What is the minimum value of \(\left( \frac{a^2 + 3a + 1}{a} \right) \left( \frac{b^2 + 3b + 1}{b} \right)\) for \(a, b > 0\)?
Minimum Value Calculation for Algebraic Expression
The problem asks for the minimum value of the expression \(\left( \frac{a^2 + 3a + 1}{a} \right) \left( \frac{b^2 + 3b + 1}{b} \right)\), given that \(a\) and \(b\) are positive real numbers (i.e., \(a > 0\) and \(b > 0\)).
Simplifying the Expression
First, let's simplify the term inside each parenthesis. We can divide each term in the numerator by the variable in the denominator:
\( \frac{x^2 + 3x + 1}{x} = \frac{x^2}{x} + \frac{3x}{x} + \frac{1}{x} = x + 3 + \frac{1}{x} \)Applying this simplification to both factors in the expression, we get:
\( \left( a + 3 + \frac{1}{a} \right) \left( b + 3 + \frac{1}{b} \right) \)Applying the AM-GM Inequality
To find the minimum value, we can use the Arithmetic Mean - Geometric Mean (AM-GM) inequality. For any non-negative numbers \(x\) and \(y\), the AM-GM inequality states:
\( \frac{x+y}{2} \ge \sqrt{xy} \)Which can also be written as:
\( x + y \ge 2\sqrt{xy} \)Let's consider the term \(a + \frac{1}{a}\). Since \(a > 0\), both \(a\) and \(\frac{1}{a}\) are positive. Applying the AM-GM inequality:
\( a + \frac{1}{a} \ge 2 \sqrt{a \cdot \frac{1}{a}} \) \( a + \frac{1}{a} \ge 2 \sqrt{1} \) \( a + \frac{1}{a} \ge 2 \)Similarly, for the term \(b + \frac{1}{b}\):
\( b + \frac{1}{b} \ge 2 \sqrt{b \cdot \frac{1}{b}} \) \( b + \frac{1}{b} \ge 2 \)Finding the Minimum Value of Each Factor
Now, let's look at the individual factors in our simplified expression:
- The first factor is \(a + 3 + \frac{1}{a}\). We can rewrite this as \(\left( a + \frac{1}{a} \right) + 3\). Since we know \(a + \frac{1}{a} \ge 2\), the minimum value of this factor is \(2 + 3 = 5\).
- The second factor is \(b + 3 + \frac{1}{b}\). Similarly, this can be written as \(\left( b + \frac{1}{b} \right) + 3\). Since \(b + \frac{1}{b} \ge 2\), the minimum value of this factor is \(2 + 3 = 5\).
Calculating the Minimum of the Product
The minimum value of the entire expression is the product of the minimum values of each factor, because \(a\) and \(b\) are independent positive variables.
\( \text{Minimum Value} = (\text{Minimum value of the first factor}) \times (\text{Minimum value of the second factor}) \) \( \text{Minimum Value} = 5 \times 5 \) \( \text{Minimum Value} = 25 \)Condition for Minimum Value
The AM-GM inequality \(x + \frac{1}{x} \ge 2\) achieves equality (i.e., equals 2) when \(x = \frac{1}{x}\), which means \(x^2 = 1\). Since \(a\) and \(b\) must be positive (\(a > 0, b > 0\)), the equality holds when \(a = 1\) and \(b = 1\). These values satisfy the condition \(a, b > 0\).
Let's check the value of the expression when \(a=1\) and \(b=1\):
\( \left( \frac{1^2 + 3(1) + 1}{1} \right) \left( \frac{1^2 + 3(1) + 1}{1} \right) = \left( \frac{1 + 3 + 1}{1} \right) \left( \frac{1 + 3 + 1}{1} \right) = (5)(5) = 25 \)This confirms that the minimum value is indeed 25.
Conclusion
The minimum value of the expression \(\left( \frac{a^2 + 3a + 1}{a} \right) \left( \frac{b^2 + 3b + 1}{b} \right)\) for \(a, b > 0\) is 25.
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