Is it true that $\ln(x)+\frac{1}{\ln(x)}>\ln(x+1)$ for $x>m$

Question

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.

Answer 1

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 $\ln x<x$?

Answer 2

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.

Answer 3

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:

$\dfrac{\log x}{t} \lt \dfrac{ \log x }{x} <1$.

Answer 4

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.