Estimating the natural logarithm from both sides: 1 / ( a + 1 ) &lt; ln ⁡ ( ( 1 + a ) / a ) &lt; 1 / aI must prove that for all a &gt; 0 1 a + 1 &lt; ln ⁡ 1 + a a &lt; 1 a Can anyone help me?

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Estimating the natural logarithm from both sides:  1      /    (  a  +  1  )  &amp;lt;  ln  ⁡  (  (  1  +  a  )      /    a  )  &amp;lt;  1      /    aI must prove that for all   a  &amp;gt;  0      1          a      +      1        &amp;lt;  ln  ⁡                    1        +        a            a        &amp;lt;      1    a  Can anyone help me?

nifeonibonitozg · Accepted Answer

Consider   f  (  x  )  =  log  ⁡  (  a  +  x  ),   x  ∈  [  0  ,  1  ]. Then  log  ⁡            1      +      a        a    =  log  ⁡  (  1  +  a  )  −  log  ⁡  a  =  f  (  1  )  −  f  (  0  )  .But      f    ′    (  x  )  =      1          a      +      x        ,and by the mean value theorem there is   c  ∈  (  0  ,  1  ) such that   f  (  1  )  −  f  (  0  )  =      f    ′    (  c  ). Now,      1          a      +      1        &amp;lt;      1          a      +      c        &amp;lt;      1    a    ,and you conclude.

Jorge Lawson · Accepted Answer

ln  ⁡            a      +      1        a    =  ln  ⁡  (  a  +  1  )  −  ln  ⁡  (  a  )  =  ln  ⁡  (  1  +      1    a    )  &amp;gt;  0  ln  ⁡      1          a      +      1        =  ln  ⁡  (  1  )  −  ln  ⁡  (  a  +  1  )  &amp;lt;  0  ln  ⁡      1    a    =  ln  ⁡      1    −  ln  ⁡      a    &amp;gt;  ln  ⁡  1  −  ln  ⁡  (  a  +  1  )  ln  ⁡  x  &amp;gt;     ln  ⁡  y  →  x  &amp;gt;  yUsing the last statement that log is monotonically increasing, and the previous three, you should be able to deduce your inequality.

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