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Apr 22, 2016 at 17:33 comment added Xiaosheng Mu @FanZheng Thanks for your advice :)
Apr 22, 2016 at 13:11 comment added Gypsum One can usually extend this approach a certain amount, but not as far as the limit. The fact that equality is achieved at $x = (5+\sqrt{5})/2$ with the optimal bound is related to the fact that $x - 1$ and $x - 2$ are both units. But there is no reason this had to happen. BTW, to improve the bound beyond this point (with the exceptions $x = 2$ and $(5 \pm \sqrt{5})/2$, you simply need to add a very small multiple of the term $\log(x^2 - 5 x + 5)$.
Apr 22, 2016 at 6:25 comment added Fan Zheng Welcome to mathoverflow! This is really a great question, but as you are new here, please be informed that the answers are just "answers"; new questions included in answers may not get enough attention. If you want to probe further, please post a separate question; you may include a link back to this question.
Apr 22, 2016 at 4:44 history answered Xiaosheng Mu CC BY-SA 3.0