<p>Find the greatest value of the product of three positive numbers if the sum of their products taking two at a time is 12.</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality strategically by relating the constraint (sum of pairwise products) to the objective (product of three numbers). If xy + yz + zx = 12, then maximizing xyz requires recognizing that equality in AM-GM occurs when all three numbers are equal.
<p><strong>Step 1:</strong> Let the three positive numbers be x, y, z with constraint xy + yz + zx = 12.</p><p><strong>Step 2:</strong> By AM-GM inequality: (xy + yz + zx)/3 ≥ ∛(xy·yz·zx) = ∛(x²y²z²) = (xyz)^(2/3)</p><p><strong>Step 3:</strong> Substituting the constraint: 12/3 ≥ (xyz)^(2/3), which gives 4 ≥ (xyz)^(2/3)</p><p><strong>Step 4:</strong> Cubing both sides: 64 ≥ (xyz)². Therefore xyz ≤ 8 (since x,y,z are positive).</p><p><strong>Step 5:</strong> Equality in AM-GM holds when xy = yz = zx, which means x = y = z. Substituting into constraint: 3x² = 12, so x = 2.</p><p><strong>Step 6:</strong> When x = y = z = 2, the product xyz = 8, and the constraint is satisfied: 2·2 + 2·2 + 2·2 = 12 ✓</p><p>∴ <strong>Answer: 8</strong></p>
Correct Answer: 8