<Question>
Let $a,$ $b,$ and $c$ be positive real numbers.  Find the minimum value of
\[\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}.\]
</Question>

<think>
<Plan_1>
Okay, so I need to find the minimum value of this expression: \(\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}\), where \(a\), \(b\), and \(c\) are positive real numbers. Hmm, this looks like an optimization problem involving three variables. I remember that for such problems, techniques like the AM-GM inequality or maybe calculus (using partial derivatives) can be helpful. Let me think through this step by step.

First, let me try to understand the structure of the expression. The numerator is \((a + b + c)\) multiplied by the sum of two squares: \((a + b)^2\) and \((a + b + 4c)^2\). The denominator is the product \(abc\). Since all variables are positive, I don't have to worry about division by zero or negative numbers, which simplifies things a bit.

Maybe I can simplify the expression by substituting variables. Let me set \(x = a + b\) and \(y = c\). Then, the numerator becomes \((x + y)[x^2 + (x + 4y)^2]\), and the denominator is \(ab y\). But wait, \(ab\) is still in terms of \(a\) and \(b\), which sum to \(x\). Perhaps I can express \(ab\) in terms of \(x\)?

<prompt_1.1> Substitute variables to simplify the expression as per the above plan. </prompt_1.1>
-----

Alternatively, maybe I can use substitution variables from the start. Let me set t = a/b. Then a = tb.

<prompt_1.2> Use substitution variables to find the minimum value as per the above plan. </prompt_1.2>
-----

Alternatively, another approach might be to use the AM-GM inequality directly on the terms in the numerator. Let's see.

<prompt_1.3> Explore another approach using the AM-GM inequality as per the above plan. </prompt_1.3>

But just to ensure there's no oversight, let's consider another angle. Suppose we use the method of Lagrange multipliers. Let's set up the function:

<prompt_1.4> Verify the result using the method of Lagrange multipliers as per the above plan. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Hmm, the maximum of \(ab\) given \(a + b = x\) is when \(a = b = x/2\), right? Because for fixed sum, the product is maximized when the numbers are equal. So, the maximum \(ab\) is \((x/2)^2 = x^2/4\). But here, in the denominator, we have \(ab y\), so if I want to minimize the entire expression, since the numerator is fixed for given \(x\) and \(y\), I need to maximize the denominator \(ab y\). Therefore, to minimize the expression, I should maximize \(ab\) given \(a + b = x\), which as I said is \(x^2/4\). Therefore, substituting \(a = b = x/2\) would give the maximum \(ab\), leading to the minimal value of the original expression for fixed \(x\) and \(y\).

So maybe this substitution can help. Let me proceed with that.
</execution_1.1>

<execution_1.2>
Hmm, not sure if that helps. Alternatively, perhaps set variables such that we can apply AM-GM. Let me see.

Alternatively, since the expression is homogeneous in \(a\), \(b\), \(c\), maybe we can assume some normalization, like \(abc = 1\), to simplify. Wait, but since the expression is homogeneous of degree 3 in the numerator and degree 3 in the denominator (since \(abc\) is degree 3), so scaling variables would preserve the ratio. Therefore, we can set \(abc = k\), but maybe setting one variable to 1? Alternatively, set \(c = 1\) and express \(a\) and \(b\) in terms of \(c\). Wait, but since all variables are positive, maybe substituting variables as ratios. Let me try setting \(a = pc\) and \(b = qc\), where \(p\) and \(q\) are positive real numbers. Then, since \(c\) is positive, we can express everything in terms of \(p\) and \(q\).

So, substituting \(a = pc\), \(b = qc\), then \(a + b = (p + q)c\), and the numerator becomes:

\(( (p + q)c + c ) [ ( (p + q)c )^2 + ( (p + q)c + 4c )^2 ] = ( (p + q + 1)c ) [ ( (p + q)c )^2 + ( (p + q + 4)c )^2 ]\)

Factoring out \(c\) terms:

= \( (p + q + 1)c \times [ c^2 (p + q)^2 + c^2 (p + q + 4)^2 ] \)
= \( (p + q + 1)c \times c^2 [ (p + q)^2 + (p + q + 4)^2 ] \)
= \( (p + q + 1) c^3 [ (p + q)^2 + (p + q + 4)^2 ] \)

The denominator \(abc = pc \times qc \times c = pq c^3\).

