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 November 16th, 2017, 11:25 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 256 Thanks: 28 Prod inequality For $\displaystyle \; Y_1 , Y_2 , ... , Y_{2017} \in (0,1)$ Confirm $\displaystyle \; \sqrt[2016]{\prod_{j=1}^{2017} Y_j } +\sqrt[2016]{\prod_{j=1}^{2017} (1- Y_j ) } < 1$
 November 16th, 2017, 12:21 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,169 Thanks: 1141 This isn't true. The less than should be less than or equal to. Just note that the triangle inequality holds for $\Large \sqrt[2016]{x}$ and apply it. Thanks from idontknow
November 17th, 2017, 12:22 AM   #3
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Quote:
 Originally Posted by romsek This isn't true. The less than should be less than or equal to. Just note that the triangle inequality holds for $\Large \sqrt[2016]{x}$ and apply it.
Can you post in short terms , using triangle inequality ?

November 17th, 2017, 12:40 AM   #4
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Quote:
 Originally Posted by idontknow Can you post in short terms , using triangle inequality ?
$\large \sqrt[2016]{\prod \limits _{j=1}^{2017}Y_j} + \sqrt[2016]{\prod \limits _{j=1}^{2017}(1-Y_j)} \leq \sqrt[2016]{\prod \limits _{j=1}^{2017}(Y_j+(1-Y_j)}=\sqrt[2016]{\prod \limits _{j=1}^{2017} 1} = 1$

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