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Math Help - how to prove these innequalities

  1. #1
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    how to prove these innequalities

    first:
     |x - y| \ge |\arctan x - \arctan y|{\rm{ }}\forall {\rm{x,y}} \in {\rm{R}}
    second:
     {\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:
     |x - y| \ge |\arctan x - \arctan y|{\rm{ }}\forall {\rm{x,y}} \in {\rm{R}}
    second:
     {\rm{1 - }}\frac{1}{x} < \ln x < x - 1{\rm{ }}\forall {\rm{ x > 1}}
    We can use the MVT for the first one with f(z) = \tan^{-1} z
    \tan^{-1}x - \tan^{-1}y = \frac{x-y}{1+c^2} where x \le c \le y or vice-versa

    then
    \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:
    <br />
   f'(c) = {1 \over {1 + c^2 }} = {{f(x) - f(y)} \over {x - y}} = {{\tan ^{ - 1} x - \tan ^{ - 1} y} \over {x - y}}  <br />
    <br />
    <br />
   \tan ^{ - 1} x - \tan ^{ - 1} y = {{x - y} \over {1 + c^2 }} <br />

    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:
    <br />
f'(c) = {1 \over {1 + c^2 }} = {{f(x) - f(y)} \over {x - y}} = {{\tan ^{ - 1} x - \tan ^{ - 1} y} \over {x - y}} <br />
    <br />
    <br />
\tan ^{ - 1} x - \tan ^{ - 1} y = {{x - y} \over {1 + c^2 }} <br />

    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 a = b then certainly |a| = |b|. The converse is not true. For the second try using a MVT again with

    f(x) = \ln x
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  5. #5
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    x<c<y
    <br /> <br />
   f(x) = \ln x<br />
    <br />
   f'(c) = {1 \over {\ln c}} <br />
      <br />
   {1 \over c} = {{\ln x - \ln y} \over {x - y}}   <br /> <br />

    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
    <br /> <br />
f(x) = \ln x<br />
    <br />
f'(c) = {1 \over {\ln c}} <br />
     <br />
{1 \over c} = {{\ln x - \ln y} \over {x - y}} <br /> <br />

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

     <br />
{1 \over c} = {{\ln x - \ln 1} \over {x - 1}} <br />
or \frac{x-1}{c} = \ln x

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

     \ln x = \frac{x-1}{c} < x-1
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  7. #7
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    Here's another way. Let f(x) = \ln x -x + 1. Now f(1) = 0 and f'(x) = \frac{1}{x} - 1. For x > 1\;\;\; \text{then}\;\;\; f' < 0 (f is decresing from 0) so \ln x - x + 1 < 0 or \ln x < x -1
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