Question

If$\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}$ are different positive real numbers ,then $[a+b+c]\left[\left(\frac{1}{a}\right)+\left(\frac{1}{b}\right)+\left(\frac{1}{c}\right)\right]$ is

• A. $<9$
• B. $>9$
• C. $>0$
• D. $>24$

Answer 1

The correct option is B

$>9$

Explanation for the correct option:

Applying AM-GM inequality:

$[a+b+c]\left[\left(\frac{1}{a}\right)+\left(\frac{1}{b}\right)+\left(\frac{1}{c}\right)\right]$

$\because AM>GM$

$\Rightarrow$$\frac{[a+b+c]}{3}>{\left(abc\right)}^{\frac{1}{3}}$

$\Rightarrow$ $[a+b+c]>3{\left(abc\right)}^{\frac{1}{3}}$ — (1)

Similarly, $\frac{\left[{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}\right]}{3}>{\left[\left(\frac{1}{a}\right)\left(\frac{1}{b}\right)\left(\frac{1}{c}\right)\right]}^{\frac{1}{3}}$

$\Rightarrow$ $\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]>3{\left[\frac{1}{abc}\right]}^{\frac{1}{3}}$ — (2)

Multiply equation (1) and (2),

$\Rightarrow$$[a+b+c]\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]>9$

Hence, option ‘B’ is correct.