Prove that Geometric mean is less than Arithmetic mean using Jensen's inequality












1












$begingroup$


So in class we solved the following exersice: State the Jensen’s Inequality for a convex function and use it to show that for a finite set of real numbers the geometric mean is less than or equal to the arithmetic mean. Unfortunately I lost my notes and cannot recall the proof at all.



The Jensen's inequality there is in Roydens book is the following:

Let $phi$ be a convex function on $mathbb{R}$, $f$ an integrable function over $[0,1]$ and $phicirc f$ integrable on $[0,1]$



Then $phi(int_0^1 fdx)=int_0^1 phicirc f dx$



This equation seems that to concern the Lebesque measure however I remember that in class to prove the result we had introduced the counting measure defined on a set. So I guess maybe the above Jensen's inequality is different than the one we used in class.



So does anyone know a proof of the result using the counting measure defined on a set and using some sort of Jensen's inequality possibly different than the one I have above?



Thanks in advance






measure-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    So in class we solved the following exersice: State the Jensen’s Inequality for a convex function and use it to show that for a finite set of real numbers the geometric mean is less than or equal to the arithmetic mean. Unfortunately I lost my notes and cannot recall the proof at all.



    The Jensen's inequality there is in Roydens book is the following:

    Let $phi$ be a convex function on $mathbb{R}$, $f$ an integrable function over $[0,1]$ and $phicirc f$ integrable on $[0,1]$



    Then $phi(int_0^1 fdx)=int_0^1 phicirc f dx$



    This equation seems that to concern the Lebesque measure however I remember that in class to prove the result we had introduced the counting measure defined on a set. So I guess maybe the above Jensen's inequality is different than the one we used in class.



    So does anyone know a proof of the result using the counting measure defined on a set and using some sort of Jensen's inequality possibly different than the one I have above?



    Thanks in advance






    measure-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So in class we solved the following exersice: State the Jensen’s Inequality for a convex function and use it to show that for a finite set of real numbers the geometric mean is less than or equal to the arithmetic mean. Unfortunately I lost my notes and cannot recall the proof at all.



      The Jensen's inequality there is in Roydens book is the following:

      Let $phi$ be a convex function on $mathbb{R}$, $f$ an integrable function over $[0,1]$ and $phicirc f$ integrable on $[0,1]$



      Then $phi(int_0^1 fdx)=int_0^1 phicirc f dx$



      This equation seems that to concern the Lebesque measure however I remember that in class to prove the result we had introduced the counting measure defined on a set. So I guess maybe the above Jensen's inequality is different than the one we used in class.



      So does anyone know a proof of the result using the counting measure defined on a set and using some sort of Jensen's inequality possibly different than the one I have above?



      Thanks in advance






      measure-theory







      share|cite|improve this question











      $endgroup$




      So in class we solved the following exersice: State the Jensen’s Inequality for a convex function and use it to show that for a finite set of real numbers the geometric mean is less than or equal to the arithmetic mean. Unfortunately I lost my notes and cannot recall the proof at all.



      The Jensen's inequality there is in Roydens book is the following:

      Let $phi$ be a convex function on $mathbb{R}$, $f$ an integrable function over $[0,1]$ and $phicirc f$ integrable on $[0,1]$



      Then $phi(int_0^1 fdx)=int_0^1 phicirc f dx$



      This equation seems that to concern the Lebesque measure however I remember that in class to prove the result we had introduced the counting measure defined on a set. So I guess maybe the above Jensen's inequality is different than the one we used in class.



      So does anyone know a proof of the result using the counting measure defined on a set and using some sort of Jensen's inequality possibly different than the one I have above?



