Question

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

• A. $2n-\alpha$
• B. $\frac{3n}{\alpha }$
• C. $\frac{n(n-1)}{\alpha }$
• D. $\frac{n^2}{\alpha }$

Answer 1

The correct option is D $\frac{n^2}{\alpha }$

Given that ${\sum }_{i=1}^n{a}_i=\alpha ⟹a_1+a_2+...+a_n=\alpha$

we know that, $A.M.\ge H.M.$

A.M. of $a_1,a_2,...,a_n$ is $\frac{a_1+a_2+...+a_n}{n}=\frac{\alpha }{n}$

H.M. of $a_1,a_2,...,a_n$ is $\frac{n}{({\frac{1}{a}}_1+{\frac{1}{a}}_2+...+{\frac{n}{a}}_n)}=\frac{n}{{\sum }_{i=1}^n{\frac{1}{a}}_i}=\frac{n}{{\sum }_{i=1}^n{a}_i^{-1}}$

Therefore, $A.M.\ge H.M.⟹\frac{\alpha }{n}\ge \frac{n}{{\sum }_{i=1}^n{a}_i^{-1}}$

$⟹{\sum }_{i=1}^n{a}_i^{-1}\le \frac{n^2}{\alpha }$

therefore, the least value of ${\sum }_{i=1}^n{a}_i^{-1}$ is $\frac{n^2}{\alpha }$