Is $|f(b)-f(a)| > |b-a|$ true for $f(x)=x+(1+e^x)^{-1}$? I'd like to use this as part of a proof, but I couldn't realize how to show this (and if it) is true. The function is: $f(x)=x+(1+e^x)^{-1}$
 A: $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 $.
A: $$
f'(x) = 1 - \frac{e^x}{(1+e^x)^2}, \text{ so } f'(0) = \frac 3 4 = 0.75.
$$
Since $f'$ is continuous, $f'(x)$ remains close to $0.75$ if $x$ is close enough to $0$; thus we can guarantee that $0.7<f'(x)<0.8$ by making $x$ close enough to $0$.
So let $a,b$ be two number that are that close to $0$. By the mean value theorem,
$$
\frac{f(a)-f(b)}{a-b} = f'(c) \in(0.7,\,0.8) \text{ for some } c \text{ between } a \text{ and } b.
$$
Thus $|f(a)-f(b)|< |a-b|$ if $a,b$ are close enough to $0$.
A: $\displaystyle f(b)-f(a)=b-a+(\frac{1}{1+e^b}-\frac{1}{1+e^a})=b-a+\frac{e^a-e^b}{(1+e^a)(1+e^b)}=b-a-\frac{e^b-e^a}D$
We have $f(b)-f(a)=(b-a)-C$ with $\displaystyle C=\frac{e^b-e^a}D,\quad D>0$.
Since $\exp$ is an increasing function then $b>a\implies e^b>e^a$ 
So $C$ is the same sign than $b-a$.
$f(b)-f(a)=b-a-\operatorname{sgn}(b-a)|C|\implies|f(b)-f(a)|\le|b-a|$

On your question about the fixed point theorem:
First a remark, to apply the theorem you need to prove $f$ is k-lipschitzian with $k<1$.
I.e. $|f(b)-f(a)|\le k|b-a|$ this is a stronger constraint than just $|f(b)-f(a)|<|b-a|$ for $a\neq b$.
But in the present case, this is not the problem, $f'(x)\to 1$ in infinity and on any compact we can find a $k<1$ such that $f'(x)\le k$.
Your function is not much different from $g(x)=x+\frac 1x$, which shares the contraction inequality (at least for $x$ far enough from $0$) but does not have a fixed point.
The problem is in fact that the theorem require $f:X\to X$.
But since $f(x)>x$ for all $x$ this condition cannot be met and $f(X)\not\subset X$ when $X$ is bounded.
If we try to fulfil this condition then we need to operate on $X=[0,+\infty[$ at the expense that the fixed point is then rejected at infinity. (i.e. $\lim\limits_{x\to+\infty}f(x)=+\infty$).
In conclusion beware of the exact requirements before applying a theorem.
A: Consider $a=1, b=0$.
$$|f(a) - f(b)|  =  \left|\left(1+\frac{1}{1+e}\right) - \frac{1}{2}\right| = \left|\frac{1}{2} + \frac{1}{1+e}\right| < \left|\frac{1}{2} + \frac{1}{2}\right| = 1 = |a-b|$$
Hence its not true.
