prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$

Solution 1:

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.

Solution 2:

$$ \frac{d^2}{dx^2}\left[-2+x+(2+x)e^{-x}\right] = x e^{-x} $$ 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 $\mathbb{R}^+$.

Solution 3:

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.