For any n positive numbers `a_(1),a_(2),…,a_(n)` such that
`sum_(i=1)^(n) a_(i)=alpha`, the least value of `sum_(i=1)^(n) a_(i)^(-1)`, is

Options

A. `2n-alpha`
B. `(3n)/(alpha)`
C. `(n(n-1))/(alpha)`
D. `(n^(2))/(alpha)`

Answer

D

Explanation

We have,
`(underset(i=1)overset(n)suma_(i))(underset(i=1)overset(n)suma_(i)^(-1))`
`=n+((a_(1))/(a_(2))+(a_(2))/(a_(1)))+((a_(1))/(a_(3))+(a_(3))/(a_(1)))+...+((a_(1))/(a_(n))+(a_(n))/(a_(1)))`
`+((a_(2))/(a_(3))+(a_(3))/(a_(2)))+((a_(2))/(a_(4))+(a_(4))/(a_(2)))+...+((a_(2))/(a_(n))+(a_(n))/(a_(2)))`
`+....+....+....+((a_(n-1))/(a_(n))+(a_(n))/(a_(n-1)))`
`len+2(n-1+n-2+...+1)" "[becausex+(1)/(x)ge2" for "xgt0]`
`len^(2)`
`implies" "(underset(i=1)overset(n)suma_(i)^(-1))ge(n^(2))/(alpha)`
Hence, the minimum value of `underset(i=1)overset(n)suma_(i)^(-1)" is "(n^(2))/(alpha)`