Question

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

• A. $3$
• B. $9$
• C. $\frac{27}{4}$
• D. $Noneofthese$

Answer 1

The correct option is B $9$

Since a,b,c are three positive numbers. As we know that $AM\ge GM$
$\Rightarrow \frac{a+b+c}{3}\ge \sqrt[3]{3abc}$
Their reciprocal are $\frac{1}{a},\frac{1}{b},\frac{1}{c}$
Again applying
$AM\ge GM$
$\Rightarrow \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \sqrt[3]{3\frac{1}{abc}}$
Multiplying eqs. (1) and (2), we get
$(a+b+c)(\frac{1}{a},\frac{1}{b},\frac{1}{c})\ge 9$