Question

If a,b,c are 3 unequal positive numbers, then (a+b+c) $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ is

• A. > 9
• B. < 9
• C. > 27
• D. 27

Answer 1

The correct option is A

> 9

Since A.M of a,b,c > G.M of a,b,c

$\frac{a+b+c}{3}$ > $(abc{)}^{\frac{1}{3}}$

(a+b+c) > 3$(abc{)}^{\frac{1}{3}}$ -------------(1)

Again, AM of $\frac{1}{a}$,$\frac{1}{b}$,$\frac{1}{c}$ > GM of $\frac{1}{a}$,$\frac{1}{b}$,$\frac{1}{c}$

$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}$ > ${[\frac{1}{a}.\frac{1}{b}.\frac{1}{c}]}^{\frac{1}{3}}$

$[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]$ > $\frac{3}{(abc{)}^{\frac{1}{3}}}$ --------------(2)

Multiplying of corresponding sides of (1) and (2), we get

(a + b + c) $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$> 3$(abc{)}^{\frac{1}{3}}$.$\frac{3}{(abc{)}^{\frac{1}{3}}}$

Hence, (a+b+c) $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ > 9