Inequality Proof Given that $$x>1$$Prove$$\dfrac{x^{\dfrac{1}{\ln 2}}\ln \left(1+\dfrac{1}{x}\right)}{\ln (1+x)}>1$$
I found that $\ln 2$ is the largest number for this inequality to hold via some numerical verifications. However couln't find a way to prove it. Derivatives are really messy. Much appreciated if anyone can give me an idea on the proof.
 A: By setting $x=e^t$, we want to show that for any $t>0$
$$ e^{\frac{t}{\log 2}} \log(1+e^{-t})> \log(1+e^t) \tag{1}$$
or
$$ \frac{t}{\log 2}>\log\log(1+e^{t})-\log\log(1+e^{-t}).\tag{2} $$
In order to prove $(2)$ it is enough to show that 
$$ \forall t>0,\qquad \frac{1}{1+e^t}\cdot\frac{1}{\log(1+e^{-t})}+\frac{e^t}{1+e^t}\cdot\frac{1}{\log(1+e^t)}<\frac{1}{\log 2}\tag{3} $$
then integrate with respect to $t$. On the other hand $(3)$ is equivalent to:
$$ \forall u>1,\qquad \frac{1}{1+u}\cdot\frac{1}{\log\left(1+\frac{1}{u}\right)}+\frac{u}{1+u}\cdot\frac{1}{\log(1+u)}<\frac{1}{\log 2}\tag{4} $$
or, by setting $g(u)=\frac{u}{(1+u)\log(u+1)}$, to:
$$ \forall u>1,\qquad g(u)+g\left(\frac{1}{u}\right) < \frac{1}{\log 2}\tag{5} $$
that can be written in  the following nice form:
$$  \forall u>1,\qquad \int_{0}^{1}\frac{ds}{(1+u)^s}+\int_{0}^{1}\frac{ds}{(1+\frac{1}{u})^s}=\int_{0}^{1}\frac{1+u^s}{(1+u)^s}\,ds<\frac{1}{\log 2}\tag{6} $$
that is straightforward to prove: for any $s\in(0,1)$, the function $h_s(u)=\frac{1+u^s}{(1+u)^s}$ is decreasing on the interval $u\in[1,+\infty)$, hence:
$$ \int_{0}^{1}h_s(u)\,ds < \int_{0}^{1}h_s(1)\,ds = \int_{0}^{1}\frac{ds}{2^{s-1}}=\frac{1}{\log 2}\tag{7}$$
as wanted.
