Question

If a, b, c are positive real numbers, the least value of $(a+b+c)(\frac{1}{a}+\frac{1}{b},\frac{1}{c})$ is :

• A. 1
• B. 9
• C. 12
• D. None of these

Answer 1

The correct option is B 9

$AM\ge GM\Rightarrow \frac{a+b+c}{3}\ge (abc{)}^{\frac{1}{3}}$
(AM $\to$ Arithmetic mean, GM $\to$ Geometric mean)
and $\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge (\frac{1}{abc}{)}^{\frac{1}{3}}$
$\Rightarrow \frac{1}{3}(a+b+c)\times \frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge \left(abc{)}^{\frac{1}{3}}\right(\frac{1}{abc}{)}^{\frac{1}{3}}=1\Rightarrow (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 9$
Putting a = b = c = 1, expression takes the value 9, which therefore, its least value.