      Thanks in advance






      measure-theory




      measure-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 22:20









      Masacroso

      13.1k41747




      13.1k41747










      asked Dec 4 '18 at 22:02









      TheGeometerTheGeometer

      937619




      937619






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Forget about measure theory; I'm pretty sure your class only wants you to use the conceptually simpler result $f(sum_i w_i x_i)le sum_i w_i f(x_i)$, with non-negative $w_i$ summing to $1$. The choice $f(x):=-ln x,,w_i=frac{1}{n}$ gives $-lnfrac{1}{n}sum_ix_ile-frac{1}{n}sum_iln x_i$. Applying $xmapstoexp -x$, which reverses order, completes the proof.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            The Jensen's inequality is the result of applying the inequality



            $$f(x(1-t)+yt)le (1-t)f(x)+t f(y)tag1$$



            for some convex function $f$ and $tin[0,1]$. This can be generalized to



            $$fleft(sum_{k=1}^n a_k x_kright)lesum_{k=1}^n a_k f(x_k)tag2$$



            for $sum_{k=1}^n a_k=1$ and each $a_kin[0,1]$. The inequalities above are reverted if $f$ is concave instead of convex.



            This can be extended also to series and integrals, in the latter case we can derive the Jensen's inequality.



            In your case, for $c_k>0$, we have that



            $$frac1nsum_{k=1}^n c_kgesqrt[n]{prod_{k=1}^n c_k}iff logleft(sum_{k=1}^n frac{c_k}nright)ge sum_{k=1}^nfrac1nlog(c_k)tag3$$



            what holds because $log$ is concave.






            share|cite|improve this answer











            $endgroup$













              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',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              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%2f3026244%2fprove-that-geometric-mean-is-less-than-arithmetic-mean-using-jensens-inequality%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Forget about measure theory; I'm pretty sure your class only wants you to use the conceptually simpler result $f(sum_i w_i x_i)le sum_i w_i f(x_i)$, with non-negative $w_i$ summing to $1$. The choice $f(x):=-ln x,,w_i=frac{1}{n}$ gives $-lnfrac{1}{n}sum_ix_ile-frac{1}{n}sum_iln x_i$. Applying $xmapstoexp -x$, which reverses order, completes the proof.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Forget about measure theory; I'm pretty sure your class only wants you to use the conceptually simpler result $f(sum_i w_i x_i)le sum_i w_i f(x_i)$, with non-negative $w_i$ summing to $1$. The choice $f(x):=-ln x,,w_i=frac{1}{n}$ gives $-lnfrac{1}{n}sum_ix_ile-frac{1}{n}sum_iln x_i$. Applying $xmapstoexp -x$, which reverses order, completes the proof.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Forget about measure theory; I'm pretty sure your class only wants you to use the conceptually simpler result $f(sum_i w_i x_i)le sum_i w_i f(x_i)$, with non-negative $w_i$ summing to $1$. The choice $f(x):=-ln x,,w_i=frac{1}{n}$ gives $-lnfrac{1}{n}sum_ix_ile-frac{1}{n}sum_iln x_i$. Applying $xmapstoexp -x$, which reverses order, completes the proof.






                  share|cite|improve this answer









                  $endgroup$



                  Forget about measure theory; I'm pretty sure your class only wants you to use the conceptually simpler result $f(sum_i w_i x_i)le sum_i w_i f(x_i)$, with non-negative $w_i$ summing to $1$. The choice $f(x):=-ln x,,w_i=frac{1}{n}$ gives $-lnfrac{1}{n}sum_ix_ile-frac{1}{n}sum_iln x_i$. Applying $xmapstoexp -x$, which reverses order, completes the proof.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 22:13









                  J.G.J.G.

                  28.1k22844




                  28.1k22844























                      1












                      $begingroup$

                      The Jensen's inequality is the result of applying the inequality



                      $$f(x(1-t)+yt)le (1-t)f(x)+t f(y)tag1$$



                      for some convex function $f$ and $tin[0,1]$. This can be generalized to



                      $$fleft(sum_{k=1}^n a_k x_kright)lesum_{k=1}^n a_k f(x_k)tag2$$



                      for $sum_{k=1}^n a_k=1$ and each $a_kin[0,1]$. The inequalities above are reverted if $f$ is concave instead of convex.



