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If a(1)gt0 for all i=1,2,…..,n. Then, th...

If `a_(1)gt0` for all `i=1,2,…..,n`. Then, the least value of
`(a_(1)+a_(2)+……+a_(n))((1)/(a_(1))+(1)/(a_(2))+…+(1)/(a_(n)))`, is

A

`n^(2)`

B

2n

C

n

D

`(1)/(n)`

Text Solution

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The correct Answer is:
To find the least value of the expression \[ E = (a_1 + a_2 + \ldots + a_n) \left( \frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n} \right), \] given that \( a_i > 0 \) for all \( i = 1, 2, \ldots, n \), we will use the Cauchy-Schwarz inequality. ### Step 1: Apply the Cauchy-Schwarz Inequality According to the Cauchy-Schwarz inequality, we have: \[ \left( \sum_{i=1}^{n} a_i \right) \left( \sum_{i=1}^{n} \frac{1}{a_i} \right) \geq n^2. \] ### Step 2: Set Up the Expression Thus, we can write: \[ E = (a_1 + a_2 + \ldots + a_n) \left( \frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n} \right) \geq n^2. \] ### Step 3: Conclusion The least value of \( E \) is therefore: \[ \text{Minimum value of } E = n^2. \]
To find the least value of the expression \[ E = (a_1 + a_2 + \ldots + a_n) \left( \frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n} \right), \] given that \( a_i > 0 \) for all \( i = 1, 2, \ldots, n \), we will use the Cauchy-Schwarz inequality. ...
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