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

Options

A. 1
B. 9
C. 12
D. none of these

Answer

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

### Step 1: Rewrite the expression
We start by rewriting the expression:
\[ (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = (a+b+c) \left(\frac{bc + ac + ab}{abc}\right) \]
This simplifies to:
\[ \frac{(a+b+c)(bc + ac + ab)}{abc} \]

### 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
\[ \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \geq \sqrt[3]{\frac{1}{abc}} \]

### Step 3: Combine the inequalities
Multiplying these two inequalities gives:
\[ \left(\frac{a+b+c}{3}\right)\left(\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3}\right) \geq \sqrt[3]{abc} \cdot \sqrt[3]{\frac{1}{abc}} = 1 \]
Thus,
\[ \frac{(a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)}{9} \geq 1 \]
This implies:
\[ (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq 9 \]

### Step 4: Determine when equality holds
Equality in the AM-GM inequality holds when \(a = b = c\). Therefore, if we set \(a = b = c\), we can find the least value:
Let \(a = b = c = k\), then:
\[ (a+b+c) = 3k \quad \text{and} \quad \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = \frac{3}{k} \]
Thus,
\[ (3k)\left(\frac{3}{k}\right) = 9 \]

### Conclusion
The least value of \((a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\) is:
\[ \boxed{9} \]