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November 16th, 2017, 10:25 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27  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, 11:21 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,039 Thanks: 1063 
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. 
November 16th, 2017, 11:22 PM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27  
November 16th, 2017, 11:40 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,039 Thanks: 1063  $\large \sqrt[2016]{\prod \limits _{j=1}^{2017}Y_j} + \sqrt[2016]{\prod \limits _{j=1}^{2017}(1Y_j)} \leq \sqrt[2016]{\prod \limits _{j=1}^{2017}(Y_j+(1Y_j)}=\sqrt[2016]{\prod \limits _{j=1}^{2017} 1} = 1$


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inequality, prod 
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