Why is $x(\log(x)-1)+1\geq \frac{1}{4}(x-1)^2$ in a neighborhood of $x=1$?

Question

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.

Answer 1

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.

Answer 2

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

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