# Proving the inequality: $x\exp(-x^2/4)(\exp(x)+\exp(-x)) \leq 1000 \exp(-x)$

Doing some tests with Maple I "guessed" the following inequality with exponential function (for $x\geq 0$)

$$x\exp(-x^2/4)(\exp(x)+\exp(-x)) \leq 1000 \exp(-x).$$

Is there an easy proof?

Can one improve the "constant" $1000$?

One can probably give a very ugly proof as follows.

It suffices to show that $$x\exp(-x^2/4) \exp(x) \leq 500 \exp(-x).$$ This inequality holds for $x=0$. The maximum value $M$ of the LHS can be calculated explicitly and one can show that the RHS is bigger than $M$ for $0\leq x \leq x_1$, where $x_1$ is some explicit real number. Then, we just compute the derivatives and we show that they satisfy a certain inequality. (This becomes messy.)

Any suggestions?

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Take the logarithm on both sides before starting with derivatives. Then finding the minimum of the differences will also show you how much you can adjust the $1000$. –  Henning Makholm Mar 10 '12 at 15:47
Taking logarithms is very efficient. I can show that one can replace 1000 by 12 and the proof is easy. Thanks alot! –  seporhau Mar 10 '12 at 16:56

The inequality $$x\exp(-x^2/4)(\exp(x)+\exp(-x)) \leq c\; \exp(-x)$$ can be written as $$x\exp(-x^2/4+x)(\exp(x)+\exp(-x)) \leq c.$$ Plotting this function with Mathematica one can see that the optimum $c$ lies around $231$ (probably marginally beyond), and that the derivative has a single root, which is the unique local maximum. But, of course, that's no proof.

Now, if we write \begin{eqnarray} x\exp(-x^2/4+x)(\exp(x)+\exp(-x))&=&x\exp(-x^2/4+x)\exp(x)+x\exp(-x^2/4+x)\exp(-x) \\ &=&x\exp(-x^2/4+2x)+x\exp(-x^2/4), \end{eqnarray} it is very easy to see that the second term is always less that $1$. For the first term, it is also easy to check that its only maximum occurs at $x=2+\sqrt6$.

Then $$x\exp(-x^2/4+x)(\exp(x)+\exp(-x))\leq (2+\sqrt6)\exp(-(2+\sqrt6)^2/4+2\sqrt6)+1\leq 231+1=232.$$ So the constant $232$ works, but the actual optimal constant is likely very near $231$. This can be improved a little by playing more carefully with where the second term achieves its maximum.

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Are you sure the optimal constant is very near 231? I thought I was able to show it was about 12 as follows. For $r \geq 0$, we have $r \exp(-r^2/4)(\exp(r) + \exp(-r)) \leq 2r \exp(-r^2/4) \exp(r) \leq 12 \exp(-r)$. Take logarithms. We have to show that $\log(r) -r^2/4 + 2r \leq 6$. This is equivalent to $\log(r) \leq r^2/4 -2r +6 = 1/4(r^2-8r) + 6= 1/4( (r-4)^2-16) +6 = (r-4)^2/4 +2$. The inequality $\log (r) \leq (r-4)^2/4 +2$ is easily verified and also shows that $12$ is very close to the optimal choice. In fact, the value $11$ doesn't work. –  seporhau Mar 10 '12 at 17:24
Just take the LHS and the RHS in your inequality, and evaluate both at $2+\sqrt6$. You'll get roughly $2.69912$ for the LHS, and $0.0116845$ for the RHS. The quotient between these two values is around $231$. –  Martin Argerami Mar 10 '12 at 17:35
Numerically: the maximum is $230.999548$ and occurs at $4.4489933$. –  GEdgar Mar 10 '12 at 17:45
Sorry guys, you're right. I forgot to take the logarithm on the RHS in my "proof". –  seporhau Mar 10 '12 at 17:58

Here is an idea which might work.

$$x\exp(-x^2/4) \exp(x) \leq 500 \exp(-x)$$

is equivalent to

$$x\exp(-x^2/4) \exp(2x) \leq 500$$

or

$$x\exp(-x^2+8x/4) \leq 500 \,.$$

$$\frac{x}{\exp(x^2-8x/4)} \leq 500 \,.$$

Now using the standard $exp(y) \geq 1+y$ you get:

$$\frac{x}{\exp(x^2-8x/4)}\leq \frac{x}{x^2-8x+1}$$

so it suffices to prove that $\frac{x}{x^2-8x+1} \leq 500$, which if true is easy to show.

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Thanks for the suggestion. The inequality $\exp(y) \geq 1+y$ is too rough though and the last inequality you need is not true. That is, $x \leq 500(x^2-8x+1)$ does not hold for all $x\geq 0$. –  seporhau Mar 10 '12 at 16:31