Question

If \[{x_i} > 0\], \[i = 1,2,3, \cdots ,n\] then find the minima value of \[\left( {{x_1} + {x_2} + \cdots + {x_n}} \right)\left( {\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}} \right)\].
A. \[{n^2}\]
B. \[ \ge {n^2}\]
C. \[ \le {n^2}\]
D. None of these

Answer 1

Hint:We know the arithmetic mean of \[{x_1},{x_2}, \cdots ,{x_n}\] is \[\dfrac{{{x_1} + {x_2} + \cdots + {x_n}}}{n}\] and the harmonic mean of \[\dfrac{1}{{{x_1}}},\dfrac{1}{{{x_2}}}, \cdots ,\dfrac{1}{{{x_n}}}\] is \[\dfrac{n}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}}}\]. Then put the value of arithmetic mean and harmonic mean in \[A.M \ge H.M\]. From the above relation we can get value of \[\left( {{x_1} + {x_2} + \cdots + {x_n}} \right)\left( {\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}} \right)\].

Formula Used:
The arithmetic mean of \[{x_1},{x_2}, \cdots ,{x_n}\] is \[\dfrac{{{x_1} + {x_2} + \cdots + {x_n}}}{n}\].
The harmonic mean of \[\dfrac{1}{{{x_1}}},\dfrac{1}{{{x_2}}}, \cdots ,\dfrac{1}{{{x_n}}}\] is \[\dfrac{n}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}}}\].
The relation between the arithmetic mean and harmonic mean is \[A.M \ge H.M\].

Complete step by step solution:
The arithmetic mean of \[{x_1},{x_2}, \cdots ,{x_n}\] is \[\dfrac{{{x_1} + {x_2} + \cdots + {x_n}}}{n}\].
The harmonic mean of \[\dfrac{1}{{{x_1}}},\dfrac{1}{{{x_2}}}, \cdots ,\dfrac{1}{{{x_n}}}\] is \[\dfrac{n}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}}}\].
Now putting the value of arithmetic mean and harmonic mean in \[A.M \ge H.M\]

\[\dfrac{{{x_1} + {x_2} + \cdots + {x_n}}}{n} \ge \dfrac{n}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}}}\]
Cross multiply
\[ \Rightarrow \left( {{x_1} + {x_2} + \cdots + {x_n}} \right)\left( {\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}} \right) \ge n \cdot n\]

\[ \Rightarrow \left( {{x_1} + {x_2} + \cdots + {x_n}} \right)\left( {\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \cdots + \dfrac{1}{{{x_n}}}} \right) \ge {n^2}\]

Hence the correct option is option B.

Note:Students often confused with the relation \[A.M \ge G.M \ge H.M\] and \[A.M > G.M > H.M\]. We apply the formula \[A.M \ge G.M \ge H.M\] if all numbers are positive and apply the formula \[A.M > G.M > H.M\], if all numbers distinct.