For positive numbers a, b, c the least value of `(a+b+c)((1)/(a)+(1)/(b)+(1)/(c))` is
(a)3
(b)9
(c)27/4
(d)None of these

Options

A. 3
B. 9
C. `(27)/(4)`
D. None of these

Answer

To find the least value of the expression \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) for positive numbers \(a\), \(b\), and \(c\), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ E = (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \]
This can be expanded to:
\[ E = (a+b+c)\left(\frac{bc + ac + ab}{abc}\right) \]

### Step 2: Apply the AM-GM Inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that:
\[ \frac{a+b+c}{3} \geq \sqrt[3]{abc} \]
and
\[ \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \geq \sqrt[3]{\frac{1}{abc}} \]
Multiplying these two inequalities gives:
\[ \left(\frac{a+b+c}{3}\right)\left(\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3}\right) \geq \sqrt[3]{abc} \cdot \sqrt[3]{\frac{1}{abc}} = 1 \]

### Step 3: Combine the Inequalities
Thus, we have:
\[ E \geq 9 \]
This indicates that the minimum value of \(E\) is at least 9.

### Step 4: Check if the Minimum is Achievable
To check if this minimum can be achieved, we set \(a = b = c = 1\):
\[ E = (1 + 1 + 1)\left(\frac{1}{1} + \frac{1}{1} + \frac{1}{1}\right) = 3 \cdot 3 = 9 \]
This shows that the least value of the expression is indeed 9.

### Conclusion
Thus, the least value of \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) for positive numbers \(a\), \(b\), and \(c\) is:
\[ \boxed{9} \]