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Question Number 36217 by Rio Mike last updated on 30/May/18

$Suppose a_{1}, . . ., a_{n}, are non - negative$ $reals such that S = a_{1} + . . . + a_{n} <$ $p r o o f t h a t$ $1 + S ⩽ (1 + a_{1}) ._{. . .} . (1 + a_{n}) ⩽ \frac{1}{1 - s}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18

	$(1 + a_{1}) (1 + a_{2}) = 1 + a_{1} + a_{2} + a_{1} a_{2} > 1 + a_{1} + a_{2}$ $(1 + a_{1}) (1 + a_{2}) (1 + a_{3}) > 1 + a_{1} + a_{2} + a_{3}$ $. . . . .$ $. . . . . (1 + a_{1}) (1 + a_{2}) . . . . (1 + a_{n}) > 1 + a_{1} + a_{2} + . . + a_{n}$ $w h e n a_{1} + a_{2} + a_{3} + . . . + a_{n} = S$ $(1 + a_{1}) (1 + a_{2}) . . (1 + a_{n}) > 1 + S p r o v e d$ $c o n t d$ $l e t s_{2} = a_{1} + a_{2}$ $(1 + a_{1}) (1 + a_{2}) = 1 + a_{1} + a_{2} + a_{1} a_{2} = 1 + s_{2} + a_{1} a_{2}$ $(1 - a_{1}) (1 - a_{2}) = 1 - (a_{1} + a_{2}) + a_{2} a_{2} = 1 - s_{2} + a_{1} a_{2}$ $1 + s_{2} > 1 - s_{2}$ $w a i t p l s$
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