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Jan 31, 2017 at 20:29 comment added Pietro Majer On the technical side, the idea for the construction is: if you have two or more solutions of the weak inequalities ($\ge$), maybe not decreasing, you can sum them, and if one of them verifies a strict inequality at $\bf i$, so does the sum. Also, any affine sequence is a trivial solution of the weak inequalities (with $=$), but it can make the sum monotone if its slope is large enough. This way the problem is somehow reduced to easier ones.
Jan 31, 2017 at 20:28 comment added Pietro Majer You're welcome! Actually, I don't think I understand it more than you (I was amazed by the regular structure revealed by what seemed just a technical question; to me this is a clear sign that it comes from beautiful and deep mathematics :) )
S Jan 31, 2017 at 20:01 history suggested Abhishek Parab CC BY-SA 3.0
Corrected a typo.
Jan 31, 2017 at 19:47 review Suggested edits
S Jan 31, 2017 at 20:01
Jan 31, 2017 at 19:28 history bounty ended Abhishek Parab
Jan 31, 2017 at 19:28 vote accept Abhishek Parab
Jan 31, 2017 at 19:28 comment added Abhishek Parab This is an amazing answer! I feel somewhat like Cantor - I see the proof but I don't understand how or why. Your answer is insightful but it'll take me time to absorb it and attempt it in other cases. This question was for $SL(n)$. Thank you!
Jan 31, 2017 at 14:07 comment added Pietro Majer $\phantom{...}$...done!
Jan 31, 2017 at 14:06 history edited Pietro Majer CC BY-SA 3.0
added 1397 characters in body
Jan 30, 2017 at 21:48 comment added Pietro Majer of course, let me just write it down nicely ;)
Jan 30, 2017 at 17:14 comment added Abhishek Parab Thank you for the answer. I made a mistake (explained above). I will award your bounty after few days so that the question will get attention. I would really appreciate if you could try this inequality too. The $b_i$'s will depend on $\tau$ now.
Jan 29, 2017 at 14:37 history edited Pietro Majer CC BY-SA 3.0
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Jan 29, 2017 at 11:13 history answered Pietro Majer CC BY-SA 3.0