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







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  • 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






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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






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oldest

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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





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

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        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






























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