How to prove...












2












$begingroup$


Question:




let $a,b,c>0$,show that:



$$sqrt{frac{ab}{2a^2+bc+ca}}+sqrt{frac{bc}{2b^2+ca+ab}}+sqrt{frac{ca}{2c^2+ab+bc}}gefrac{81}{2}cdotfrac{abc}{(a+b+c)^3}$$




maybe this inequality can Use Holder inequality to solve it
$$left(sum_{cyc}sqrt{dfrac{ab}{2a^2+bc+ca}}right)^2sum_{cyc}(2a^2+bc+ca)ge(sum_{cyc}sqrt[3]{ab})^3$$



then I can't prove it






inequality a.m.-g.m.-inequality cauchy-schwarz-inequality uvw







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Question:




    let $a,b,c>0$,show that:



    $$sqrt{frac{ab}{2a^2+bc+ca}}+sqrt{frac{bc}{2b^2+ca+ab}}+sqrt{frac{ca}{2c^2+ab+bc}}gefrac{81}{2}cdotfrac{abc}{(a+b+c)^3}$$




    maybe this inequality can Use Holder inequality to solve it
    $$left(sum_{cyc}sqrt{dfrac{ab}{2a^2+bc+ca}}right)^2sum_{cyc}(2a^2+bc+ca)ge(sum_{cyc}sqrt[3]{ab})^3$$



    then I can't prove it






    inequality a.m.-g.m.-inequality cauchy-schwarz-inequality uvw







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Question:




      let $a,b,c>0$,show that:



      $$sqrt{frac{ab}{2a^2+bc+ca}}+sqrt{frac{bc}{2b^2+ca+ab}}+sqrt{frac{ca}{2c^2+ab+bc}}gefrac{81}{2}cdotfrac{abc}{(a+b+c)^3}$$




      maybe this inequality can Use Holder inequality to solve it
      $$left(sum_{cyc}sqrt{dfrac{ab}{2a^2+bc+ca}}right)^2sum_{cyc}(2a^2+bc+ca)ge(sum_{cyc}sqrt[3]{ab})^3$$



      then I can't prove it






      inequality a.m.-g.m.-inequality cauchy-schwarz-inequality uvw







      share|cite|improve this question











      $endgroup$




      Question:




      let $a,b,c>0$,show that:



      $$sqrt{frac{ab}{2a^2+bc+ca}}+sqrt{frac{bc}{2b^2+ca+ab}}+sqrt{frac{ca}{2c^2+ab+bc}}gefrac{81}{2}cdotfrac{abc}{(a+b+c)^3}$$




      maybe this inequality can Use Holder inequality to solve it
      $$left(sum_{cyc}sqrt{dfrac{ab}{2a^2+bc+ca}}right)^2sum_{cyc}(2a^2+bc+ca)ge(sum_{cyc}sqrt[3]{ab})^3$$



      then I can't prove it






      inequality a.m.-g.m.-inequality cauchy-schwarz-inequality uvw




      inequality a.m.-g.m.-inequality cauchy-schwarz-inequality uvw






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 18:53









      Michael Rozenberg

      106k1893198




      106k1893198










      asked May 18 '14 at 8:53









      china mathchina math

      10.2k631118




      10.2k631118






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          We can use also, AM-GM, C-S and uvw.



          Indeed, $$sum_{cyc}sqrt{frac{ab}{2a^2+bc+ca}}=sum_{cyc}frac{4ab}{2sqrt{4ab(2a^2+bc+ca)}}geq$$
          $$geqsum_{cyc}frac{4ab}{2a^2+bc+ca+4ab}=sum_{cyc}frac{4b^2}{frac{b}{a}(2a^2+bc+ca+4ab)}geq$$
          $$geqfrac{4(a+b+c)^2}{sumlimits_{cyc}frac{b}{a}(2a^2+bc+ca+4ab)}.$$
          Thus, it's enough to prove that
          $$8(a+b+c)^5geq81sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or
          $$8(a+b+c)^5geq81sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or
          $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81sum_{cyc}a^3c^2geq0.$$
          But by Rearrangement
          $$sum_{cyc}a^3c^2=a^2b^2c^2sum_{cyc}frac{a}{b^2}geq a^2b^2c^2sum_{cyc}left(acdotfrac{1}{a^2}right)=sum_{cyc}a^2b^2c.$$
          Id east, it's enough to prove that
          $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)geq0.$$
          Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



          Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove



          the last inequality for an extreme value of $w^3,$ which happens in the following cases.





          1. $w^3rightarrow0^+$.


          Let $crightarrow0^+$ and $b=1$.



          Thus, we need to prove that
          $$8(a+1)^5geq81a^2(a+1),$$ which is true by AM-GM:
          $$8(a+1)^5=8(a+1)^4(a+1)geq8left(2sqrt{a}right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$
          2. $b=c=1$.