So the entire expression becomes:

\[ \frac{(p + q + 1) c^3 [ (p + q)^2 + (p + q + 4)^2 ]}{pq c^3} = \frac{(p + q + 1)[ (p + q)^2 + (p + q + 4)^2 ]}{pq} \]

So now, the expression is independent of \(c\), which confirms the homogeneity. Therefore, we can set \(c = 1\) without loss of generality, and just consider \(a\) and \(b\) as \(p\) and \(q\).
</execution_1.2>

<execution_1.3>
The expression is \(\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}\). Let's denote \(S = a + b + c\), which is the sum of the variables. Then, the numerator is \(S[(x)^2 + (x + 4c)^2]\) where \(x = a + b\). Maybe expanding the numerator will help.

Expanding the squares:

\((a + b)^2 = x^2\)

\((a + b + 4c)^2 = x^2 + 8xc + 16c^2\)

So the sum of the squares is \(x^2 + x^2 + 8xc + 16c^2 = 2x^2 + 8xc + 16c^2\)

Therefore, the numerator becomes \(S(2x^2 + 8xc + 16c^2)\). Since \(S = x + c\), substituting that in:

\((x + c)(2x^2 + 8xc + 16c^2)\)

Let me expand this:

First, multiply \(x\) by each term in the quadratic:

\(x * 2x^2 = 2x^3\)

\(x * 8xc = 8x^2c\)

\(x * 16c^2 = 16xc^2\)

Then multiply \(c\) by each term:

\(c * 2x^2 = 2x^2c\)

\(c * 8xc = 8xc^2\)

\(c * 16c^2 = 16c^3\)

Adding all these up:

2x³ + 8x²c + 16xc² + 2x²c + 8xc² + 16c³

Combine like terms:

2x³ + (8x²c + 2x²c) + (16xc² + 8xc²) + 16c³

Which simplifies to:

2x³ + 10x²c + 24xc² + 16c³
</execution_1.3>

<execution_1.4>
\(f(a, b, c) = \frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}\)

We need to find the minimum of f(a, b, c) for \(a, b, c > 0\). Let's set up the Lagrangian:

\(\mathcal{L} = \frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc} + \lambda (abc - k)\)

Wait, actually, since we are not given a constraint other than positivity, maybe Lagrange multipliers aren't necessary here. Alternatively, we can consider taking partial derivatives with respect to a, b, c, set them to zero, and solve. However, that might be algebraically intensive. But given that we already found a critical point via substitution and AM-GM gives the same bound, it's likely that this is indeed the minimum.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me try that substitution.
So, \(x = a + b\), \(y = c\). Then, \(ab \leq (x/2)^2\) by AM-GM. So, if I substitute \(a = b = x/2\), then \(ab = x^2/4\), which is the maximum possible for given \(x\). Therefore, the original expression becomes:

