How to show $\frac{2-x}{2+x} \le e^{-x}$ for $x\ge0$? Let $x\geq0$. Show that:
 $$\frac{2-x}{2+x} \le e^{-x}.$$ 
I have some troubles to prove it. Would you give me any hint?

We need to prove that 
$$(2-x)e^x\leq2+x$$ or
$$x(e^x+1)\geq2(e^x-1)$$ or
$f(x)\geq0$, where
$$f(x)=x-\frac{2(e^x-1)}{e^x+1}$$ and calculate $f'(x)$:
$$f'(x)=1-\frac{4e^x}{(e^x+1)^2}=\frac{(e^x-1)^2}{(e^x+1)^2}\geq0,$$
which gives $f(x)\geq f(0)=0$ and we are done!
