<p>If \(l, m, n\) are the three positive roots of the equation \(x^3 - ax^2 + bx - 48 = 0\), then the minimum value of \(\dfrac{1}{l} + \dfrac{2}{m} + \dfrac{3}{n}\) equals</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to establish constraints (lm + mn + nl = b and lmn = 48), then apply AM-GM inequality strategically by weighting the terms according to their coefficients in the target expression.
<p><strong>Step 1:</strong> From Vieta's formulas for roots l, m, n:</p><ul><li>lmn = 48</li><li>lm + mn + nl = b</li></ul><p><strong>Step 2:</strong> Apply weighted AM-GM inequality. For the expression 1/l + 2/m + 3/n, we construct the inequality strategically:</p><p>Consider that by AM-GM:</p><p>$$\frac{1}{l} + \frac{2}{m} + \frac{3}{n} \geq 3\sqrt[3]{\frac{1}{l} \cdot \frac{2}{m} \cdot \frac{3}{n}} = 3\sqrt[3]{\frac{6}{lmn}}$$</p><p><strong>Step 3:</strong> Substitute lmn = 48:</p><p>$$\frac{1}{l} + \frac{2}{m} + \frac{3}{n} \geq 3\sqrt[3]{\frac{6}{48}} = 3\sqrt[3]{\frac{1}{8}} = 3 \cdot \frac{1}{2} = \frac{3}{2}$$</p><p><strong>Step 4:</strong> Equality holds when:</p><p>$$\frac{1}{l} = \frac{2}{m} = \frac{3}{n}$$</p><p>Combined with lmn = 48, solving gives l = 2, m = 4, n = 6 (which satisfies all conditions).</p><p><strong>Step 5:</strong> Verify: 1/2 + 2/4 + 3/6 = 1/2 + 1/2 + 1/2 = 3/2</p><p>∴ Answer: <strong>D</strong> (minimum value = <strong>3/2</strong>)</p>
Correct Answer: D