Is it true that $\ln(x)+\frac{1}{\ln(x)}>\ln(x+1)$ for $x>m$ I was wondering if it is true for $x>m$  where $m$ is a constant, we have:
$\ln(x)+\frac{1}{\ln x} >\ln(x+1)$
If we plot the figure $\ln(​x)(\ln(​x+​1)-​\ln(​x))$ in google we see:

As you can see the line is decreasing when $x>6$ which highlights the fact that:
$$1>\ln(​x)(\ln(​x+​1)-​\ln(​x))$$
I want a formal proof for this fact.
Thanks.
 A: Hint for large enough (and positive) $x$
$$\ln{(x+1)}=\ln{x}+\ln{\left(1+\frac{1}{x}\right)}=\ln{x}+\frac{\ln{\left(1+\frac{1}{x}\right)^{x}}}{x}\sim \\
\ln{x}+\frac{1}{x}<\ln{x}+\frac{1}{\ln{x}}$$
With more technical details, we know that $\lim\limits_{x\rightarrow\infty}\ln{\left(1+\frac{1}{x}\right)^{x}}=1 \Rightarrow \ln{\left(1+\frac{1}{x}\right)^{x}}<1+\varepsilon$, for all the $x$ from some value onwards, thus
$$\ln{x}+\frac{\ln{\left(1+\frac{1}{x}\right)^{x}}}{x}<\ln{x}+\frac{1+\varepsilon}{x}$$
and because $\lim\limits_{x\rightarrow\infty}\frac{x}{\ln{x}}\rightarrow\infty$ then $\frac{x}{\ln{x}} > 1+\varepsilon$ from some $x$ onwards.
A: You are looking for an $m$ such that if $x > m$ one has
$${1 \over \ln x} > \ln(x+1) - \ln(x)$$
By the mean value theorem, $\ln(x + 1) - \ln (x) = 1/y$ for some $y$ between $x$ and $x + 1$. So we have
$$\ln(x + 1) - \ln (x) < {1 \over x}$$
Hence it suffices to find an $m$ such that if $x > m$ one has
$${1 \over \ln x} > {1 \over x}$$
Equivalently (assuming $m > 1$), one wants
$$\ln x < x$$
This holds for all $x > 1$, which can be shown by various elementary means. See this answer for details: How to prove $\ln x<x$?
A: Attempt of a proof:
Let x > 1.
MVT:
$\dfrac{\log (x+1)-\log (x)}{1}=$
$(\log)'(t) =1/t$,  where $x <t<x+1$.
The inequality reads:
$(\log x)^{-1}> 1/t$, or
$ \dfrac{\log x}{t}<1$.
We have:
$\dfrac{\log x}{t} \lt \dfrac{ \log x }{x} <1$.
A: For $x<e^2-1$ we have $$\ln(1+x)<\ln(1+e^2-1)=2\le \ln x+\dfrac{1}{\ln x}$$For $x\ge e^2-1$ we know that $\ln x<x$ and $\ln(1+u)<u$ therefore by substituting $u=\dfrac{1}{x}$ we obtain $$\dfrac{1}{\ln x}>\dfrac{1}{x}>\ln(1+\dfrac{1}{x})=\ln(1+x)-\ln x$$or $$\dfrac{1}{\ln x}+\ln x>\ln (x+1)$$so our proof is complete.
