What is the minimum value of \(\frac{(a^8+a^4+1)(b^8+b^4+1)}{a^4b^4}\) where \(a > 0, b > 0\) ?
Understanding the Minimum Value Problem
The question asks for the minimum possible value of the expression \(\frac{(a^8+a^4+1)(b^8+b^4+1)}{a^4b^4}\), given that both \(a\) and \(b\) are positive numbers (\(a > 0, b > 0\)).
Simplifying the Algebraic Expression
To make the expression easier to work with, let's introduce substitutions. Let \(x = a^4\) and \(y = b^4\). Since \(a\) and \(b\) are positive, \(a^4\) and \(b^4\) will also be positive. So, \(x > 0\) and \(y > 0\). The expression can be rewritten in terms of \(x\) and \(y\) as:
\( \frac{(x^2+x+1)(y^2+y+1)}{xy} \)
We can separate this expression into the product of two fractions:
\( \left(\frac{x^2+x+1}{x}\right) \left(\frac{y^2+y+1}{y}\right) \)
Analyzing the Term \(\frac{x^2+x+1}{x}\)
Let's simplify the first part of the product:
\( \frac{x^2+x+1}{x} = \frac{x^2}{x} + \frac{x}{x} + \frac{1}{x} = x + 1 + \frac{1}{x} \)
Now, we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For any positive number \(x\), the AM-GM inequality states that the arithmetic mean is greater than or equal to the geometric mean:
\( \frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} \)
Simplifying the right side:
\( \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1 \)
So, we have:
\( \frac{x + \frac{1}{x}}{2} \ge 1 \)
Multiplying both sides by 2 gives:
\( x + \frac{1}{x} \ge 2 \)
Now, substitute this back into our simplified term:
\( x + 1 + \frac{1}{x} \ge 2 + 1 = 3 \)
The minimum value of \(x + 1 + \frac{1}{x}\) is 3. This minimum occurs when the equality in the AM-GM inequality holds, which is when \(x = \frac{1}{x}\). This leads to \(x^2 = 1\). Since we know \(x > 0\), the minimum occurs at \(x = 1\).
Applying AM-GM to the Term \(\frac{y^2+y+1}{y}\)
We can perform the exact same analysis for the second part of the product involving \(y\). The term is:
\( \frac{y^2+y+1}{y} = y + 1 + \frac{1}{y} \)
Using the AM-GM inequality for \(y > 0\):
\( y + \frac{1}{y} \ge 2 \)
Therefore:
\( y + 1 + \frac{1}{y} \ge 2 + 1 = 3 \)
The minimum value of this term is also 3, occurring when \(y = 1\).
Finding the Overall Minimum Value
The original expression is the product of the two terms we analyzed:
\( \text{Expression} = \left(a^4 + 1 + \frac{1}{a^4}\right) \left(b^4 + 1 + \frac{1}{b^4}\right) \)
We found that the minimum value of the first factor \((a^4 + 1 + \frac{1}{a^4})\) is 3, and the minimum value of the second factor \((b^4 + 1 + \frac{1}{b^4})\) is 3.
The minimum value of the product is the product of their individual minimum values:
\( \text{Minimum Value} = (\text{Min of first factor}) \times (\text{Min of second factor}) \)
\( \text{Minimum Value} = 3 \times 3 = 9 \)
This minimum value of 9 is achieved when both factors are at their minimum, meaning \(a^4 = 1\) and \(b^4 = 1\). Since \(a > 0\) and \(b > 0\), this occurs when \(a=1\) and \(b=1\).
Conclusion
The minimum value of the given expression \(\frac{(a^8+a^4+1)(b^8+b^4+1)}{a^4b^4}\) for \(a > 0, b > 0\) is 9.
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