                      This can be extended also to series and integrals, in the latter case we can derive the Jensen's inequality.



                      In your case, for $c_k>0$, we have that



                      $$frac1nsum_{k=1}^n c_kgesqrt[n]{prod_{k=1}^n c_k}iff logleft(sum_{k=1}^n frac{c_k}nright)ge sum_{k=1}^nfrac1nlog(c_k)tag3$$



                      what holds because $log$ is concave.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        The Jensen's inequality is the result of applying the inequality



                        $$f(x(1-t)+yt)le (1-t)f(x)+t f(y)tag1$$



                        for some convex function $f$ and $tin[0,1]$. This can be generalized to



                        $$fleft(sum_{k=1}^n a_k x_kright)lesum_{k=1}^n a_k f(x_k)tag2$$



                        for $sum_{k=1}^n a_k=1$ and each $a_kin[0,1]$. The inequalities above are reverted if $f$ is concave instead of convex.



                        This can be extended also to series and integrals, in the latter case we can derive the Jensen's inequality.



                        In your case, for $c_k>0$, we have that



                        $$frac1nsum_{k=1}^n c_kgesqrt[n]{prod_{k=1}^n c_k}iff logleft(sum_{k=1}^n frac{c_k}nright)ge sum_{k=1}^nfrac1nlog(c_k)tag3$$



                        what holds because $log$ is concave.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The Jensen's inequality is the result of applying the inequality



                          $$f(x(1-t)+yt)le (1-t)f(x)+t f(y)tag1$$



                          for some convex function $f$ and $tin[0,1]$. This can be generalized to



                          $$fleft(sum_{k=1}^n a_k x_kright)lesum_{k=1}^n a_k f(x_k)tag2$$



                          for $sum_{k=1}^n a_k=1$ and each $a_kin[0,1]$. The inequalities above are reverted if $f$ is concave instead of convex.



                          This can be extended also to series and integrals, in the latter case we can derive the Jensen's inequality.



                          In your case, for $c_k>0$, we have that



                          $$frac1nsum_{k=1}^n c_kgesqrt[n]{prod_{k=1}^n c_k}iff logleft(sum_{k=1}^n frac{c_k}nright)ge sum_{k=1}^nfrac1nlog(c_k)tag3$$



                          what holds because $log$ is concave.






                          share|cite|improve this answer











                          $endgroup$



                          The Jensen's inequality is the result of applying the inequality



                          $$f(x(1-t)+yt)le (1-t)f(x)+t f(y)tag1$$



                          for some convex function $f$ and $tin[0,1]$. This can be generalized to



                          $$fleft(sum_{k=1}^n a_k x_kright)lesum_{k=1}^n a_k f(x_k)tag2$$



                          for $sum_{k=1}^n a_k=1$ and each $a_kin[0,1]$. The inequalities above are reverted if $f$ is concave instead of convex.



                          This can be extended also to series and integrals, in the latter case we can derive the Jensen's inequality.



                          In your case, for $c_k>0$, we have that



                          $$frac1nsum_{k=1}^n c_kgesqrt[n]{prod_{k=1}^n c_k}iff logleft(sum_{k=1}^n frac{c_k}nright)ge sum_{k=1}^nfrac1nlog(c_k)tag3$$



                          what holds because $log$ is concave.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 4 '18 at 22:52

























                          answered Dec 4 '18 at 22:28









                          MasacrosoMasacroso

                          13.1k41747




                          13.1k41747






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026244%2fprove-that-geometric-mean-is-less-than-arithmetic-mean-using-jensens-inequality%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              How to send String Array data to Server using php in android

                              Image
                              0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
                              Read more

                              Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                              Image
                              up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                              Read more

                              Is anime1.com a legal site for watching anime?

                              Image
                              up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
                              Read more