Question

If a, b, c are three positive real numbers,
then the least value of
(1+a+a?) (1 + b + b2 ) (1+c+c2)/abc

is
abc
(A) 3
(B) 9
(C) 27
(D) None of these​

Answer 1

Answer:

(C) 27

Step-by-step explanation:

(1 + a + a²) ( 1 + b + b²) ( 1 + c + c² ) / abc

= ( 1/a + 1 + a ) ( 1/b + 1 + b ) ( 1/c + 1 + c)

By the AM-GM inequality, we know that

 ( 1/a + a ) / 2 ≥ √( 1/a × a ) = 1, with equality when a=1,

so ( 1/a + 1 + a ) ≥ 3.

[ Alternatively, use the fact that ( √a - 1/√a )² ≥ 0, which upon expansion becomes  a + 1/a - 2 ≥ 0 . ]

Similarly, ( 1/b + 1 + b ) ≥ 3 and ( 1/c + 1 + c ) ≥ 3, with equality when b=c=1.

All together, this means

(1 + a + a²) ( 1 + b + b²) ( 1 + c + c² ) / abc

= ( 1/a + 1 + a ) ( 1/b + 1 + b ) ( 1/c + 1 + c)

≥ 3 × 3 × 3 = 27,

when equality when a = b = c = 1.

Therefore 27 is the least value.

Hope this helps!