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In https://arxiv.org/pdf/0904.2952.pdf they say on p. 25 that $x(\log(x)-1)+1\geq \frac{1}{5}(x-1)^2$ in a neighborhood of $x=1$. In another paper that I don't have the link to they say that it's $x(\log(x)-1)+1\geq \frac{1}{4}(x-1)^2$ and it follows from a Taylor expansion. I can't figure out why.

mlstudent
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  • If you differentiate the function $f(x) = x(\log(x)-1)+1-\frac{1}{4}(x-1)^2$ twice and substitute $1$, you will find that $1$ is a local minima for this function. Thus the inequality holds locally. – fGDu94 Jan 26 '20 at 02:10

2 Answers2

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Let $f(x)=x(\ln x-1)+1-\frac{1}{4}(x-1)^2$. Then,

$$f(x)=(1+t)\ln(1+t)-t-\frac14t^2$$

where $t=x-1$. Taylor expand $\ln(1+t) = t -\frac12t^2$ to express $f(x)$ around $t= 0$,

$$f(x) = (1+t)(t-\frac12t^2)-t-\frac14t^2=\frac14t^2=\frac14(x-1)^2\ge0$$

Thus, in the neighborhood of $x=1$, $f(x)\ge0$, or

$$x(\ln x-1)+1\ge\frac{1}{4}(x-1)^2$$

Quanto
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Doing a Taylor series expansion around $1$, we have $$ x(\log x - 1)+1 = \frac{1}{2}(x-1)^2 + o((x-1)^2)) $$ Put differently, $$ \lim_{x\to 1} \frac{x(\log x - 1)+1}{(x-1)^2} = \frac{1}{2} $$ so that, for every $\varepsilon > 0$, there exists $\delta>0$ such that $$ \frac{x(\log x - 1)+1}{(x-1)^2} > \frac{1}{2} - \varepsilon $$ as long as $|x-1|< \delta$. This implies both statements you mention.

Clement C.
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  • Can you show how you do that Taylor series expansion? I am getting something different using https://math.stackexchange.com/a/585189/56922. In particular I don't have the $\frac{1}{2}$ coefficient and . – mlstudent Jan 26 '20 at 04:11
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    @mlstudent Are you sure you are doing it for $x$ around $1$, not $0$? A good way to proceed is to set $h=x-1$ (so that $h\to0$) and do the Taylor expansion of $$(1+h)(\log(1+h)-1)+1$$ for $h$ around $0$. – Clement C. Jan 26 '20 at 06:14