Dividing x² by x gives x. Multiply x + 1 by x to get x² + x. Subtract that from the numerator: (x² + 7x + 10) - (x² + x) = 6x + 10. Now divide 6x by x to get 6. Multiply x + 1 by 6 to get 6x + 6. Subtract that from the remainder: (6x + 10) - (6x + 6) = 4. So the division gives x + 6 with a remainder of 4. Therefore, the function simplifies to x + 6 + 4/(x + 1). 

So now the function is f(x) = x + 6 + 4/(x + 1). This looks easier to handle. Maybe taking the derivative now would help find the minimum. Since we need to find the minimum for x > 0, calculus seems like a good approach here.

Let's compute the derivative f'(x). The derivative of x is 1, the derivative of 6 is 0, and the derivative of 4/(x + 1) requires the chain rule. The derivative of 4/(x + 1) is -4/(x + 1)². So putting it all together, f'(x) = 1 - 4/(x + 1)².

To find the critical points, set f'(x) equal to zero: 1 - 4/(x + 1)² = 0. Solving for x:

1 = 4/(x + 1)²  
Multiply both sides by (x + 1)²:  
(x + 1)² = 4  
Take square roots:  
x + 1 = ±2  
But since x > 0, x + 1 must be positive, so x + 1 = 2  
Therefore, x = 1.

So the critical point is at x = 1. Now we need to check if this is a minimum. We can use the second derivative test. Let's compute the second derivative f''(x). The first derivative was f'(x) = 1 - 4/(x + 1)². The derivative of 1 is 0, and the derivative of -4/(x + 1)² is 8/(x + 1)³ (using the chain rule again: derivative of (x + 1)^(-2) is -2(x + 1)^(-3) * 1, so multiplying by -4 gives 8/(x + 1)^3). So f''(x) = 8/(x + 1)³.

Evaluate f''(1): 8/(1 + 1)³ = 8/(2)³ = 8/8 = 1. Since f''(1) is positive, the function is concave up at x = 1, which means this critical point is a local minimum. 

Therefore, the minimum value occurs at x = 1. Let's compute f(1): (1 + 5)(1 + 2)/(1 + 1) = (6)(3)/2 = 18/2 = 9. So the minimum value is 9.

Wait a second, let me double-check my calculations. When I simplified the original function, I got x + 6 + 4/(x + 1). Plugging x = 1 into that gives 1 + 6 + 4/2 = 7 + 2 = 9. Yep, same result. 

Is there another way to verify this? Maybe using the AM-GM inequality? Let me see. The original function is (x + 5)(x + 2)/(x + 1). Let me try to express this in terms that might allow AM-GM. Let's set t = x + 1, so x = t - 1. Then the function becomes:

[(t - 1) + 5][(t - 1) + 2]/t = (t + 4)(t + 1)/t. Wait, that's (t² + 5t + 4)/t = t + 5 + 4/t. So f(t) = t + 5 + 4/t. Now this is a function of t, where t = x + 1 > 1 (since x > 0). 

Now, to minimize t + 5 + 4/t. Let's set g(t) = t + 4/t + 5. The minimum of g(t) can be found by taking its derivative: g'(t) = 1 - 4/t². Setting to zero: 1 - 4/t² = 0 → t² = 4 → t = 2 (since t > 1). Then g(2) = 2 + 4/2 + 5 = 2 + 2 + 5 = 9. So the minimum is 9 when t = 2, which corresponds to x = t - 1 = 1. 

So both methods give the same result, which is reassuring. Therefore, the minimum value is indeed 9. 

Just to ensure there are no mistakes, let's check the derivative approach again. The original function was f(x) = (x + 5)(x + 2)/(x + 1). The derivative, using the quotient rule: [(x + 1)(2x + 7) - (x + 5)(x + 2)]/(x + 1)².

Let me compute the numerator:

First term: (x + 1)(2x + 7) = 2x² + 7x + 2x + 7 = 2x² + 9x + 7  
Second term: (x + 5)(x + 2) = x² + 2x + 5x + 10 = x² + 7x + 10  
Subtracting the second term from the first: (2x² + 9x + 7) - (x² + 7x + 10) = x² + 2x - 3

So f'(x) = (x² + 2x - 3)/(x + 1)². Setting numerator equal to zero: x² + 2x - 3 = 0. Solving quadratic equation: x = [-2 ± sqrt(4 + 12)]/2 = [-2 ± sqrt(16)]/2 = [-2 ± 4]/2. So x = (2)/2 = 1 or x = (-6)/2 = -3. Since x > 0, x = 1 is the critical point. Therefore, same result. 

So all methods confirm that the minimum occurs at x = 1 with value 9. Seems solid. 

Another check: plug in values near x = 1 to see if the function is indeed minimized there. Let's try x = 0.5: (0.5 + 5)(0.5 + 2)/(0.5 + 1) = (5.5)(2.5)/1.5 = 13.75/1.5 ≈ 9.17, which is higher than 9. x = 1.5: (1.5 + 5)(1.5 + 2)/(1.5 + 1) = (6.5)(3.5)/2.5 = 22.75/2.5 = 9.1, still higher. x = 2: (7)(4)/3 ≈ 28/3 ≈ 9.33. So yes, 9 is indeed the minimum. 

Therefore, the answer is 9.
Final Answer: The minimum value is \boxed{9}.
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