If a, b, c are positive such that a b2 c3=64 then least value of ((1/a)+(2/b)+(3/c)) is -

Question

If a, b, c are positive such that a b2 c3=64 then least value of ((1/a)+(2/b)+(3/c)) is - (A) 6 (B) 2 (C) 3 (D) 32

Tardigrade Admin · Accepted Answer

G.M. ≥ H.M. (a.b.b.c.c.c) (1/6) ≥ (6/(1) a +(1) b +(1/ b )+(1/ c )+(1/ c )+(1/ c ) ⇒(64)/1 / 6) ≥ (6/(1) a +(2) b +(3/ c ) ⇒ (1/ a )+(2/ b )+(3/ c ) ≥ 3

Q. If $a, b, c$ are positive such that $a b^{2} c^{3} = 64$ then least value of $(\frac{1}{a} + \frac{2}{b} + \frac{3}{c})$ is -

6

2

3

32

G.M. $\geq$ H.M.
$(a . b . b . c . c . c)^{\frac{1}{6}} \geq \frac{6}{\frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{c} + \frac{1}{c}}$
$\Rightarrow (64)^{1/6} \geq \frac{6}{\frac{1}{a} + \frac{2}{b} + \frac{3}{c}}$
$\Rightarrow \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \geq 3$

Q. If a,b,c are positive such that ab2c3=64 then least value of (a1​+b2​+c3​) is -

6

2

3

32

G.M. ≥ H.M. (a.b.b.c.c.c)61​≥a1​+b1​+b1​+c1​+c1​+c1​6​ ⇒(64)1/6≥a1​+b2​+c3​6​ ⇒a1​+b2​+c3​≥3

Q. If $a, b, c$ are positive such that $a b^{2} c^{3} = 64$ then least value of $(\frac{1}{a} + \frac{2}{b} + \frac{3}{c})$ is -

G.M. $\geq$ H.M.
$(a . b . b . c . c . c)^{\frac{1}{6}} \geq \frac{6}{\frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{c} + \frac{1}{c}}$
$\Rightarrow (64)^{1/6} \geq \frac{6}{\frac{1}{a} + \frac{2}{b} + \frac{3}{c}}$
$\Rightarrow \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \geq 3$