<s>[INST] prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$ I need to prove that 
$ -2 + x + (2+x)e^{-x}>0   \quad \forall x>0$
If I define $f(x) = -2 + x + (2+x)e^{-x}$ and plot it I can see it's a monotonously growing function, and f(0)=0. Then $f(x)>0$ if $x>0$.
However I can't find the way to prove this.
Ideally I would like to prove it without deriving the function and by using some inequalities, however I don't know if it is possible. Any hint is really appreciated.
 [/INST] This approach requires an extra condition $x\lt2.$
\begin{align}\ln\left(1+\dfrac{x}{2}\right)-\ln\left(1-\dfrac{x}{2}\right)&=\left[\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2+\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3+\cdots\right]-\left[-\dfrac{x}{2}-\dfrac{1}{2}\left(\dfrac{x}{2}\right)^2-\dfrac{1}{3}\left(\dfrac{x}{2}\right)^3-\cdots\right]\\&=x+\dfrac{2}{3}\left(\dfrac{x}{2}\right)^3+\cdots\\&\gt x\quad\forall x\lt2\cdots(*)\\\end{align}
$\forall x\gt0,\dfrac{2+x}{2-x}\gt1\implies\ln\left(\dfrac{2+x}{2-x}\right)\gt0$
$\therefore0\lt x\lt\ln\left(\dfrac{2+x}{2-x}\right)\implies e^{-x}\gt\dfrac{2-x}{2+x}\implies -2+x+(2+x)e^{-x}\gt0$
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