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.
 A: We have that $f(0)=0$ and
$$f(x)=-2 + x + (2+x)e^{-x}\implies f'(x)=e^{-x}(e^x-1-x)$$
As alternative note that for $x\ge2$ the inequality is trivially satisfied then consider $0<x<2$ and we have
$$-2 + x + (2+x)e^{-x}>0 \iff e^x<\frac{2+x}{2-x}$$
and by $x=\frac2y$ with $y>1$ we obtain
$$f(y)=\left(\frac{y+1}{y-1}\right)^y=\left(1+\frac{2}{y-1}\right)^y>e^2$$
which, for my knowledge, can’t be easily proved for  $y\in\mathbb{R}$ without derivatives, namely showing that $f(y)$ is monotonic.
The monotonicity of $f(y)$ can be easily proved for $y\in\mathbb{N}$ and extended to the real case assuming that $f(y)$ is convex. We could also try to extend the result to reals through rationals.
A: $$ \frac{d^2}{dx^2}\left[-2+x+(2+x)e^{-x}\right] = x e^{-x} $$
hence $f(x)=-2+x+(2+x)e^{-x}$ is a convex function on $\mathbb{R}^+$.Since $f'(0)=f(0)=0$, $f(x)$ is increasing and positive on $\mathbb{R}^+$.
A: We have to prove that
$$e^{-x}>{2-x\over 2+x}\qquad(x>0)\ .\tag{1}$$
This is obvious when $x\geq2$. For $0<x<2$ replace the claim $(1)$ by
$$e^x<{2+x\over2-x}=1+{x\over 1-{x\over2}}\qquad(0<x<2)\ .$$This means that we have to prove
$$1+x+{x^2\over2!}+{x^3\over3!}+{x^4\over4!}+\ldots<1+x+{x^2\over2}+{x^3\over2^2}+{x^4\over 2^3}+\ldots\qquad(0<x<2) ,$$
which follows immediately by comparing terms.
A: 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}
$\forall x\gt0,\dfrac{2+x}{2-x}\gt1\implies\ln\left(\dfrac{2+x}{2-x}\right)\gt0$
$\therefore0\lt x\lt\ln\left(\dfrac{2+x}{2-x}\right)\implies e^{-x}\gt\dfrac{2-x}{2+x}\implies -2+x+(2+x)e^{-x}\gt0$
A: A variation on Christian Blatter's solution: 
$e^{-x} (2+x) +x-2 = e^{-x} ( 1+ \frac{x-2}{x+2})  $ 
As $e^{-x} \gt \  0 \ \forall \ x \in \mathbb{R} $, it's enough to check if, 
$(1+\frac{x-2}{x+2}) \gt 0 $ for $x\gt 0$
However, now we have: $1+\frac{x-2}{x+2} = 1+\frac{x-2+2-2}{x+2} = 2-\frac{4}{x+2}$ which is strictly increasing for $x\gt 0$ and bounded below by $ 0 = e^{-x} ( 1+ \frac{x-2}{x+2}$ ) for $x=0$
A: Another alternative that I found but is not as elegant as the answer marked as the accepted one is expanding the exponential and rearranging the terms as follows:
$
 \left(e^{x } (x -2 )+2 +x \right) >0  \\
\iff 
   (1+x+\frac{x^2}{2!}++\frac{x^3}{3!}+\dots) (x -2 )+2 +x  >0  \\
   \iff 
  \sum_{n=3}^{\infty} x^n(\frac{1}{(n-1)!}-\frac{2}{(n-1)!n})>0
$
