The minimum value of `((a^2 +3a+1)(b^2+3b + 1)(c^2+ 3c+ 1))/(abc)`The minimum value of , where `a, b, c in R^+` is

Options

A. `(11^(3))/(2^(3))`
B. 125
C. 25
D. 27

Answer

B

Explanation

Let `A=((a^(2)+3a+1)(b^(2)+3b+1)(c^(2)+3c+1))/(abc)`
`=((a^(2)+3a+1)/(a))((b^(2)+3b+1)/(b))((c^(2)+3c+1)/(c))`
`=(a+3+(1)/(a))(b+3+(1)/(b))(c+3+(1)/(c))`
where `a,b,c in R^(+)`.
Applying `Am ge GM` on a and `(1)/(a)`
`a+(1)/(a)ge 2 " " implies a+(1)/(b)+3ge 5`
Similarly, `b+(1)/(b)ge 2 " " implies b+(1)/(b)+3ge 5`
and `c+(1)/(c)ge 2 " " implies c+(1)/(c)+3ge 5`
`:.(a+(1)/(a)+3)(b+(1)/(b)+3)(c+(1)/(c)+3)ge 125`
So, `Age 5*5*5 " " implies Age 125`
minimum value of A is 125.