1. ## Proving logarithmic inequality

Hello, everyone!
I'm stuck at the following problem.

Let us have an equation that I intend to solve with some numerical method:

$\displaystyle |x (e^{1/x} - 1) - 1| = \epsilon$, where $\displaystyle \epsilon$ is a small number and $\displaystyle x$ is positive.

First off, I want to get rid of the modulus sign to avoid solving a system of equations. To do so, I need to show that
$\displaystyle \forall x > 0: x (e^{1/x} - 1) > 1$
(This fact I established empirically and want to prove. The plot can be seen here: 1 is actually a limit).

The inequality is equivalent to the following one:

$\displaystyle e^{1/x} > \frac{1}{x} + 1$
$\displaystyle {\frac{1}{x}} > \ln{\left(\frac{1}{x} + 1\right)}$

Denote $\displaystyle 1 + \frac{1}{x} = u$ ($\displaystyle u > 1$), then

$\displaystyle u > 1 + \ln{u}$

And here's where I'm out of ideas. How do I analytically prove the last inequality for all $\displaystyle u > 1$?

Any thoughts would be very appreciated!

PS I probably should have posted this in Calculus subforum, my bad.

2. Originally Posted by fiontie
Hello, everyone!
I'm stuck at the following problem.

Let us have an equation that I intend to solve with some numerical method:

$\displaystyle |x (e^{1/x} - 1) - 1| = \epsilon$, where $\displaystyle \epsilon$ is a small number and $\displaystyle x$ is positive.

First off, I want to get rid of the modulus sign to avoid solving a system of equations. To do so, I need to show that
$\displaystyle \forall x > 0: x (e^{1/x} - 1) > 1$
(This fact I established empirically and want to prove. The plot can be seen here: 1 is actually a limit).

The inequality is equivalent to the following one:

$\displaystyle e^{1/x} > \frac{1}{x} + 1$
$\displaystyle {\frac{1}{x}} > \ln{\left(\frac{1}{x} + 1\right)}$

Denote $\displaystyle 1 + \frac{1}{x} = u$ ($\displaystyle u > 1$), then

$\displaystyle u > 1 + \ln{u}$

And here's where I'm out of ideas. How do I analytically prove the last inequality for all $\displaystyle u > 1$?

Any thoughts would be very appreciated!

PS I probably should have posted this in Calculus subforum, my bad.
Are familiar with Taylor's theroem? Because if so it's a direct application of it to show $\displaystyle e^{1/x} > \frac{1}{x} + 1$.

3. Originally Posted by fiontie
Hello, everyone!
I'm stuck at the following problem.

Let us have an equation that I intend to solve with some numerical method:

$\displaystyle |x (e^{1/x} - 1) - 1| = \epsilon$, where $\displaystyle \epsilon$ is a small number and $\displaystyle x$ is positive.

First off, I want to get rid of the modulus sign to avoid solving a system of equations. To do so, I need to show that
$\displaystyle \forall x > 0: x (e^{1/x} - 1) > 1$
(This fact I established empirically and want to prove. The plot can be seen here: 1 is actually a limit).

The inequality is equivalent to the following one:

$\displaystyle e^{1/x} > \frac{1}{x} + 1$

$\displaystyle {\frac{1}{x}} > \ln{\left(\frac{1}{x} + 1\right)}$

Denote $\displaystyle 1 + \frac{1}{x} = u$ ($\displaystyle u > 1$), then

$\displaystyle u > 1 + \ln{u}$

And here's where I'm out of ideas. How do I analytically prove the last inequality for all $\displaystyle u > 1$?

Any thoughts would be very appreciated!

PS I probably should have posted this in Calculus subforum, my bad.

Define $\displaystyle f(u)=\ln u-u+1$ $\displaystyle \Longrightarrow f'(u)=\frac{1}{u}-1<0\,\,\,\forall u>1\Longrightarrow$ the function is monotone decreasing , so $\displaystyle 0=f(1)> f(u)\,\,\forall u>1$ and we're done

Tonio

Ps. Very nice and good idea to define u as you did.

4. @tonio: elegance is easy

Thank you guys for the quick responses, they saved my day!