Basic Mathematics & Logarithm
Inequalities involving means
Grade 11
Question:
<p>If \(a, b, c \in R^+\), then \((a+b+c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)\) is always</p>
<p>(1) \(\geq 12\)</p>
<p>(2) \(\geq 9\)</p>
<p>(3) \(\leq 12\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Apply the Cauchy-Schwarz inequality or expand the expression directly to show that the product of a sum and the sum of reciprocals creates cross terms that guarantee a minimum value of 9.
<p><strong>Step 1:</strong> Expand the expression:</p><p>$(a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = 1 + \frac{b}{a} + \frac{c}{a} + \frac{a}{b} + 1 + \frac{c}{b} + \frac{a}{c} + \frac{b}{c} + 1$</p><p><strong>Step 2:</strong> Rewrite as:</p><p>$= 3 + \left(\frac{a}{b} + \frac{b}{a}\right) + \left(\frac{b}{c} + \frac{c}{b}\right) + \left(\frac{c}{a} + \frac{a}{c}\right)$</p><p><strong>Step 3:</strong> Apply AM-GM inequality: For positive $x$, $x + \frac{1}{x} \geq 2$</p><p>Therefore: $\frac{a}{b} + \frac{b}{a} \geq 2$, $\frac{b}{c} + \frac{c}{b} \geq 2$, $\frac{c}{a} + \frac{a}{c} \geq 2$</p><p><strong>Step 4:</strong> Adding all inequalities:</p><p>$(a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq 3 + 2 + 2 + 2 = 9$</p><p>Equality holds when $a = b = c$</p><p>∴ Answer: <strong>Always ≥ 9</strong></p>
Correct Answer: B