          We need to prove that
          $$8(a+2)^5geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or
          $$(a-1)^2(4a^3+48a^2+9a+47)geq0.$$
          Done!






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            By Hölder's inequality we have $$left(sum_{cyc}sqrt{frac{a}{2bc + ca + abc}}right)^2 cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right) geq 27$$ Hence it suffices to prove that $$27 geq frac{81}{2} cdot frac{abc}{(a + b + c)^3} cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right)$$ which is obvious after homogenization and expansion






            share|cite|improve this answer









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              $begingroup$

              We can use also, AM-GM, C-S and uvw.



              Indeed, $$sum_{cyc}sqrt{frac{ab}{2a^2+bc+ca}}=sum_{cyc}frac{4ab}{2sqrt{4ab(2a^2+bc+ca)}}geq$$
              $$geqsum_{cyc}frac{4ab}{2a^2+bc+ca+4ab}=sum_{cyc}frac{4b^2}{frac{b}{a}(2a^2+bc+ca+4ab)}geq$$
              $$geqfrac{4(a+b+c)^2}{sumlimits_{cyc}frac{b}{a}(2a^2+bc+ca+4ab)}.$$
              Thus, it's enough to prove that
              $$8(a+b+c)^5geq81sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or
              $$8(a+b+c)^5geq81sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or
              $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81sum_{cyc}a^3c^2geq0.$$
              But by Rearrangement
              $$sum_{cyc}a^3c^2=a^2b^2c^2sum_{cyc}frac{a}{b^2}geq a^2b^2c^2sum_{cyc}left(acdotfrac{1}{a^2}right)=sum_{cyc}a^2b^2c.$$
              Id east, it's enough to prove that
              $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)geq0.$$
              Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



              Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove



              the last inequality for an extreme value of $w^3,$ which happens in the following cases.





              1. $w^3rightarrow0^+$.


              Let $crightarrow0^+$ and $b=1$.



              Thus, we need to prove that
              $$8(a+1)^5geq81a^2(a+1),$$ which is true by AM-GM:
              $$8(a+1)^5=8(a+1)^4(a+1)geq8left(2sqrt{a}right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$
              2. $b=c=1$.



              We need to prove that
              $$8(a+2)^5geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or
              $$(a-1)^2(4a^3+48a^2+9a+47)geq0.$$
              Done!






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                We can use also, AM-GM, C-S and uvw.



                Indeed, $$sum_{cyc}sqrt{frac{ab}{2a^2+bc+ca}}=sum_{cyc}frac{4ab}{2sqrt{4ab(2a^2+bc+ca)}}geq$$
                $$geqsum_{cyc}frac{4ab}{2a^2+bc+ca+4ab}=sum_{cyc}frac{4b^2}{frac{b}{a}(2a^2+bc+ca+4ab)}geq$$
                $$geqfrac{4(a+b+c)^2}{sumlimits_{cyc}frac{b}{a}(2a^2+bc+ca+4ab)}.$$
                Thus, it's enough to prove that
                $$8(a+b+c)^5geq81sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or
                $$8(a+b+c)^5geq81sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or
                $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81sum_{cyc}a^3c^2geq0.$$
                But by Rearrangement
                $$sum_{cyc}a^3c^2=a^2b^2c^2sum_{cyc}frac{a}{b^2}geq a^2b^2c^2sum_{cyc}left(acdotfrac{1}{a^2}right)=sum_{cyc}a^2b^2c.$$
                Id east, it's enough to prove that
                $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)geq0.$$
                Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



                Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove



                the last inequality for an extreme value of $w^3,$ which happens in the following cases.





                1. $w^3rightarrow0^+$.


                Let $crightarrow0^+$ and $b=1$.



                Thus, we need to prove that
                $$8(a+1)^5geq81a^2(a+1),$$ which is true by AM-GM:
                $$8(a+1)^5=8(a+1)^4(a+1)geq8left(2sqrt{a}right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$
                2. $b=c=1$.



                We need to prove that
                $$8(a+2)^5geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or
                $$(a-1)^2(4a^3+48a^2+9a+47)geq0.$$
                Done!






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  We can use also, AM-GM, C-S and uvw.



