If a, b, c are positive real numbers, then the least value of `(a+b+c)((1)/(a)+(1)/(b)+(1)/( c ))`, is

Options

A. 9
B. 3
C. `(10)/(3)`
D. none of these

Explanation

To find the least value of the expression \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) for positive real numbers \(a\), \(b\), and \(c\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Apply AM-GM Inequality**:
We know from the AM-GM inequality that for any positive real numbers \(x_1, x_2, x_3\):
\[ \frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3} \]
Let's apply this to \(a\), \(b\), and \(c\):
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
This implies:
\[ a + b + c \geq 3\sqrt[3]{abc} \]

2. **Apply AM-GM to the reciprocals**:
Now, apply AM-GM to the numbers \(\frac{1}{a}\), \(\frac{1}{b}\), and \(\frac{1}{c}\):
\[ \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \geq \sqrt[3]{\frac{1}{abc}} \]
This implies:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}} \]

3. **Combine the results**:
Now, multiply the two inequalities:
\[ (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq \left(3\sqrt[3]{abc}\right)\left(\frac{3}{\sqrt[3]{abc}}\right) \]
Simplifying the right-hand side:
\[ (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq 9 \]

4. **Conclusion**:
The least value of \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) is 9, which occurs when \(a = b = c\).

### Final Answer:
The least value of \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) is **9**.

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