Let a_i >0, ∀_i =1, 2, 3, …2016 If (a_1 a_2 …a_(2016) )^(1/(2016)) =2 then (1+a_1 )(1+a_2 )…(1+a


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Question Number 55904 by gunawan last updated on 06/Mar/19

$Let a_{i} > 0, \forall_{i} = 1, 2, 3, \dots 2016$ $If {(a_{1} a_{2} \dots a_{2016})}^{\frac{1}{2016}} = 2$ $then$ $(1 + a_{1}) (1 + a_{2}) \dots (1 + a_{2016}) ⩾ . . .$
	Answered by mr W last updated on 06/Mar/19

	$f o r a, b > 0 w e h a v e \frac{a + b}{2} ⩾ \sqrt{a b}$ $(1 + a_{1}) (1 + a_{2}) \dots (1 + a_{2016})$ $⩾ 2 \sqrt{a_{1}} \times 2 \sqrt{a_{2}} \times . . . \times 2 \sqrt{a_{2016}}$ $= 2^{2016} \sqrt{a_{1} a_{2} . . . a_{2016}}$ $= 2^{2016} {(2^{2016})}^{\frac{1}{2}}$ $= 2^{2016} 2^{1008}$ $= 2^{3024}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19

	$a_{1} a_{2} a_{3} . . . a_{n} = 2^{n}$ $n = 2016$ $\frac{1 + a_{1}}{2} ⩾ \sqrt{1 \times a_{1}}$ $(1 + a_{1}) (1 + a_{2}) . . . (1 + a_{n}) ⩽ 2^{} \times {(a_{1})}^{\frac{1}{2}} \times 2 {(a_{2})}^{\frac{1}{2}} . . 2 \times {(a_{n})}^{\frac{1}{n}}$ $d o ⩾ 2^{n} {(a_{1} a_{2} . . . a_{n})}^{\frac{1}{2}}$ $d o ⩾ 2^{n} {(2^{n})}^{\frac{1}{2}}$ $d 0 ⩾ 2^{n + \frac{n}{2}}$ $s o r e q u i r e d a n s w e r i s 2^{2016 + 1008}$ $(1 + a_{1}) (1 + a_{2}) . . . (1 + a_{2016}) ⩾ 2^{3024}$
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