Let $a;b;c>0$. Prove : $frac{a^{2}}{b+c}+frac{b^{2}}{c+a}+frac{c^{2}}{a+b}geq...












1












$begingroup$


Let $a;b;c>0$. Prove :



$frac{a^{2}}{b+c}+frac{b^{2}}{c+a}+frac{c^{2}}{a+b}geq frac{b^{2}}{b+c}+frac{c^{2}}{c+a}+frac{a^{2}}{a+b}$



P/s : Only use AM-GM and Cauchy-Schwarz



In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!






inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What are you summing over?
    $endgroup$
    – David H
    Dec 2 '13 at 15:03
















1












$begingroup$


Let $a;b;c>0$. Prove :



$frac{a^{2}}{b+c}+frac{b^{2}}{c+a}+frac{c^{2}}{a+b}geq frac{b^{2}}{b+c}+frac{c^{2}}{c+a}+frac{a^{2}}{a+b}$



P/s : Only use AM-GM and Cauchy-Schwarz



In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!






inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What are you summing over?
    $endgroup$
    – David H
    Dec 2 '13 at 15:03














1












1








1


2



$begingroup$


Let $a;b;c>0$. Prove :



$frac{a^{2}}{b+c}+frac{b^{2}}{c+a}+frac{c^{2}}{a+b}geq frac{b^{2}}{b+c}+frac{c^{2}}{c+a}+frac{a^{2}}{a+b}$



P/s : Only use AM-GM and Cauchy-Schwarz



In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!






inequality







share|cite|improve this question











$endgroup$




Let $a;b;c>0$. Prove :



$frac{a^{2}}{b+c}+frac{b^{2}}{c+a}+frac{c^{2}}{a+b}geq frac{b^{2}}{b+c}+frac{c^{2}}{c+a}+frac{a^{2}}{a+b}$



P/s : Only use AM-GM and Cauchy-Schwarz



In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '13 at 15:08









Michael Joyce

12.4k21939




12.4k21939










asked Dec 2 '13 at 15:01









Lê Tấn KhangLê Tấn Khang

608310




608310








  • 1




    $begingroup$
    What are you summing over?
    $endgroup$
    – David H
    Dec 2 '13 at 15:03














  • 1




    $begingroup$
    What are you summing over?
    $endgroup$
    – David H
    Dec 2 '13 at 15:03








1




1




$begingroup$
What are you summing over?
$endgroup$
– David H
Dec 2 '13 at 15:03




$begingroup$
What are you summing over?
$endgroup$
– David H
Dec 2 '13 at 15:03










3 Answers
3






active

oldest

votes


















1












$begingroup$

I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
$$ frac{a^2}{1-a} + frac{b^2}{1-b} + frac{c^2}{1-c}
ge frac{b^2}{1-a} + frac{c^2}{1-b} + frac{a^2}{1-c} $$
Since $xmapsto x^2$ and $xmapstofrac1{1-x}$ are increasing functions for $xin(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(frac1{1-a},frac1{1-b},frac1{1-c})$ are in the same order (e.g., if $age bge c$ then $a^2ge b^2ge c^2$ and $frac1{1-a}gefrac1{1-b}gefrac1{1-c}$), so the desired inequality follows from the rearrangement inequality.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We need to prove that $$sumlimits_{cyc}frac{a^2-b^2}{b+c}geq0$$ or $$sumlimits_{cyc}(a^2-b^2)(a+b)(a+c)geq0$$ or
    $$sumlimits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)geq0$$ or
    $$sumlimits_{cyc}(a^2-b^2)a^2geq0$$ or
    $$sumlimits_{cyc}left(frac{a^4+b^4}{2}-a^2b^2right)geq0,$$
    which is true by AM-GM.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You can just do it in one step (without any computation) if you use the rearrangement inequality.



      Notice that for any positive reals a, b and c,
      $(a^2,b^2,c^2)$ and $(1/b+c,1/c+a,1/a+c)$ are similarly sorted. So,








      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%2f589678%2flet-abc0-prove-fraca2bc-fracb2ca-fracc2ab-ge%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
        $$ frac{a^2}{1-a} + frac{b^2}{1-b} + frac{c^2}{1-c}
        ge frac{b^2}{1-a} + frac{c^2}{1-b} + frac{a^2}{1-c} $$
        Since $xmapsto x^2$ and $xmapstofrac1{1-x}$ are increasing functions for $xin(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(frac1{1-a},frac1{1-b},frac1{1-c})$ are in the same order (e.g., if $age bge c$ then $a^2ge b^2ge c^2$ and $frac1{1-a}gefrac1{1-b}gefrac1{1-c}$), so the desired inequality follows from the rearrangement inequality.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
          $$ frac{a^2}{1-a} + frac{b^2}{1-b} + frac{c^2}{1-c}
          ge frac{b^2}{1-a} + frac{c^2}{1-b} + frac{a^2}{1-c} $$
          Since $xmapsto x^2$ and $xmapstofrac1{1-x}$ are increasing functions for $xin(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(frac1{1-a},frac1{1-b},frac1{1-c})$ are in the same order (e.g., if $age bge c$ then $a^2ge b^2ge c^2$ and $frac1{1-a}gefrac1{1-b}gefrac1{1-c}$), so the desired inequality follows from the rearrangement inequality.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
            $$ frac{a^2}{1-a} + frac{b^2}{1-b} + frac{c^2}{1-c}
            ge frac{b^2}{1-a} + frac{c^2}{1-b} + frac{a^2}{1-c} $$
            Since $xmapsto x^2$ and $xmapstofrac1{1-x}$ are increasing functions for $xin(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(frac1{1-a},frac1{1-b},frac1{1-c})$ are in the same order (e.g., if $age bge c$ then $a^2ge b^2ge c^2$ and $frac1{1-a}gefrac1{1-b}gefrac1{1-c}$), so the desired inequality follows from the rearrangement inequality.






