Basic Mathematics & Logarithm
Inequalities involving means
Grade 11
Question:
<p>If the product of \(n\) positive numbers is \(n^n\), then their sum is</p>
<p>(1) a positive integer</p>
<p>(2) divisible by \(n\)</p>
<p>(3) equal to \(n + 1/n\)</p>
<p>(4) never less than \(n^2\)</p>
Step-by-Step Solution
Key Concept: By AM-GM inequality, for positive numbers with fixed product, the sum is minimized when all numbers are equal. Since the product equals n^n and there are n numbers, each number must be n, making the minimum sum n·n = n².
<p><strong>Step 1:</strong> Apply AM-GM Inequality</p><p>For n positive numbers a₁, a₂, ..., aₙ:</p><p>$$\frac{a_1 + a_2 + ... + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot ... \cdot a_n}$$</p><p><strong>Step 2:</strong> Substitute the given condition (product = n^n)</p><p>$$\frac{S}{n} \geq \sqrt[n]{n^n} = n$$</p><p>where S is the sum of the n numbers.</p><p><strong>Step 3:</strong> Solve for S</p><p>$$S \geq n \cdot n = n^2$$</p><p><strong>Step 4:</strong> Check equality condition</p><p>Equality in AM-GM holds when all terms are equal: a₁ = a₂ = ... = aₙ = n</p><p>Verify: Product = n · n · ... · n (n times) = n^n ✓</p><p>Sum = n + n + ... + n (n times) = n²</p><p>∴ Answer: The sum is <strong>n²</strong> (Option D)</p>
Correct Answer: D