Inequality of exponential function compared to its rational (Padé) approximation

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In the context of studying the convexity of the real function (which is not DCP-convex but really "looks" convex)

$$g(x) = \frac{1}{1-\exp(-1/x)}, \text{for } x\geq 0,$$

after some manipulation of the second derivative, I end up having to study the sign of the expression:

$$e^{-z} (z+2) + z-2$$

which should be positive iff $g$ is convex. However, I wasn't able to bound it directly. Also, the positivity of that expression is equivalent to the following inequality which compares the $\exp$ function with its rational approximation (more precisely its Padé approximant of order $(1,1)$):

$$ e^{-x} \geq \frac{2-x}{2+x} = \frac{1-x/2}{1+x/2}, x\geq 0$$

Visually, I have no doubts about the validity of this inequality (see graph below) on the positive domain, but I cannot find how to proove it. I haven't find it in the list of inequalities involving the Exp function on functions.wolfram.com.

Perhaps I should compare the series expansions?

Plot of the Exp function compared to its Padé approximant