# Show that this is less than that

1. Nov 11, 2013

### ArcanaNoir

1. The problem statement, all variables and given/known data
I would like to prove that when x is between 1 and 2 $$\frac{k^{0.5}x^x}{1+x^{2k}} < \frac{1}{\sqrt{x-1}}$$ for all $k\in \mathbb{N}$

2. Relevant equations

3. The attempt at a solution

This is for my analysis class. I am algebraist, so please take pity :)
I have learned that if I were to show that the numerator of the first fraction is less than the numerator of the second and the denominator of the first fraction is greater than the denominator of the second then the first fraction would be smaller than the second. That technique doesn't seem to be working though. If the second fraction needs tweaking to make this work then that is fine. I'm just trying to bound the first fraction with something integrable. I think my bound works based on Maple graphs.

2. Nov 11, 2013

### LCKurtz

Does the bound have to be independent of $k$?

3. Nov 11, 2013

### ArcanaNoir

Yes it does.

4. Nov 11, 2013

### dirk_mec1

Can't be true, if I let k-> inf surely it will not be valid.

5. Nov 11, 2013

### ArcanaNoir

alas! you may be right. I have just received word that a classmate may have a proof that the sequence cannot be bounded by an integrable function.

6. Nov 11, 2013

### D H

Staff Emeritus
Sure it is. The left hand side goes to zero as k→∞ for all x in (1,∞), and 0<1/sqrt(x-1) for all x in (1,2).

7. Nov 11, 2013

### dirk_mec1

Yes you're right. Then how will the proof read?

8. Nov 11, 2013

### I like Serena

Here's a possible outline of a proof.

Take the natural logarithm of the left hand side.
\begin{aligned}\ln \frac{k^{0.5}x^x}{1+x^{2k}}
&= 0.5\ln k + x\ln x - \ln(1+x^{2k}) \\
&< 0.5\ln k + x\ln x - \ln(x^{2k}) \\
&= 0.5\ln k + x\ln x - 2k \ln x
\end{aligned}
Find the maximum value of this last expression with respect to $k$, treating it as a real number.
The result is $k=1/\ln x$ with the maximum $\frac 1 2 \ln(1/\ln x) - 1$.

Therefore an upper bound of the original left hand side is:
$$\frac 1 {e\sqrt{\ln x}}$$

What is left is the proof that this is less than the right hand side.
Substitute $x=u+1$ and make a first order Taylor expansion of $\ln$.

And.... we're done.

Last edited: Nov 11, 2013
9. Nov 11, 2013

### ArcanaNoir

Thank you so much ILS. This analysis makes me want to cry!

10. Nov 11, 2013

Cry? Why?