Question

If a, b, c are positive numbers then value of \((a+b+c) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\), is

• A. \(\geq 10\)
• B. \(\geq 9\)
• C. \(\geq 12\)
• D. None of these

Hint:

This is a direct application of the AM-HM inequality for \(n\) variables: \(\frac{x_1 + x_2 + \dots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}}\). For \(n=3\), moving the denominators yields \((a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 3 \times 3 = 9\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks for the minimum value of a specific algebraic expression involving three positive real numbers $a, b,$ and $c$. This is a classic problem in the study of inequalities. The expression is the product of the sum of the variables and the sum of their reciprocals. Because the variables are restricted to being positive, we can apply various mean-related inequalities to establish a lower bound.
Step 2: Key Formulas and approach:
Two common approaches can be used for this problem:
1. The Arithmetic Mean-Harmonic Mean (AM-HM) Inequality: $\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$. Rearranging this directly gives $(a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$.
2. Algebraic Expansion: Multiplying the two brackets out and using the property that for any positive $x$, $x + \frac{1}{x} \geq 2$.
We will use the expansion method as it provides a clear step-by-step derivation of why the value 9 is the absolute minimum.
Step 3: Detailed Explanation:

We start by expanding the product $(a+b+c) \times (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$:

Term-by-term multiplication: $a(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + b(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + c(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$.

Resulting terms: $(1 + \frac{a}{b} + \frac{a}{c}) + (\frac{b}{a} + 1 + \frac{b}{c}) + (\frac{c}{a} + \frac{c}{b} + 1)$.

Combining the constant integers: $1 + 1 + 1 = 3$.

Now group the remaining reciprocal pairs: $3 + (\frac{a}{b} + \frac{b}{a}) + (\frac{b}{c} + \frac{c}{b}) + (\frac{a}{c} + \frac{c}{a})$.

Using the AM-GM inequality for two positive terms, we know $\frac{x}{y} + \frac{y}{x} \geq 2$.

Applying this to our pairs:

$(\frac{a}{b} + \frac{b}{a}) \geq 2$

$(\frac{b}{c} + \frac{c}{b}) \geq 2$

$(\frac{a}{c} + \frac{c}{a}) \geq 2$

Summing these lower bounds: Value $\geq 3 + 2 + 2 + 2$.

Value $\geq 9$.

Note that equality occurs when $a = b = c$, in which case the expression becomes $(3a)(3/a) = 9$.

Step 4: Final Answer:
The minimum value of the expression for positive numbers is 9. Therefore, the correct option is (B).

Answer 2

The Arithmetic Mean - Harmonic Mean (AM-HM) inequality is a standard result for positive numbers, and applying it directly here skips the expansion step entirely.

1. Stating AM-HM: For any positive numbers \(a,b,c\), the arithmetic mean is always at least as large as the harmonic mean: \(\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\).
2. Rearranging: Cross-multiplying both sides by the positive quantities 3 and \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) gives \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9\), exactly the expression in the question.
3. Why AM-HM holds: AM-HM itself follows from applying AM-GM twice, once to \(a,b,c\) and once to \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\), and combining the two resulting inequalities; it is a standard result that can be applied directly here without re-deriving it from scratch.
4. Checking the boundary case: When \(a=b=c\), both sides of AM-HM become equal, so the original expression equals exactly 9, confirming that 9 is the tightest possible lower bound, not just any valid bound.

Since the expression matches the AM-HM inequality rearranged, and equality is achievable at \(a=b=c\), the bound of 9 is both valid and the best possible one.

Therefore, the correct answer is \(\geq9\).