$f(x)=x+\frac{1}{e^x+1}$. Prove that for any $x,y$ : $|f(x)-f(y)|\leq|x-y|$

Question

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.

Answer 1

Hint: Use first the Mean value theorem: $$\frac{f(x)-f(y)}{x-y} = f'(c)$$ ant then take: $|\cdot|$.

Answer 2

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 $f=g+h$).

Answer 3

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 $\forall x: f'(x)\leq1$.