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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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    Taking logarithms is very efficient. I can show that one can replace$1000$by 12 and the proof is easy. Thanks alot!2012-03-10

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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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    Sorry guys, you're right. I forgot to take the logarithm on the RHS in my "proof".2012-03-10
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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 $x\geq 0$.2012-03-10