# Exponential inequalities

• Sep 8th 2008, 07:36 AM
bkarpuz
Exponential inequalities
Dear friends,

I could not obtain proof for either one of the following inequalities:
$\displaystyle \mathrm{e}^{rx}\geq x+\frac{\mathrm{ln}(r)+1}{r}\text{ for }x\geq0\text{ and }r>0$
or
$\displaystyle \mathrm{e}^{rx}\geq x+\frac{\mathrm{ln}(\mathrm{e}r)}{r}\text{ for }x\geq0\text{ and }r>0.$

I tried to prove it by using their tangents, but it seems not to be so simple.
Any ideas?

Thanks.
• Sep 8th 2008, 09:23 AM
Laurent
There's probably a nicer proof, but here is one anyway:

Let $\displaystyle r>0$. The derivative of the function $\displaystyle f:x\mapsto e^{rx}-x$ is $\displaystyle f':x\mapsto re^{rx}-1$, which is equal to 0 at $\displaystyle x=-\frac{\ln r}{r}$, and respectively negative and positive at the left and right of this point. As a consequence, for every $\displaystyle x\in\mathbb{R},$ $\displaystyle f(x)\geq f(-\frac{\ln r}{r})=\frac{1}{r}+\frac{\ln r}{r}$, which is exactly what we want.

Laurent.
• Sep 8th 2008, 09:45 AM
bkarpuz
Thanks, Laurent.
You have any ideas for the second one?
I guess that it is a little bit harder because your $\displaystyle f$ function will be same for this inequality too.
To be honest, I attempted to solve the latter one at first...
• Sep 8th 2008, 10:10 AM
wingless
What do you mean? That second one is the same as the first one.. Or are you talking about another question somewhere else?
• Sep 8th 2008, 10:16 AM
bkarpuz
Quote:

Originally Posted by wingless
What do you mean? That second one is the same as the first one.. Or are you talking about another question somewhere else?

Selam wingless, I am looking for answers for the following question, I think the proof for this one is different than the one given by Laurent:

$\displaystyle \mathrm{e}^{rx}\geq x+\frac{\mathrm{ln}(\mathrm{e}r)}{r}\text{ for }x\geq0\text{ and }r>0.$
• Sep 8th 2008, 10:21 AM
wingless
Selam =)

$\displaystyle e^{rx} \geq x + \frac{\ln (er)}{r}$

$\displaystyle e^{rx} \geq x + \frac{\ln e + \ln (r)}{r}$

$\displaystyle e^{rx} \geq x + \frac{1 + \ln (r)}{r}$

So both inequalities are actually identical.

Also notice that the question says 'or' between the inequalities...
• Sep 8th 2008, 10:36 AM
bkarpuz
I have to say, that seems to be a joke :S
How can i miss that thing!!!

By the way, I put the 'or' there...
• Sep 10th 2008, 10:37 PM
bkarpuz
Now, I see that I always had mistakes in the calculation. (Worried)
I again checked my way for the last time and saw that it works.

Let me explain it below.

Set $\displaystyle f(x):=\mathrm{e}^{rx}$, we clearly know that $\displaystyle f(x)\geq f^{\prime}(x_{0})(x-x_{0})+f(x_{0})=f^{\prime}(x_{0})x+f(x_{0})-f^{\prime}(x_{0})x_{0}$ for every $\displaystyle x_{0}\geq0$ since $\displaystyle f$ is convex.
What I need to find is that
$\displaystyle f^{\prime}(x_{0})=1\qquad(1)$
and
$\displaystyle f(x_{0})-f^{\prime}(x_{0})x_{0}=f(x_{0})-x_{0}=\frac{1}{r}\mathrm{ln}(\mathrm{e}r).\qquad(2 )$
Solving (1), we get
$\displaystyle x_{0}=\frac{1}{r}\mathrm{ln}\bigg(\frac{1}{r}\bigg )\qquad(3).$
Clearly, (3) satisfies (2), and thus the solution is complete.

I really still don't know where I had mistakes all the time. (Headbang)