Is there a clean way to show $\frac{\log(x-\tfrac12\log x)}{\log(x+\tfrac12\log x)}>1-\tfrac1x$ for all $x>1$?

Question

In writing a paper I had to show that the inequality holds for large enough $x$, which is easy, but I ended up being pretty sure it holds for all $x>1$, so I would like to include the proof of the complete result. However, it seems stronger than $\log(1-\varepsilon)>-\varepsilon-\frac{\varepsilon^2}{2(1-\varepsilon)}$ ($0<\varepsilon<1$). Many manipulations and the substitution $y=\tfrac{\log x}{x}$ do not seem to simplify the task, and I wouldn't want to complicate the proof unnecessarily. Can some trick or general inequality wrap this up?

Answer 1

If we use composition of Taylor series, for large values of $x$, we have $$\frac{\log \left(x-\frac{1}{2}\log (x)\right)}{\log \left(x+\frac{1}{2}\log (x)\right)}=1-\frac{1}{x}+\frac{1}{2\, x^2}-\frac{2 \log ^2\left({x}\right)+3 \log \left({x}\right)+6}{24\, x^3}+O\left(\frac{1}{x^4}\right)$$ More difficult to see is when $x$ is close to $1$ since $$\frac{\log \left(x-\frac{1}{2}\log (x)\right)}{\log \left(x+\frac{1}{2}\log (x)\right)}=\frac{1}{3}+\frac{7 }{18}(x-1)-\frac{13}{72} (x-1)^2+\frac{125 }{2592}(x-1)^3+O\left((x-1)^4\right)$$ but $$\Delta=\frac{\log \left(x-\frac{1}{2}\log (x)\right)}{\log \left(x+\frac{1}{2}\log (x)\right)}-\left(1-\frac 1x\right)=\frac{1}{3}-\frac{11 }{18}(x-1)+\frac{59}{72} (x-1)^2-\frac{2467 }{2592}(x-1)^3+O\left((x-1)^4\right)$$ Expanding the rhs, we get $$\Delta=\frac{580601}{155520}-\frac{89189 }{9720}x+\frac{253511 }{25920}x^2-\frac{97633 }{19440}x^3+\cdots$$ which shows a real root around $x \approx 0.825$.

Answer 2

The following approach is not particularly elegant, but it should work. It is based on the inequalities \begin{align} \frac{t-1}{\log(t)} &> \frac{1+t-(t-1)^2}{2} \, , \, 1<t<2 \, , \tag{1} \\ \frac{1}{x} &> \left(x^{1/x}-1\right)^2 \, , \, x > 1 \, , \tag{2} \end{align} which are not too hard to verify.

We take $x>1$, let $t=x^{1/x} \in (1,\mathrm{e}^{1/\mathrm{e}}]$ in $(1)$ and use $(2)$ to get $$ x \frac{x^{1/x}-1}{\log(x)} > \frac{1+x^{1/x}-(x^{1/x}-1)^2}{2} > \frac{1+x^{1/x}-\frac{1}{x}}{2} \, . $$ Rearranging we find that $$ x - \frac{\log(x)}{2} > x^{-1/x} \frac{(x-1)\left(x+\frac{\log(x)}{2}\right)+x}{x} > \left(x+\frac{\log(x)}{2}\right)^{1-\frac{1}{x}} $$ holds. The last step follows from the weighted AM-GM inequality $$ \left(a^{x-1} b\right)^{1/x} < \frac{(x-1) a + b}{x} \, , \, a,b>0 \, $$ or simply Bernoulli's inequality. We can now take the logarithm of both sides to conclude the proof.