Question

If $x_i>0,i=1,2,3,...n$ then $\left(x_1+x_2+...+x_n\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}\right)$ is equal to

• A. $n^2$
• B. $\ge n^2$
• C. $\le n^2$
• D. None of these

Answer 1

The correct option is B

$\ge n^2$

Explanation for the correct option.

Find the inequality:

The arithmetic mean of the terms $x_1,x_2,...,x_n$ is given as: $A=\frac{x_1+x_2+...+x_n}{n}$.

The harmonic mean of the terms $x_1,x_2,...,x_n$ is given as: $H=\frac{n}{{\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}+...+{\displaystyle \frac{1}{x_n}}}$.

Now, it is known that arithmetic mean is always greater than or equal to harmonic mean. So $A\ge H$.

$$\begin{gathered} \frac{x_1+x_2+...+x_n}{n}\ge \frac{n}{{\displaystyle \frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}} \\ \Rightarrow \left(x_1+x_2+...+x_n\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}\right)\ge n^2 \end{gathered}$$

Thus the value of $\left(x_1+x_2+...+x_n\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}\right)$ is greater than or equal to $n^2$.

Hence, the correct option is B.