how to prove $f'(x) \leq \frac{f(x) - f(y)}{x - y} \leq f'(y) $

Question

$f(x) = x - \arctan(\ln(x)) $ on the interval $[0,+\infty[$. How to use the Mean value theorem to show that $f'(x) \leq \frac{f(x) - f(y)}{x - y} \leq f'(y) $, I know that according to theorem ,
$\exists c \in ]x,y[, f'(c) = \frac{f(y) - f(x)}{y-x} $
$ \frac{1}{1+c^2}= \frac{f(y) - f(x)}{y -x} $
All help is appreciated

Thanks in advance :)

Answer 1

hint Here is the derivative of $f$ :

$$f'(x)=1-\frac{1}{x(1+\ln^2(x))}.$$ Observe that $$x\mapsto x(1+\ln^2(x))$$ is increasing at $(0,+\infty)$ as a product of two increasing functions. hence,

$f'$ is increasing thus...

$$f'(x)\le f'(c)=\frac{f(x)-f(y)}{x-y} \le f'(y)$$