<prompt_2.1> Simplify the expression using the substitution \(x = a + b\) and \(y = c\). </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
\[
\frac{(x + y)[x^2 + (x + 4y)^2]}{ (x/2)^2 y } = \frac{(x + y)[x^2 + x^2 + 8xy + 16y^2]}{ (x^2/4) y } = \frac{(x + y)(2x^2 + 8xy + 16y^2)}{ (x^2 y)/4 }
\]
Simplifying the denominator: \( (x^2 y)/4 \), so the entire expression becomes:
\[
\frac{4(x + y)(2x^2 + 8xy + 16y^2)}{x^2 y}
\]
Let me factor out the terms in the numerator. First, notice that \(2x^2 + 8xy + 16y^2 = 2(x^2 + 4xy + 8y^2)\). So:
\[
\frac{4(x + y) \cdot 2(x^2 + 4xy + 8y^2)}{x^2 y} = \frac{8(x + y)(x^2 + 4xy + 8y^2)}{x^2 y}
\]
Now, let's make a substitution here to reduce variables. Let me set \(t = x/y\), which is the ratio of \(x = a + b\) to \(y = c\). Since \(a\), \(b\), \(c\) are positive, \(t\) is positive as well. Then, \(x = ty\). Substituting into the expression:
\[
\frac{8(ty + y)( (ty)^2 + 4(ty)y + 8y^2 ) }{(ty)^2 y} = \frac{8y(t + 1) [ t^2 y^2 + 4t y^2 + 8y^2 ] }{t^2 y^3}
\]
Factor out \(y^2\) from the terms inside the brackets:
\[
= \frac{8y(t + 1) y^2 (t^2 + 4t + 8) }{t^2 y^3} = \frac{8(t + 1)(t^2 + 4t + 8)}{t^2}
\]
So now, the expression simplifies to:
\[
\frac{8(t + 1)(t^2 + 4t + 8)}{t^2}
\]
Now, we need to find the minimum of this expression with respect to \(t > 0\).
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me denote this function as \(f(t)\):
\[
f(t) = \frac{8(t + 1)(t^2 + 4t + 8)}{t^2}
\]
To find the minimum, we can take the derivative of \(f(t)\) with respect to \(t\), set it to zero, and solve for \(t\). Let's compute \(f'(t)\).

<prompt_3.1> Compute the derivative of the function \(f(t)\) and find critical points. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First, expand the numerator to make differentiation easier. Let's compute \((t + 1)(t^2 + 4t + 8)\):

Multiply \(t\) by each term: \(t \cdot t^2 = t^3\), \(t \cdot 4t = 4t^2\), \(t \cdot 8 = 8t\)

Multiply \(1\) by each term: \(1 \cdot t^2 = t^2\), \(1 \cdot 4t = 4t\), \(1 \cdot 8 = 8\)

Adding these up: \(t^3 + 4t^2 + 8t + t^2 + 4t + 8 = t^3 + 5t^2 + 12t + 8\)

So, \(f(t) = 8 \cdot \frac{t^3 + 5t^2 + 12t + 8}{t^2}\)

Simplify the fraction by dividing each term by \(t^2\):
\[
f(t) = 8 \left( t + 5 + \frac{12}{t} + \frac{8}{t^2} \right )
\]
Now, take the derivative \(f'(t)\):
\[
f'(t) = 8 \left( 1 - \frac{12}{t^2} - \frac{16}{t^3} \right )
\]
Set \(f'(t) = 0\):
\[
1 - \frac{12}{t^2} - \frac{16}{t^3} = 0
\]
Multiply both sides by \(t^3\) to eliminate denominators:
\[
t^3 - 12t - 16 = 0
\]
So, we need to solve the cubic equation \(t^3 - 12t - 16 = 0\).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let me try to find rational roots using the Rational Root Theorem. Possible rational roots are factors of 16 divided by factors of 1, so ±1, ±2, ±4, ±8, ±16. Let's test t=4:
4^3 -12*4 -16 = 64 -48 -16 = 0. Hey, t=4 is a root!

Therefore, we can factor out (t - 4) from the cubic. Let's perform polynomial division or use synthetic division.

<prompt_4.1> Factor the cubic equation and find all real roots. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Using synthetic division:

Divide t^3 -12t -16 by (t - 4). Coefficients: 1 (t^3), 0 (t^2), -12 (t), -16 (constant).

Bring down 1.

Multiply 1 by 4: 4. Add to next coefficient: 0 + 4 = 4.

Multiply 4 by 4: 16. Add to next coefficient: -12 +16 = 4.

Multiply 4 by 4: 16. Add to last coefficient: -16 +16 = 0. So, the cubic factors as (t - 4)(t^2 + 4t + 4) = 0. Therefore, roots are t = 4 and roots of t^2 +4t +4=0, which are t = -2 (double root). But since t > 0, the only valid root is t = 4.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Therefore, the critical point is at t = 4. Now, we need to check if this is a minimum. Let's compute the second derivative or check the sign changes of the first derivative.

