# Thread: how to prove these innequalities

1. ## how to prove these innequalities

first:
$\displaystyle |x - y| \ge |\arctan x - \arctan y|{\rm{ }}\forall {\rm{x,y}} \in {\rm{R}}$
second:
$\displaystyle {\rm{1 - }}\frac{1}{x} < \ln x < x - 1{\rm{ }}\forall {\rm{ x > 1}}$

2. Originally Posted by transgalactic
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
$\displaystyle \left|\tan^{-1}x - \tan^{-1}y \right|= \frac{\left| x-y \right| }{1+c^2} \le \left| x-y \right|$

3. 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)

4. Originally Posted by transgalactic
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)
if $\displaystyle a = b$ then certainly $\displaystyle |a| = |b|$. The converse is not true. For the second try using a MVT again with

$\displaystyle f(x) = \ln x$

5. x<c<y
$\displaystyle f(x) = \ln x$
$\displaystyle f'(c) = {1 \over {\ln c}}$
$\displaystyle {1 \over c} = {{\ln x - \ln y} \over {x - y}}$

what does it prove??
what to do now?

6. Originally Posted by transgalactic
x<c<y
$\displaystyle f(x) = \ln x$
$\displaystyle f'(c) = {1 \over {\ln c}}$
$\displaystyle {1 \over c} = {{\ln x - \ln y} \over {x - y}}$

what does it prove??
what to do now?
Choose $\displaystyle y = 1$ so

$\displaystyle {1 \over c} = {{\ln x - \ln 1} \over {x - 1}}$ or $\displaystyle \frac{x-1}{c} = \ln x$

Since x > 1, then c > 1 so $\displaystyle \frac{1}{c} < 1$

$\displaystyle \ln x = \frac{x-1}{c} < x-1$

7. 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 $\displaystyle \ln x < x -1$