If ai-0 - i-1-2-n- Then- the least value of a1-a2-an 1a1- 1a2- 1an- is

The correct option is C n^2 Since AM≥ HM so we apply AM and HM on (a_1,a_2,.....a_n) ⇒ a_1+a_2+......+a_nn≥n1a_1+1a_2+.....+1a_n ...

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Standard XII

Mathematics

AM,GM,HM Inequality

If ai>0 ∀ i...

Question

If $a_{i} > 0 \forall i = 1, 2, . . . ., n$ . Then, the least value of $(a_{1} + a_{2} . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . + \frac{1}{a_{n}})$ , is

2 n

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n^{2}

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n

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\frac{1}{n}

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Solution

The correct option is C $n^{2}$
Since $A M \geq H M$
so we apply AM and HM on $(a_{1}, a_{2}, . . . . . a_{n})$
$\Rightarrow$ $\frac{a_{1} + a_{2} + . . . . . . + a_{n}}{n} \geq \frac{n}{\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . . . + \frac{1}{a_{n}}}$

$\Rightarrow$ $(a_{1} + a_{2} + . . . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . . . + \frac{1}{a_{n}}) \geq n^{2}$

Hence, option B is correct.

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If ai>0∀i=1,2,....,n. Then, the least value of (a1+a2...+an)(1a1+1a2+...+1an), is

The correct option is C n2Since AM≥HMso we apply AM and HM on (a1,a2,.....an)⇒ a1+a2+......+ann≥n1a1+1a2+.....+1an⇒(a1+a2+.....+an)(1a1+1a2+.....+1an)≥n2Hence, option B is correct.

If $a_{i} > 0 \forall i = 1, 2, . . . ., n$ . Then, the least value of $(a_{1} + a_{2} . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . + \frac{1}{a_{n}})$ , is