Suppose-a-1-a-n-are-non-negative-reals-such-that-S-a-1-a-n-lt-proof-that-1-S-1-a-1-1-a-n-1-1-s-

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

Suppose a_{1}, \dots, a_{n}, are non - negative

reals such that S = a_{1} + \dots + a_{n} <

p r o o f t h a t

1 + S ⩽ (1 + a_{1}) ._{\dots} . (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}

\dots . .

\dots . . (1 + a_{1}) (1 + a_{2}) \dots . (1 + a_{n}) > 1 + a_{1} + a_{2} + . . + a_{n}

w h e n a_{1} + a_{2} + a_{3} + \dots + 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