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:
We are asked to find the minimum possible value or the lower bound of an algebraic expression involving three positive numbers \(a\), \(b\), and \(c\).


Step 2: Key Formula or Approach:
This problem can be elegantly solved using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Alternatively, algebraic expansion paired with the property that \(x + \frac{1}{x} \geq 2\) for positive \(x\) works perfectly.


Step 3: Detailed Explanation:
Let's expand the given expression: \[ (a+b+c) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \] \[ = a\left(\frac{1}{a}\right) + a\left(\frac{1}{b}\right) + a\left(\frac{1}{c}\right) + b\left(\frac{1}{a}\right) + b\left(\frac{1}{b}\right) + b\left(\frac{1}{c}\right) + c\left(\frac{1}{a}\right) + c\left(\frac{1}{b}\right) + c\left(\frac{1}{c}\right) \] \[ = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 \] Grouping reciprocal terms together: \[ = 3 + \left(\frac{a}{b} + \frac{b}{a}\right) + \left(\frac{b}{c} + \frac{c}{b}\right) + \left(\frac{c}{a} + \frac{a}{c}\right) \] By the AM-GM inequality, for any two positive numbers \(x\) and \(y\): \[ \frac{x}{y} + \frac{y}{x} \geq 2\sqrt{\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)} = 2\sqrt{1} = 2 \] Applying this to each paired term in our expansion: \[ \frac{a}{b} + \frac{b}{a} \geq 2 \] \[ \frac{b}{c} + \frac{c}{b} \geq 2 \] \[ \frac{c}{a} + \frac{a}{c} \geq 2 \] Substituting these minimum values back into the grouped expression: \[ \text{Value} \geq 3 + 2 + 2 + 2 \] \[ \text{Value} \geq 9 \] Equality holds only when \(a = b = c\).


Step 4: Final Answer:
The value is \(\geq 9\).

Answer 2

A clean alternate way to establish this inequality is to apply the Cauchy-Schwarz inequality directly, instead of expanding the product and pairing terms with AM-GM.

The Cauchy-Schwarz inequality states that for real numbers \(x_1,x_2,x_3\) and \(y_1,y_2,y_3\):

\[ (x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2) \geq (x_1y_1+x_2y_2+x_3y_3)^2 \]

Choose \(x_1=\sqrt{a}, x_2=\sqrt{b}, x_3=\sqrt{c}\) and \(y_1=\frac{1}{\sqrt{a}}, y_2=\frac{1}{\sqrt{b}}, y_3=\frac{1}{\sqrt{c}}\), which is valid since \(a,b,c\) are positive.

Then \(x_1^2+x_2^2+x_3^2=a+b+c\) and \(y_1^2+y_2^2+y_3^2=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\), while each product \(x_iy_i\) simplifies to exactly 1, since \(\sqrt{a}\times\frac{1}{\sqrt{a}}=1\), and similarly for the other two terms.

So \(x_1y_1+x_2y_2+x_3y_3=1+1+1=3\), and the Cauchy-Schwarz inequality directly gives:

\[ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 3^2 = 9 \]

This confirms the inequality without any expansion or pairwise AM-GM steps at all. Equality holds precisely when \(a=b=c\), matching the equality condition found by the expansion method as well.

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