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?
 A: 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.
A: 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.
