Question

11.
The minimum value of the quantity
(a^2 + 3a +1)(b^2 + 4b + 1)(c ^2+ 5c +1)
, where a, b and care positive integers,
abc
is -
(D)11.13.15/2^3
(A) 125
(B) 210
(C) 60​

Answer 1

Answer:

125

Step-by-step explanation:

by equating all the terms....i.e. a=b=c

we will get f=(a^2+3a+1)=(b^2+3b+1)=(c^2+3c+1)

or

dividing by a^3 because a=b=c

f(a)= (a^2+3a+1)^3/a^3

f(a)= (a+1/a+3)

min. value of a+1/a is 2 i.e. for a=1

f(a)= (a+1/a+3)^3

f(1)=(1+1/1+3)^3

f(1)=(2+3)^3

f(1)=(5)^3

f(1)=125