            share|cite|improve this answer









            $endgroup$



            I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
            $$ frac{a^2}{1-a} + frac{b^2}{1-b} + frac{c^2}{1-c}
            ge frac{b^2}{1-a} + frac{c^2}{1-b} + frac{a^2}{1-c} $$
            Since $xmapsto x^2$ and $xmapstofrac1{1-x}$ are increasing functions for $xin(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(frac1{1-a},frac1{1-b},frac1{1-c})$ are in the same order (e.g., if $age bge c$ then $a^2ge b^2ge c^2$ and $frac1{1-a}gefrac1{1-b}gefrac1{1-c}$), so the desired inequality follows from the rearrangement inequality.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '13 at 23:32







            user21467






























                0












                $begingroup$

                We need to prove that $$sumlimits_{cyc}frac{a^2-b^2}{b+c}geq0$$ or $$sumlimits_{cyc}(a^2-b^2)(a+b)(a+c)geq0$$ or
                $$sumlimits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)geq0$$ or
                $$sumlimits_{cyc}(a^2-b^2)a^2geq0$$ or
                $$sumlimits_{cyc}left(frac{a^4+b^4}{2}-a^2b^2right)geq0,$$
                which is true by AM-GM.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  We need to prove that $$sumlimits_{cyc}frac{a^2-b^2}{b+c}geq0$$ or $$sumlimits_{cyc}(a^2-b^2)(a+b)(a+c)geq0$$ or
                  $$sumlimits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)geq0$$ or
                  $$sumlimits_{cyc}(a^2-b^2)a^2geq0$$ or
                  $$sumlimits_{cyc}left(frac{a^4+b^4}{2}-a^2b^2right)geq0,$$
                  which is true by AM-GM.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    We need to prove that $$sumlimits_{cyc}frac{a^2-b^2}{b+c}geq0$$ or $$sumlimits_{cyc}(a^2-b^2)(a+b)(a+c)geq0$$ or
                    $$sumlimits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)geq0$$ or
                    $$sumlimits_{cyc}(a^2-b^2)a^2geq0$$ or
                    $$sumlimits_{cyc}left(frac{a^4+b^4}{2}-a^2b^2right)geq0,$$
                    which is true by AM-GM.






                    share|cite|improve this answer









                    $endgroup$



                    We need to prove that $$sumlimits_{cyc}frac{a^2-b^2}{b+c}geq0$$ or $$sumlimits_{cyc}(a^2-b^2)(a+b)(a+c)geq0$$ or
                    $$sumlimits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)geq0$$ or
                    $$sumlimits_{cyc}(a^2-b^2)a^2geq0$$ or
                    $$sumlimits_{cyc}left(frac{a^4+b^4}{2}-a^2b^2right)geq0,$$
                    which is true by AM-GM.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 13 '16 at 15:04









                    Michael RozenbergMichael Rozenberg

                    105k1892198




                    105k1892198























                        0












                        $begingroup$

                        You can just do it in one step (without any computation) if you use the rearrangement inequality.



                        Notice that for any positive reals a, b and c,
                        $(a^2,b^2,c^2)$ and $(1/b+c,1/c+a,1/a+c)$ are similarly sorted. So,








                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You can just do it in one step (without any computation) if you use the rearrangement inequality.



                          Notice that for any positive reals a, b and c,
                          $(a^2,b^2,c^2)$ and $(1/b+c,1/c+a,1/a+c)$ are similarly sorted. So,








                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You can just do it in one step (without any computation) if you use the rearrangement inequality.



                            Notice that for any positive reals a, b and c,
                            $(a^2,b^2,c^2)$ and $(1/b+c,1/c+a,1/a+c)$ are similarly sorted. So,








                            share|cite|improve this answer









                            $endgroup$



                            You can just do it in one step (without any computation) if you use the rearrangement inequality.



                            Notice that for any positive reals a, b and c,
                            $(a^2,b^2,c^2)$ and $(1/b+c,1/c+a,1/a+c)$ are similarly sorted. So,









                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 30 '18 at 18:00









                            Shashwat1337Shashwat1337

                            789




                            789






























                                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%2f589678%2flet-abc0-prove-fraca2bc-fracb2ca-fracc2ab-ge%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

                                Bressuire

                                Image
                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                Read more

                                Cabo Verde

                                Image
                                  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                                Read more

                                Gyllenstierna

                                Image
                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                Read more