If a, b, c are any three positive numbers, then the least value of (a+b+c) ((1/a) + (1/b) + (1/c)) is

Question

If a, b, c are any three positive numbers, then the least value of (a+b+c) ((1/a) + (1/b) + (1/c)) is (A) 3 (B) 6 (C) 9 (D) none of these.

Tardigrade Admin · Accepted Answer

A.M. of any number of non-negative numbers is always greater than or equal to the H.M. between them, therefore, (a+b+c/3) ge (3/(1)a + (1)b+(1/c) ⇒ (a+b+c) ((1/a)+(1/b)+(1/c)) ge 9

Q. If a, b, c are any three positive numbers, then the least value of $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ is

3

6

9

none of these.

A.M. of any number of non-negative numbers is always greater than or equal to the H.M. between them, therefore,
$\frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$
$\Rightarrow (a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$

Q. If a, b, c are any three positive numbers, then the least value of (a+b+c)(a1​+b1​+c1​) is

3

6

9

none of these.

A.M. of any number of non-negative numbers is always greater than or equal to the H.M. between them, therefore, 3a+b+c​≥a1​+b1​+c1​3​ ⇒(a+b+c)(a1​+b1​+c1​)≥9

Q. If a, b, c are any three positive numbers, then the least value of $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ is

A.M. of any number of non-negative numbers is always greater than or equal to the H.M. between them, therefore,
$\frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$
$\Rightarrow (a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$