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Mar 11, 2016 at 17:43 vote accept Thisquestionisreallyhard
Mar 10, 2016 at 13:46 history edited Venkataramana CC BY-SA 3.0
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Mar 10, 2016 at 12:22 comment added Brendan McKay Is there a $k$ that should be an $l$ in Maclaurin's inequality?
Mar 10, 2016 at 11:00 answer added Fedor Petrov timeline score: 6
Mar 10, 2016 at 8:26 comment added Fedor Petrov In other words, if, say, $i=n$ and $A_{\lambda}=(x_1,\dots,x_n)$ we have to prove $(n-1)\sigma_k(x_1,\dots,x_{n-1})\leqslant \sigma_1(x_1,\dots,x_{n-1})\sigma_{k-1}(x_1,\dots,x_{n-1})$, this is standard if $(x_1,\dots,x_{n-1})\in \Gamma_k$, but the problem is that we are not given this.
Mar 10, 2016 at 6:30 history edited Thisquestionisreallyhard CC BY-SA 3.0
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Mar 10, 2016 at 6:24 history asked Thisquestionisreallyhard CC BY-SA 3.0