Is |f(b)−f(a)|>|b−a||f(b)-f(a)| > |b-a| true for f(x)=x+(1+ex)−1f(x)=x+(1+e^x)^{-1}?

$f$ is differentiable at $\mathbb R$, then by MVT,

$$f (a)-f (b)=(a-b)f'(c)$$

with $a <c <b$.

but $$f'(c)=1-\frac{e^c}{(1+e^c)^2}=\frac{1+e^c+e^{2c}}{(1+e^c)^2}$$

$$\implies 0 <f'(c)<1$$
thus $$|f (a)-f (b)|<|a-b|$$

your statement is not true for $f$.