first:
$\displaystyle |x - y| \ge |\arctan x - \arctan y|{\rm{ }}\forall {\rm{x,y}} \in {\rm{R}} $
second:
$\displaystyle {\rm{1 - }}\frac{1}{x} < \ln x < x - 1{\rm{ }}\forall {\rm{ x > 1}} $
We can use the MVT for the first one with $\displaystyle f(z) = \tan^{-1} z$
$\displaystyle \tan^{-1}x - \tan^{-1}y = \frac{x-y}{1+c^2}$ where $\displaystyle x \le c \le y$ or vice-versa
then
$\displaystyle \left|\tan^{-1}x - \tan^{-1}y \right|= \frac{\left| x-y \right| }{1+c^2} \le \left| x-y \right|$
i got this equation:
$\displaystyle
f'(c) = {1 \over {1 + c^2 }} = {{f(x) - f(y)} \over {x - y}} = {{\tan ^{ - 1} x - \tan ^{ - 1} y} \over {x - y}}
$
$\displaystyle
$
$\displaystyle
\tan ^{ - 1} x - \tan ^{ - 1} y = {{x - y} \over {1 + c^2 }}
$
on what basis you say that you can put absolute values on the places you put them?
how to know if it equals the equation that i got
??
(how to solve the second one)
Here's another way. Let $\displaystyle f(x) = \ln x -x + 1$. Now $\displaystyle f(1) = 0$ and $\displaystyle f'(x) = \frac{1}{x} - 1$. For $\displaystyle x > 1\;\;\; \text{then}\;\;\; f' < 0$ (f is decresing from 0) so $\displaystyle \ln x - x + 1 < 0$ or $\displaystyle \ln x < x -1$