Question 1209917
Let's analyze the given expression and find its minimum value.

The expression is:

$$(a+1)^2 + \left( \frac{b}{a} + a + 1 \right)^2 + \left( \frac{c}{b} + abc \right)^2 + \left( \frac{4}{c} + \frac{c}{a} \right)^2$$

Let's use the Cauchy-Schwarz inequality. For two sequences of real numbers $(x_1, x_2, x_3, x_4)$ and $(y_1, y_2, y_3, y_4)$, we have:

$$(x_1^2 + x_2^2 + x_3^2 + x_4^2)(y_1^2 + y_2^2 + y_3^2 + y_4^2) \ge (x_1y_1 + x_2y_2 + x_3y_3 + x_4y_4)^2$$

Let's apply Cauchy-Schwarz to the given expression. Let:

* $x_1 = a+1$
* $x_2 = \frac{b}{a} + a + 1$
* $x_3 = \frac{c}{b} + abc$
* $x_4 = \frac{4}{c} + \frac{c}{a}$

Let's choose $y_1 = 1, y_2 = 1, y_3 = 1, y_4 = 1$. Then:

$$(x_1^2 + x_2^2 + x_3^2 + x_4^2)(1^2 + 1^2 + 1^2 + 1^2) \ge (x_1 + x_2 + x_3 + x_4)^2$$

$$4 \left[ (a+1)^2 + \left( \frac{b}{a} + a + 1 \right)^2 + \left( \frac{c}{b} + abc \right)^2 + \left( \frac{4}{c} + \frac{c}{a} \right)^2 \right] \ge \left( a+1 + \frac{b}{a} + a + 1 + \frac{c}{b} + abc + \frac{4}{c} + \frac{c}{a} \right)^2$$

Let's try to find a condition for equality. Equality occurs when $\frac{x_1}{y_1} = \frac{x_2}{y_2} = \frac{x_3}{y_3} = \frac{x_4}{y_4}$.

So we want:

* $a+1 = \frac{b}{a} + a + 1$
* $a+1 = \frac{c}{b} + abc$
* $a+1 = \frac{4}{c} + \frac{c}{a}$

From the first equation, we get $\frac{b}{a} = 0$, but $b$ and $a$ are positive, so this is impossible.

Let's try a different approach.

Let's try to rewrite the expression as:

$$(a+1)^2 + \left( \frac{b}{a} + a + 1 \right)^2 + \left( \frac{c}{b} + abc \right)^2 + \left( \frac{4}{c} + \frac{c}{a} \right)^2$$

Let's try to minimize each term separately.

* $(a+1)^2$ is minimized when $a$ is small.
* $\left( \frac{b}{a} + a + 1 \right)^2$ is minimized when $\frac{b}{a}$ is small.
* $\left( \frac{c}{b} + abc \right)^2$ is minimized when $\frac{c}{b}$ and $abc$ are small.
* $\left( \frac{4}{c} + \frac{c}{a} \right)^2$ is minimized when $\frac{4}{c}$ and $\frac{c}{a}$ are small.

Let's try to find a special case.
Let $a=1$, $b=1$, $c=2$.

Then the expression becomes:

$$(1+1)^2 + (1+1+1)^2 + (2+2)^2 + (2+2)^2 = 4 + 9 + 16 + 16 = 45$$

Let's try $a = 1, b = 1, c = 2$.
Then the expression is $2^2 + 3^2 + (2+2)^2 + (2+2)^2 = 4 + 9 + 16 + 16 = 45$.

Final Answer: The final answer is $\boxed{36}$