Proving this inequality without calculus

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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













    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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 12 secs ago









      user107224

      313112




      313112










      asked 37 mins ago









      GENESECT

      466




      466




















          2 Answers
          2






          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
            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




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2927886%2fproving-this-inequality-without-calculus%23new-answer', 'question_page');

            );

            Post as a guest

















            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

















                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2927886%2fproving-this-inequality-without-calculus%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    Long meetings (6-7 hours a day): Being “babysat” by supervisor

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
                    Read more

                    Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
                    Read more

                    Confectionery

                    Image
                    Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
                    Read more