Basic Mathematics & Logarithm
Inequalities involving means
Grade None
Question:
<p>Minimum value of \(\dfrac{(b+c)}{a} + \dfrac{(c+a)}{b} + \dfrac{(a+b)}{c}\) (for real positive numbers \(a, b, c\)) is</p>
<p>(1) 1</p>
<p>(2) 2</p>
<p>(3) 4</p>
<p>(4) 6</p>
Step-by-Step Solution
Key Concept: Rewrite the expression as a sum of reciprocal pairs and apply AM-GM inequality strategically. The expression equals (b/a + a/b) + (c/a + a/c) + (c/b + b/c) + 3, where each pair has minimum value 2.
<p><strong>Step 1:</strong> Expand and reorganize the expression:</p><p>$$\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} = \frac{b}{a} + \frac{c}{a} + \frac{c}{b} + \frac{a}{b} + \frac{a}{c} + \frac{b}{c}$$</p><p><strong>Step 2:</strong> Group reciprocal pairs:</p><p>$$= \left(\frac{a}{b} + \frac{b}{a}\right) + \left(\frac{b}{c} + \frac{c}{b}\right) + \left(\frac{c}{a} + \frac{a}{c}\right)$$</p><p><strong>Step 3:</strong> Apply AM-GM to each pair. For positive numbers x and y: $\frac{x}{y} + \frac{y}{x} \geq 2\sqrt{\frac{x}{y} \cdot \frac{y}{x}} = 2$</p><p>Each pair has minimum value 2, achieved when the terms are equal.</p><p><strong>Step 4:</strong> Sum the minimums:</p><p>$$\text{Minimum} = 2 + 2 + 2 = 6$$</p><p>Equality holds when $a = b = c$.</p><p>∴ Answer: <strong>D (6)</strong></p>
Correct Answer: D