If ab 2 c 3 , a 2 b 3 c 4 , a 3 b 4 c 5 are in AP (a,b,c > o), then the minimum value of a + b + c is

Step-by-step solution

∵ AM ⁡ ≥ GM ⇒ a + b + c 3 ≥ a b c 1 / 3 or a + b + c ≥ 3 a b c 1 / 3 ......(I) But given ab 2 c 3 , a 2 b 3 c 4 , a 3 b 4 c 5 are in AP (Divide the sequence by ab 2 c 3 ) ⇒ 1,abc, a 2 b 2 c 2 are also in AP ( ∵ a b c ≠ 0 ) ⇒ 2abc = 1 + a 2 b 2 c 2 ⇒ (abc - 1) 2 = 0 ∴ abc = 1 Now from Eq. (i), a + b + c ≥ 3 1 1 / 3 or a + b + c ≥ 3 Hence, minimum value of a + b + c is 3.