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Thread: how to prove these innequalities

  1. #1
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    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}} $
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    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
    $\displaystyle \left|\tan^{-1}x - \tan^{-1}y \right|= \frac{\left| x-y \right| }{1+c^2} \le \left| x-y \right|$
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  3. #3
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    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)
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    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

    $\displaystyle f(x) = \ln x$
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  5. #5
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    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?
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    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 $
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  7. #7
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    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 $\displaystyle \ln x < x -1$
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