<prompt_5.1> Verify if the critical point at \(t = 4\) is a minimum. </prompt_5.1>
-----

But let me verify if this substitution step was valid. Earlier, I substituted \(x = a + b\) and then set \(a = b = x/2\) to maximize \(ab\). Then, the problem reduced to a single variable \(t = x/y\). Then, after further substitution, we got to \(t = 4\). But let's check if the initial substitution is valid.

<prompt_5.2> Re-evaluate the substitution and confirm the minimum value. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Let's check values around t = 4. For t < 4, say t = 3:

f'(3) = 1 - 12/(9) - 16/(27) = 1 - 4/3 - 16/27 ≈ 1 - 1.333 - 0.592 ≈ -0.925 < 0

For t > 4, say t = 5:

f'(5) = 1 - 12/25 - 16/125 ≈ 1 - 0.48 - 0.128 ≈ 0.392 > 0

Therefore, the function decreases before t=4 and increases after t=4, so t=4 is indeed a minimum.
</execution_5.1>

<execution_5.2>
If we set \(a = b = x/2\), then \(x = a + b = 2*(x/2) = x\), which is consistent. Then, \(t = x/y = (a + b)/c\). So, if \(t = 4\), then \(x = 4y\), meaning \(a + b = 4c\). Since \(a = b = x/2 = 2y\), so \(a = b = 2y\), \(c = y\). Therefore, \(a = b = 2y\), \(c = y\).
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Therefore, the minimal value occurs when \(t = 4\), which corresponds to \(x = 4y\), or \(a + b = 4c\), with \(a = b = 2c\). Now, let's compute the minimal value by plugging \(t = 4\) into \(f(t)\):

<prompt_6.1> Calculate the minimum value of the expression at \(t = 4\). </prompt_6.1>
-----

Based on execution_5.2:
Let me verify this by plugging back into the original expression.

Compute the original expression:

\[
\frac{(a + b + c)[(a + b)^2 + (a + b + 4c)^2]}{abc}
\]

Substituting \(a = 2y\), \(b = 2y\), \(c = y\):

<prompt_6.2> Verify the minimum value by substituting back into the original expression. </prompt_6.2>
</Plan_6>

<Execution_6>
<execution_6.1>
\[
f(4) = 8 \cdot \frac{(4 + 1)(4^2 + 4 \cdot 4 + 8)}{4^2} = 8 \cdot \frac{5(16 + 16 + 8)}{16} = 8 \cdot \frac{5 \cdot 40}{16}
\]
Wait, wait, let me compute that step by step. Wait, no, actually, \(f(t) = \frac{8(t + 1)(t^2 + 4t + 8)}{t^2}\). So, substituting \(t = 4\):

First compute \(t + 1 = 5\)

Compute \(t^2 + 4t + 8 = 16 + 16 + 8 = 40\)

Therefore, \(f(4) = 8 * 5 * 40 / (4^2) = 8 * 5 * 40 / 16\). Let's compute this:

8 divided by 16 is 0.5, so 0.5 * 5 * 40 = 0.5 * 200 = 100. Therefore, the minimal value is 100.
</execution_6.1>

<execution_6.2>
Numerator:

\(a + b + c = 2y + 2y + y = 5y\)

\((a + b)^2 = (4y)^2 = 16y^2\)

\(a + b + 4c = 4y + 4y = 8y\), so \((a + b + 4c)^2 = 64y^2\)

Thus, the numerator becomes \(5y(16y^2 + 64y^2) = 5y * 80y^2 = 400y^3\)

Denominator: \(abc = 2y * 2y * y = 4y^3\)

Therefore, the expression is \(400y^3 / 4y^3 = 100\).
</execution_6.2>
</Execution_6>

<Plan_7>
Based on execution_6.1:
So, the minimal value is 100.
**Final Answer**
\boxed{100}
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the minimum value of the given expression is $\boxed{100}$.
</Final_answer>
</think>