Proving logarithmic inequality

**fiontie** · April 12th 2010, 10:09 AM

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

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

$|x (e^{1/x} - 1) - 1| = \epsilon$ , where $\epsilon$ is a small number and $x$ 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
$\forall x > 0: x (e^{1/x} - 1) > 1$
(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:

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

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

$u > 1 + \ln{u}$

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

Any thoughts would be very appreciated!

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

**mathman88** · April 12th 2010, 11:32 AM

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:

$|x (e^{1/x} - 1) - 1| = \epsilon$ , where $\epsilon$ is a small number and $x$ 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
$\forall x > 0: x (e^{1/x} - 1) > 1$
(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:

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

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

$u > 1 + \ln{u}$

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

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

**tonio** · April 12th 2010, 11:36 AM

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:

$|x (e^{1/x} - 1) - 1| = \epsilon$ , where $\epsilon$ is a small number and $x$ 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
$\forall x > 0: x (e^{1/x} - 1) > 1$
(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:

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

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

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

$u > 1 + \ln{u}$

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

Any thoughts would be very appreciated!

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

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

Tonio

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

**fiontie** · April 12th 2010, 12:26 PM

@tonio: elegance is easy

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

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