Is there a clean way to show $\frac{\log(x-\tfrac12\log x)}{\log(x+\tfrac12\log x)}>1-\tfrac1x$ for all $x>1$? 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?
 A: 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 $x \approx 0.825$.
A: 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.
