<p>320. The product of \(n\) positive numbers is unity. Then their sum is</p>
<p>(a) a positive integer</p>
<p>(b) divisible by \(n\)</p>
<p>(c) equal to \(n + \frac{1}{n}\)</p>
<p>(d) never less than \(n\)</p>
Step-by-Step Solution
Key Concept: By the AM-GM inequality, for n positive numbers with product equal to 1, the arithmetic mean is minimized when all numbers are equal. Since the product is unity, each number must equal 1, giving minimum sum = n.
<p><strong>Step 1:</strong> Let the n positive numbers be a₁, a₂, ..., aₙ where a₁·a₂·...·aₙ = 1</p><p><strong>Step 2:</strong> Apply AM-GM inequality: (a₁ + a₂ + ... + aₙ)/n ≥ ⁿ√(a₁·a₂·...·aₙ)</p><p><strong>Step 3:</strong> Since the product equals 1: (a₁ + a₂ + ... + aₙ)/n ≥ ⁿ√1 = 1</p><p><strong>Step 4:</strong> Therefore: a₁ + a₂ + ... + aₙ ≥ n</p><p><strong>Step 5:</strong> Equality holds in AM-GM when all numbers are equal. If a₁ = a₂ = ... = aₙ = k, then kⁿ = 1, so k = 1</p><p><strong>Step 6:</strong> The minimum sum is n (when all numbers equal 1). The sum is always ≥ n.</p><p>∴ Answer: D (The sum is greater than or equal to n, with minimum value n)</p>
Correct Answer: D