Without use of derivatives, prove that $e^{-x}\left(1+x+\frac{x^2}{2}\right)$ is strictly decreasing 
Prove,  without use of derivatives, that function $x\mapsto e^{-x}\left(1+x+\frac{x^2}{2}\right)$ is strictly decreasing.

Attempt. Functions $x\mapsto x^ne^{-x}$ for $n=0,1,2$ are not  strictly decreasing (in order to guarantee that their sum is decreasing). Working separately on $e^{-x},~~1+x+\frac{x^2}{2}$ didn't work also, since $1+x+\frac{x^2}{2}$ is not decreasing on the reals. I am looking for an elementary way to prove it (it's derivative is $-\frac{e^{-x}x^2}{2}<0$ for $x\neq 0$).
Thanks in advance for the help.

I will assume the properties $e^x > 1 +x + \frac{x^2}{2}$ and  $e^{x+y} = e^x e^y$ are given.  
For $x+h > x > 0$, we have $\left(h+ \frac{h^2}{2}\right)\left(1+x+ \frac{x^2}{2}\right)> h + xh+\frac{h^2}{2}$, and, hence,
$$e^h > 1+ h + \frac{h^2}{2} > 1 + \frac{h + xh+\frac{h^2}{2}}{1+x+ \frac{x^2}{2}} = \frac{1 + (x+h)+\frac{(x+h)^2}{2}}{1+x+ \frac{x^2}{2}}$$
Thus,
$$1+x+ \frac{x^2}{2} > e^{-h}\left(1 + (x+h)+\frac{(x+h)^2}{2} \right)$$
Multiplying both sides by $e^{-x}$ we get
$$e^{-x}\left(1+x+ \frac{x^2}{2}\right) > e^{-(x+h)}\left(1 + (x+h)+\frac{(x+h)^2}{2} \right)$$
Something similar should work for the case $x < 0$.
