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November 16th, 2017, 11:25 AM   #1
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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 $
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November 16th, 2017, 12:21 PM   #2
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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.
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November 17th, 2017, 12:22 AM   #3
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Quote:
Originally Posted by romsek View Post
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 ?
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November 17th, 2017, 12:40 AM   #4
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Quote:
Originally Posted by idontknow View Post
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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