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 can assume that $0 \le x < 2$. Then, using the geometric series,
$$
 \frac{2+x}{2-x} = 1 + \frac{x}{1 - \frac 12 x} =  1 + \sum_{n=1}^\infty \frac{x^n}{2^{n-1}}
 \ge 1 + \sum_{n=1}^\infty \frac{x^n}{n!} = e^x \, .
$$
Remark: $(2-x)/(2+x)$ is the Padé approximant of order [1/1] for $f(x) = \exp(-x)$, 
that is the (unique) rational function of the form
$$ 
R(x) =  \frac{a_0 + a_1 x}{1 + b_1 x}
$$
such that $R(0) = f(0), R'(0) = f'(0)$ and $R''(0) = f''(0)$.
