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Dec 27, 2023 at 8:01 comment added Kalas678 Got it. Holder gives, $a \geq 2 \sqrt{a-1}$.
Dec 26, 2023 at 20:25 comment added Kalas678 Which version of Holder's inequality you are using here? @XiaoshengMu
Apr 22, 2016 at 4:44 answer added Xiaosheng Mu timeline score: 3
Apr 22, 2016 at 4:19 comment added Fan Zheng @BobbyGrizzard NOT the OP LOL
Apr 22, 2016 at 3:19 vote accept Xiaosheng Mu
Apr 21, 2016 at 11:58 comment added Bobby Grizzard @FanZheng yes. Here is one effective result: ams.org/journals/mcom/1996-65-213/S0025-5718-96-00664-3/… edit: and here is the one I was probably looking at yesterday, due to Wu and Mu (Quanwu Mu, not the OP!): sciencedirect.com/science/article/pii/S0022314X12001989
Apr 21, 2016 at 5:50 comment added Fan Zheng @BobbyGrizzard Is the "with finitely many exceptions" part effective?
Apr 20, 2016 at 22:58 comment added Gerry Myerson On the hypothesis that $\alpha$ is algebraic of degree $d$, the value $\alpha=2$ can only arise in the case $d=1$, so the equality $N(\alpha)=2^d$ is achieved only for $d=1$, $\alpha=2$.
Apr 20, 2016 at 22:53 answer added Gypsum timeline score: 5
S Apr 20, 2016 at 21:37 history suggested Bobby Grizzard
added some tags
Apr 20, 2016 at 21:15 review Suggested edits
S Apr 20, 2016 at 21:37
Apr 20, 2016 at 21:13 comment added Bobby Grizzard Interesting question. I assume you're familiar with work of Smyth, Flammang, etc. on lower bounds for the Mahler measure of totally positive algebraic integers? (These give a lower bound of $C=1.722...$ for $M(\alpha)^{1/d}$ with finitely many exceptions, but valid for all totally positive integers $\alpha$, not just those with all conjugates $>1$).
Apr 20, 2016 at 20:40 review First posts
Apr 20, 2016 at 21:32
Apr 20, 2016 at 20:35 history asked Xiaosheng Mu CC BY-SA 3.0