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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 1, 2017 at 19:23 history edited Ludwig CC BY-SA 3.0
Added link to follow-up problem.
Mar 24, 2016 at 9:48 vote accept Ludwig
Mar 23, 2016 at 11:47 answer added Fedor Petrov timeline score: 9
Mar 21, 2016 at 21:36 answer added Fedor Petrov timeline score: 6
Mar 20, 2016 at 22:07 comment added Fedor Petrov It looks probable (and holds for $n=2$).
Mar 20, 2016 at 22:02 history edited Ludwig CC BY-SA 3.0
Minor corrections
Mar 19, 2016 at 7:49 comment added Fedor Petrov For $N=2$ this reduces (If I am not mistaken) to equivalent inequality $(\sum a_i^2)(\sum b_i^2)+(\sum a_ib_i)^2\geqslant \sqrt{(\sum a_i^4)(\sum b_i^4)}+\sum a_i^2b_i^2$ for arbitrary reals $a_i,b_i$.
Mar 18, 2016 at 14:31 comment added Iosif Pinelis Without loss of generality, $X^{1/2}$ is a diagonal matrix with positive $c_1,\dots,c_n$ on the diagonal such that $\sum_{k=1}^n c_k^2=1$. Letting then $v_i=[v_{i1},\dots,v_{in}]^T$ and $y_{ik}:=v_{ik}\sqrt{c_k}(\sum_{q=1}^n c_q^2v_{iq}^2)^{-1/2}$, one rewrites $(*)$ as $\prod_{i=1}^N\sum_{j=1}^N(\sum_{k=1}^n y_{ik}y_{jk})^2\ge N^N$ given that $\sum_{k=1}^n c_k^2=\sum_{k=1}^n c_k y_{ik}^2=1$ for all $i=1,\dots,N$. Then one could perhaps use Lagrange multipliers.
Mar 18, 2016 at 11:10 history edited Ludwig CC BY-SA 3.0
Added a comment on the problem.
Mar 17, 2016 at 23:16 vote accept Ludwig
Mar 17, 2016 at 23:24
Mar 14, 2016 at 7:44 history edited Ludwig CC BY-SA 3.0
Corrected error
Mar 13, 2016 at 22:11 history edited Ludwig CC BY-SA 3.0
Corrected a typo
Mar 13, 2016 at 20:50 history asked Ludwig CC BY-SA 3.0