Proving this inequality without calculus

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For all $xin(0,1)$, prove that $ln x<x$.




My attempt:



First step:
Assume a function $f(x)=ln x-x$. Analysing tells me the function is continious and non-differentiable at $x=0$.



Second step:
$F'(x) =frac1x-1$



=$frac1-xx$



This gives me $x=1$ as extremum. From further analysis this appears to be global maximum.



So combining all the info, $x=1$ is the global maxinmum with the function decreasing at points less than it. That would mean $f(x)$ is decreasing from (0, 1] and (0, 1). Hence, this proves the inequality.



However, is there a quick non-calculus way to solve this? I thought of using Taylor series but it's not working.






calculus







share|cite|improve this question



























    up vote
    1
    down vote

    favorite













    For all $xin(0,1)$, prove that $ln x<x$.




    My attempt:



    First step:
    Assume a function $f(x)=ln x-x$. Analysing tells me the function is continious and non-differentiable at $x=0$.



    Second step:
    $F'(x) =frac1x-1$



    =$frac1-xx$



    This gives me $x=1$ as extremum. From further analysis this appears to be global maximum.



    So combining all the info, $x=1$ is the global maxinmum with the function decreasing at points less than it. That would mean $f(x)$ is decreasing from (0, 1] and (0, 1). Hence, this proves the inequality.



    However, is there a quick non-calculus way to solve this? I thought of using Taylor series but it's not working.






    calculus







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      For all $xin(0,1)$, prove that $ln x<x$.




      My attempt:



      First step:
      Assume a function $f(x)=ln x-x$. Analysing tells me the function is continious and non-differentiable at $x=0$.



      Second step:
      $F'(x) =frac1x-1$



      =$frac1-xx$



      This gives me $x=1$ as extremum. From further analysis this appears to be global maximum.



      So combining all the info, $x=1$ is the global maxinmum with the function decreasing at points less than it. That would mean $f(x)$ is decreasing from (0, 1] and (0, 1). Hence, this proves the inequality.



      However, is there a quick non-calculus way to solve this? I thought of using Taylor series but it's not working.






      calculus







      share|cite|improve this question
















      For all $xin(0,1)$, prove that $ln x<x$.




      My attempt:



      First step:
      Assume a function $f(x)=ln x-x$. Analysing tells me the function is continious and non-differentiable at $x=0$.



      Second step:
      $F'(x) =frac1x-1$



      =$frac1-xx$



      This gives me $x=1$ as extremum. From further analysis this appears to be global maximum.



      So combining all the info, $x=1$ is the global maxinmum with the function decreasing at points less than it. That would mean $f(x)$ is decreasing from (0, 1] and (0, 1). Hence, this proves the inequality.



      However, is there a quick non-calculus way to solve this? I thought of using Taylor series but it's not working.






      calculus




      calculus






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      share|cite|improve this question













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      edited 12 secs ago









      user107224

      313112




      313112










      asked 37 mins ago









      GENESECT

      466




      466




















          2 Answers
          2






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          up vote
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          down vote













          For $xin (0,1)$, $log x$ is negative (because $e^tgeq 1$ for $tgeq 0$) and so $log x<0<x$ follows. (log is base $e$, as always).






          share|cite



























            up vote
            2
            down vote













            We have



            $$ln (1+x)<xiff 1+x<e^x$$



            and by $x=frac1y$ with $y>1$



            $$1+x<e^xiff 1+frac1y<e^1/yiff left(1+frac1yright)^y<e$$



            which is true.






            share|cite|improve this answer


















            • 1




              Why does the last inquality hold? It fails for $x=frac12$, for example.
              – user10354138
              15 mins ago










            • @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
              – gimusi
              2 mins ago










            Your Answer




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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            For $xin (0,1)$, $log x$ is negative (because $e^tgeq 1$ for $tgeq 0$) and so $log x<0<x$ follows. (log is base $e$, as always).






            share|cite
























              up vote
              3
              down vote













              For $xin (0,1)$, $log x$ is negative (because $e^tgeq 1$ for $tgeq 0$) and so $log x<0<x$ follows. (log is base $e$, as always).






              share|cite






















                up vote
                3
                down vote










                up vote
                3
                down vote









                For $xin (0,1)$, $log x$ is negative (because $e^tgeq 1$ for $tgeq 0$) and so $log x<0<x$ follows. (log is base $e$, as always).






                share|cite












                For $xin (0,1)$, $log x$ is negative (because $e^tgeq 1$ for $tgeq 0$) and so $log x<0<x$ follows. (log is base $e$, as always).







                share|cite












                share|cite



                share|cite










                answered 8 mins ago









                user10354138

                1,8755




                1,8755




















                    up vote
                    2
                    down vote













                    We have



                    $$ln (1+x)<xiff 1+x<e^x$$



                    and by $x=frac1y$ with $y>1$



                    $$1+x<e^xiff 1+frac1y<e^1/yiff left(1+frac1yright)^y<e$$



                    which is true.






                    share|cite|improve this answer


















                    • 1




                      Why does the last inquality hold? It fails for $x=frac12$, for example.
                      – user10354138
                      15 mins ago










                    • @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
                      – gimusi
                      2 mins ago














                    up vote
                    2
                    down vote













                    We have



                    $$ln (1+x)<xiff 1+x<e^x$$



                    and by $x=frac1y$ with $y>1$



                    $$1+x<e^xiff 1+frac1y<e^1/yiff left(1+frac1yright)^y<e$$



                    which is true.






                    share|cite|improve this answer


















                    • 1




                      Why does the last inquality hold? It fails for $x=frac12$, for example.
                      – user10354138
                      15 mins ago










                    • @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
                      – gimusi
                      2 mins ago












                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    We have



                    $$ln (1+x)<xiff 1+x<e^x$$



                    and by $x=frac1y$ with $y>1$



                    $$1+x<e^xiff 1+frac1y<e^1/yiff left(1+frac1yright)^y<e$$



                    which is true.






                    share|cite|improve this answer














                    We have



                    $$ln (1+x)<xiff 1+x<e^x$$



                    and by $x=frac1y$ with $y>1$



                    $$1+x<e^xiff 1+frac1y<e^1/yiff left(1+frac1yright)^y<e$$



                    which is true.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 7 mins ago

























                    answered 28 mins ago









                    gimusi

                    74.5k73889




                    74.5k73889







                    • 1




                      Why does the last inquality hold? It fails for $x=frac12$, for example.
                      – user10354138
                      15 mins ago










                    • @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
                      – gimusi
                      2 mins ago












                    • 1




                      Why does the last inquality hold? It fails for $x=frac12$, for example.
                      – user10354138
                      15 mins ago










                    • @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
                      – gimusi
                      2 mins ago







                    1




                    1




                    Why does the last inquality hold? It fails for $x=frac12$, for example.
                    – user10354138
                    15 mins ago




                    Why does the last inquality hold? It fails for $x=frac12$, for example.
                    – user10354138
                    15 mins ago












                    @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
                    – gimusi
                    2 mins ago




                    @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out.
                    – gimusi
                    2 mins ago

















                     

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