$f(x)=x+\frac{1}{e^x+1}$. Prove that for any $x,y$ : $|f(x)-f(y)|\leq|x-y|$ I feel like this question is related to the Mean value theorem, but the absolute value interferes with it.
I get to:
$$\frac{|f(x)-f(y)|}{|x-y|}\leq 1$$
And from there I want to prove that the derivative is always smaller than one using a proof by contradiction and the Mean value theorem.
 A: Hint: Use first the Mean value theorem:
$$\frac{f(x)-f(y)}{x-y} = f'(c)$$
ant then take: $|\cdot|$.
A: HINT: There is no problem on using the mean value theorem and take the absolute value afterwards, since MVT is an equality. Also, you can split your function as a sum of two functions and use the linearity of the derivative for simplicity: one function has derivative equal to $1$ ($g(x)=x$), and a decreasing function ($h(x)=\frac{1}{e^x+1}$) (note that $f=g+h$).
A: OK this is my summarized solution:
$$\forall x: f'(x)\leq1$$
Proof:
$$f'(x) = 1- \frac{e^x}{{e^x}^2+2e^x+1}$$
$\frac{e^x}{{e^x}^2+2e^x+1}$ Is always positive therefore $f'(x)\leq 1$ for any x.
$$\forall x: f'(x)\geq -1$$
Proof:
$$1-\frac{e^x}{{e^x}^2+2e^x+1}\geq -1$$
$$2\geq \frac{e^x}{{e^x}^2+2e^x+1}=\frac{\frac{e^x}{e^x}}{\frac{{e^x}^2+2e^x+1}{e^x}}=\frac{1}{e^x+2+1}$$
$e^x$ is always positive so this statement is true.
$$-1\leq f'(x)\leq1 \iff |f'(x)|\leq1$$
Now we assume by contradiction that there exists $x,y$ such that $\frac{|f(x)-f(y)|}{|x-y|}=|\frac{f(x)-f(y)}{x-y}|> 1$
Using MVT we can assume that there exists a $c$ such that $f'(c) = |\frac{f(x)-f(y)}{x-y}|> 1$ which is a contradiction to $\forall x: f'(x)\leq1$.
