<Question>
Find the minimum value of
\[\frac{(x + 5)(x + 2)}{x + 1}\]for $x > 0.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the minimum value of the function (x + 5)(x + 2)/(x + 1) for x > 0. Hmm, let's start by understanding what this function looks like. Maybe I should first simplify the expression or find its derivative to locate the critical points. Since it's a rational function, calculus might be the way to go here. But before jumping into derivatives, perhaps expanding the numerator could make things easier?

Let me try expanding the numerator: (x + 5)(x + 2) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10. So the function becomes (x^2 + 7x + 10)/(x + 1). Now, maybe I can perform polynomial division here to simplify it further. Let's divide x^2 + 7x + 10 by x + 1.

Dividing x^2 by x gives x, then multiplying x + 1 by x gives x^2 + x. Subtracting that from the original numerator: (x^2 + 7x + 10) - (x^2 + x) = 6x + 10. Now, divide 6x by x to get +6.

<prompt_1.1> Perform polynomial division on the function (x^2 + 7x + 10)/(x + 1) as per the above plan. </prompt_1.1>
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Alternatively, maybe there's an algebraic way to do this without calculus. Let's see. The original function is (x + 5)(x + 2)/(x + 1). Let me try to manipulate this expression. Let's set t = x + 1, since the denominator is x + 1. Then x = t - 1. Substitute into the numerator: ( (t - 1) + 5 )( (t - 1) + 2 ) = (t + 4)(t + 1).

<prompt_1.2> Use substitution t = x + 1 to rewrite and simplify the function as per the above plan. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Multiplying x + 1 by 6 gives 6x + 6. Subtracting that from 6x + 10 gives 4. So the division result is x + 6 with a remainder of 4. Therefore, the function can be written as x + 6 + 4/(x + 1). So, f(x) = x + 6 + 4/(x + 1).
</execution_1.1>

<execution_1.2>
So that's (t + 4)(t + 1) = t^2 + 5t + 4. Therefore, the function becomes (t^2 + 5t + 4)/t = t + 5 + 4/t. So the function is t + 4/t + 5, where t = x + 1. Since x > 0, t > 1. So now we have to minimize t + 4/t + 5 for t > 1.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, this is easier to analyze. To find the minimum, maybe I can take the derivative of f(x) with respect to x. Let's compute f'(x).

Using the quotient rule: if f(x) = numerator/denominator, then f'(x) = (num' * denom - num * denom') / denom^2. Let's denote numerator as N = x^2 + 7x + 10, so N' = 2x + 7. Denominator is D = x + 1, so D' = 1.

<prompt_2.1> Compute the derivative of the simplified function f(x) = x + 6 + 4/(x + 1) using the quotient rule as per the above plan. </prompt_2.1>
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Based on execution_1.2:
But perhaps this is easier. The expression t + 4/t is a classic form where we can apply the AM-GM inequality. Recall that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, t + 4/t ≥ 2√(t * 4/t) = 2√4 = 4. The minimum of t + 4/t is 4, achieved when t = 4/t, which implies t^2 = 4, so t = 2 (since t > 1, we take the positive root). Therefore, the minimum of t + 4/t is 4 at t = 2.

<prompt_2.2> Apply the AM-GM inequality to find the minimum of the function t + 4/t as per the above plan. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Therefore, f'(x) using the quotient rule would be [(2x + 7)(x + 1) - (x^2 + 7x + 10)(1)] / (x + 1)^2. Let's compute the numerator step by step.

First, expand (2x + 7)(x + 1): 2x*x + 2x*1 + 7*x + 7*1 = 2x^2 + 2x + 7x + 7 = 2x^2 + 9x + 7. Then subtract (x^2 + 7x + 10): 2x^2 + 9x + 7 - x^2 - 7x - 10 = (2x^2 - x^2) + (9x - 7x) + (7 - 10) = x^2 + 2x - 3. So the derivative is (x^2 + 2x - 3)/(x + 1)^2.
</execution_2.1>

<execution_2.2>
Therefore, the original function f(x) = t + 4/t + 5 would have a minimum value of 4 + 5 = 9 when t = 2. Since t = x + 1, x = t - 1 = 2 - 1 = 1. So, x = 1 gives the minimum value of 9.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
To find critical points, set the numerator equal to zero: x^2 + 2x - 3 = 0. Let's solve this quadratic equation.
The discriminant is 4 + 12 = 16, so roots are x = [-2 ± √16]/2 = [-2 ± 4]/2. That gives x = (-2 + 4)/2 = 1 and x = (-2 - 4)/2 = -3. Since we are considering x > 0, we discard x = -3. So the critical point is at x = 1.

Now, we need to check if this critical point is a minimum. We can use the second derivative test or check the sign changes of the first derivative around x = 1. Let's compute the second derivative.

<prompt_3.1> Use the second derivative test to determine if the critical point at x = 1 is a minimum. </prompt_3.1>
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Based on execution_2.2:
Wait, but let me confirm this with the derivative approach as well to ensure consistency. If we go back to the original derivative approach, we found critical point at x = 1. Let's check the second derivative at x = 1 or just test points around x = 1 to see if it's a minimum.

<prompt_3.2> Confirm the solution using the derivative approach around x = 1. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Alternatively, since the denominator (x + 1)^2 is always positive for x > 0, the sign of f'(x) is determined by the numerator x^2 + 2x - 3. Let's analyze the quadratic x^2 + 2x - 3. The quadratic opens upwards (since coefficient of x^2 is positive). The critical points are at x = 1 and x = -3. For x > 1, say x = 2, the quadratic is 4 + 4 - 3 = 5 > 0. Between x = -3 and x = 1, for x = 0, it's 0 + 0 - 3 = -3 < 0. Since we are looking at x > 0, between 0 and 1, the quadratic is negative, and for x > 1, it's positive. Therefore, the derivative changes from negative to positive as x passes through 1, indicating a minimum at x = 1.
</execution_3.1>

<execution_3.2>
But since we already have the algebraic confirmation via substitution and AM-GM, and the calculus approach both lead to x = 1, and the value is 9, I think that's solid. But just to be thorough, let's check the behavior of the function as x approaches 0+ and as x approaches infinity.

As x approaches 0 from the right, the function (x + 5)(x + 2)/(x + 1) becomes (5)(2)/(1) = 10. So near x = 0, the function is 10. At x = 1, the value is 9, which is lower. As x approaches infinity, the function behaves like x^2 / x = x, so it goes to infinity. Therefore, the function decreases from x = 0 to some point and then increases towards infinity. Since we have a critical point at x = 1, which is a minimum, that makes sense. So the minimum value is indeed 9.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, the minimum 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.
-----

Based on execution_3.2:
I think this is thorough. The answer is 9, achieved at x = 1.
I don't see any mistakes in the reasoning. Both calculus and algebraic methods confirm the same result, and the function's behavior at the boundaries supports that there's no lower value than 9.

**Final Answer**
The minimum value is \boxed{9}.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the minimum value of the given expression for $x > 0$ is $\boxed{9}$.
</Final_answer>
</think>