By playing around, I seem to have come across the following inequality, valid for all $x$: $$x-(1-e^{-x}) \ge e^{-\frac{2}{x}} x$$ (The constant $2$ is not necessarily the tightest one possible.)
Is there an easy way to prove this, and if so, is this inequality known in the literature? The closest I have been able to come across is $$x-(1-e^{-x}) \ge e^{-\frac{1}{x}} x$$ valid for $x\in[0,1]$ (see e.g., Mond and Paciric, "Inequalities for exponential functions and means, II", NAW, 2000. www.nieuwarchief.nl/serie5/pdf/naw5-2000-01-1-057.pdf)
If $x<0$ the after replacing $x$ by $-x$ we need to prove that $$e^x-1-x\geq-xe^{\frac{2}{x}},$$ which is true because $$e^x-1-x>0>-xe^{\frac{2}{x}}.$$ Now, let $x>0$.
Thus, we need to prove that $f(x)>0$, where $$f(x)=e^{-x}+x-1-xe^{-\frac{2}{x}}.$$ Indeed, $$f''(x)=\frac{x^3e^{-x}-4e^{-\frac{2}{x}}}{x^3},$$ which for $x^3e^x-4e^{-\frac{2}{x}}=0$ (or $x=0.511...$) get's a maximal value of $f'(x),$ where $$f'(x)=1-e^{-x}-\frac{e^{-\frac{2}{x}}(x+2)}{x}.$$
Thus, $$f'(x)>\min\left\{\lim_{x\rightarrow0^+}f'(x),\lim_{x\rightarrow+\infty}f'(x)\right\}=0$$ because two these limits are equal to zero.
Id est, $$f(x)>\lim_{x\rightarrow0^+}f(x)=0$$ and we are done!