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**:
According to the AM-GM inequality, for any non-negative real numbers \(x_1, x_2, \ldots, x_n\):
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \]
We will apply this to both \(a, b, c\) and \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\).

2. **First Application (for \(a, b, c\))**:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
Therefore,
\[ a + b + c \geq 3\sqrt[3]{abc} \]

3. **Second Application (for \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\))**:
\[ \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \geq \sqrt[3]{\frac{1}{abc}} \]
Thus,
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}} \]

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

5. **Conclusion**:
Therefore, the least value of the expression \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) is \(9\).

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