Key Concepts and Formulas

1. Arithmetic Progression (AP): Three numbers $a, b, c$ are in AP if the difference between consecutive terms is constant, which means $b$ is the arithmetic mean of $a$ and $c$. Mathematically, this is expressed as $2b = a+c$.
2. Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For any set of $n$ non-negative numbers $x_1, x_2, \dots, x_n$, the arithmetic mean is greater than or equal to the geometric mean: $\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$. Equality holds if and only if $x_1 = x_2 = \dots = x_n$. For two positive numbers $x, y$, this simplifies to $\frac{x+y}{2} \ge \sqrt{xy}$.

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Step-by-Step Solution

Step 1: Express the AP condition mathematically. We are given that $a, b,$ and $c$ are three positive numbers in Arithmetic Progression. The definition of an AP implies that the middle term is the average of the other two. Thus, we have: $b = \frac{a+c}{2}$ Rearranging this equation to isolate the sum $a+c$: $a+c = 2b \quad \text{(Equation 1)}$ This equation relates the sum of $a$ and $c$ to the term $b$.

Step 2: Utilize the given product condition. We are also given that the product of these three numbers is 8: $abc = 8 \quad \text{(Equation 2)}$ Since $a, b, c$ are positive, $b$ cannot be zero. We can rearrange Equation 2 to express the product $ac$ in terms of $b$: $ac = \frac{8}{b} \quad \text{(Equation 3)}$ This step is crucial because it provides the product of $a$ and $c$, which, along with their sum from Equation 1, can be used with the AM-GM inequality.

Step 3: Apply the AM-GM inequality to $a$ and $c$. Since $a, b, c$ are positive numbers, $a$ and $c$ are also positive. We can apply the AM-GM inequality to $a$ and $c$. The inequality states that the arithmetic mean of $a$ and $c$ is greater than or equal to their geometric mean: $\frac{a+c}{2} \ge \sqrt{ac}$ This inequality is a powerful tool for finding minimum or maximum values when sums and products are involved.

Step 4: Substitute and derive an inequality for $b$. Now, we substitute the expressions for $a+c$ (from Equation 1) and $ac$ (from Equation 3) into the AM-GM inequality: $\frac{2b}{2} \ge \sqrt{\frac{8}{b}}$ Simplifying the left side gives: $b \ge \sqrt{\frac{8}{b}}$ To remove the square root and solve for $b$, we square both sides of the inequality. This is a valid operation because both $b$ and $\sqrt{\frac{8}{b}}$ are positive (since $b>0$). $b^2 \ge \left(\sqrt{\frac{8}{b}}\right)^2$ $b^2 \ge \frac{8}{b}$ Next, we multiply both sides by $b$. This is also valid because $b$ is positive, so the inequality direction is preserved. $b^2 \cdot b \ge \frac{8}{b} \cdot b$ $b^3 \ge 8$ Finally, we take the cube root of both sides. The cube root function is strictly increasing, so it preserves the inequality direction. $\sqrt[3]{b^3} \ge \sqrt[3]{8}$ $b \ge 2$ This inequality tells us that the value of $b$ must be greater than or equal to 2. Therefore, the minimum possible value of $b$ is 2.

Step 5: Verify the equality condition. The minimum value of $b=2$ is achieved when the equality in the AM-GM inequality holds, which is when $a=c$. If $a=c$, then from Equation 1 ($a+c=2b$), we have $a+a = 2b$, which simplifies to $2a = 2b$, so $a=b$. If $a=b$ and $a=c$, then $a=b=c$. Now, we substitute this into the product condition (Equation 2: $abc=8$): $b \cdot b \cdot b = 8$ $b^3 = 8$ $b = 2$ This confirms that when $b=2$, we can have $a=b=c=2$. Let's check if these values satisfy the original conditions:

1. $a, b, c$ are positive: $2, 2, 2$ are positive.
2. $a, b, c$ are in AP: $2, 2, 2$ form an AP with a common difference of 0.
3. $abc = 8$: $2 \times 2 \times 2 = 8$. All conditions are satisfied, so the minimum value $b=2$ is achievable.

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Common Mistakes & Tips

• AM-GM Applicability: Ensure that the numbers to which AM-GM is applied are non-negative. In this problem, $a$ and $c$ are positive, so this condition is met.
• Inequality Manipulation: When multiplying or squaring inequalities, always check the sign of the terms involved. If you multiply or square by a negative number, the inequality direction reverses. Here, $b$ is positive, so operations were straightforward.
• Achievability: Always check if the equality condition for AM-GM (or other inequalities) can be met under the problem's constraints. This confirms that the derived minimum or maximum value is actually attainable.

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Summary

The problem involves finding the minimum value of $b$ given that $a, b, c$ are positive numbers in AP and their product $abc=8$. We used the AP property $a+c=2b$ and the given product $abc=8$ to express $ac = 8/b$. Applying the AM-GM inequality to $a$ and $c$ yielded $\frac{a+c}{2} \ge \sqrt{ac}$, which, after substitution, resulted in $b \ge 2$. The minimum value of $b=2$ is achieved when $a=b=c=2$, satisfying all given conditions. This problem elegantly demonstrates the application of AM-GM inequality in conjunction with sequence properties.

The final answer is $\boxed{2}$.