                  Indeed, $$sum_{cyc}sqrt{frac{ab}{2a^2+bc+ca}}=sum_{cyc}frac{4ab}{2sqrt{4ab(2a^2+bc+ca)}}geq$$
                  $$geqsum_{cyc}frac{4ab}{2a^2+bc+ca+4ab}=sum_{cyc}frac{4b^2}{frac{b}{a}(2a^2+bc+ca+4ab)}geq$$
                  $$geqfrac{4(a+b+c)^2}{sumlimits_{cyc}frac{b}{a}(2a^2+bc+ca+4ab)}.$$
                  Thus, it's enough to prove that
                  $$8(a+b+c)^5geq81sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or
                  $$8(a+b+c)^5geq81sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or
                  $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81sum_{cyc}a^3c^2geq0.$$
                  But by Rearrangement
                  $$sum_{cyc}a^3c^2=a^2b^2c^2sum_{cyc}frac{a}{b^2}geq a^2b^2c^2sum_{cyc}left(acdotfrac{1}{a^2}right)=sum_{cyc}a^2b^2c.$$
                  Id east, it's enough to prove that
                  $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)geq0.$$
                  Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



                  Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove



                  the last inequality for an extreme value of $w^3,$ which happens in the following cases.





                  1. $w^3rightarrow0^+$.


                  Let $crightarrow0^+$ and $b=1$.



                  Thus, we need to prove that
                  $$8(a+1)^5geq81a^2(a+1),$$ which is true by AM-GM:
                  $$8(a+1)^5=8(a+1)^4(a+1)geq8left(2sqrt{a}right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$
                  2. $b=c=1$.



                  We need to prove that
                  $$8(a+2)^5geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or
                  $$(a-1)^2(4a^3+48a^2+9a+47)geq0.$$
                  Done!






                  share|cite|improve this answer









                  $endgroup$



                  We can use also, AM-GM, C-S and uvw.



                  Indeed, $$sum_{cyc}sqrt{frac{ab}{2a^2+bc+ca}}=sum_{cyc}frac{4ab}{2sqrt{4ab(2a^2+bc+ca)}}geq$$
                  $$geqsum_{cyc}frac{4ab}{2a^2+bc+ca+4ab}=sum_{cyc}frac{4b^2}{frac{b}{a}(2a^2+bc+ca+4ab)}geq$$
                  $$geqfrac{4(a+b+c)^2}{sumlimits_{cyc}frac{b}{a}(2a^2+bc+ca+4ab)}.$$
                  Thus, it's enough to prove that
                  $$8(a+b+c)^5geq81sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or
                  $$8(a+b+c)^5geq81sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or
                  $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81sum_{cyc}a^3c^2geq0.$$
                  But by Rearrangement
                  $$sum_{cyc}a^3c^2=a^2b^2c^2sum_{cyc}frac{a}{b^2}geq a^2b^2c^2sum_{cyc}left(acdotfrac{1}{a^2}right)=sum_{cyc}a^2b^2c.$$
                  Id east, it's enough to prove that
                  $$8(a+b+c)^5-81sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)geq0.$$
                  Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.



                  Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove



                  the last inequality for an extreme value of $w^3,$ which happens in the following cases.





                  1. $w^3rightarrow0^+$.


                  Let $crightarrow0^+$ and $b=1$.



                  Thus, we need to prove that
                  $$8(a+1)^5geq81a^2(a+1),$$ which is true by AM-GM:
                  $$8(a+1)^5=8(a+1)^4(a+1)geq8left(2sqrt{a}right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$
                  2. $b=c=1$.



                  We need to prove that
                  $$8(a+2)^5geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or
                  $$(a-1)^2(4a^3+48a^2+9a+47)geq0.$$
                  Done!







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 18:52









                  Michael RozenbergMichael Rozenberg

                  106k1893198




                  106k1893198























                      0












                      $begingroup$

                      By Hölder's inequality we have $$left(sum_{cyc}sqrt{frac{a}{2bc + ca + abc}}right)^2 cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right) geq 27$$ Hence it suffices to prove that $$27 geq frac{81}{2} cdot frac{abc}{(a + b + c)^3} cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right)$$ which is obvious after homogenization and expansion






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        By Hölder's inequality we have $$left(sum_{cyc}sqrt{frac{a}{2bc + ca + abc}}right)^2 cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right) geq 27$$ Hence it suffices to prove that $$27 geq frac{81}{2} cdot frac{abc}{(a + b + c)^3} cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right)$$ which is obvious after homogenization and expansion






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          By Hölder's inequality we have $$left(sum_{cyc}sqrt{frac{a}{2bc + ca + abc}}right)^2 cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right) geq 27$$ Hence it suffices to prove that $$27 geq frac{81}{2} cdot frac{abc}{(a + b + c)^3} cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right)$$ which is obvious after homogenization and expansion






                          share|cite|improve this answer









                          $endgroup$



                          By Hölder's inequality we have $$left(sum_{cyc}sqrt{frac{a}{2bc + ca + abc}}right)^2 cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right) geq 27$$ Hence it suffices to prove that $$27 geq frac{81}{2} cdot frac{abc}{(a + b + c)^3} cdot left(sum_{cyc}frac{2bc + ca + abc}{a}right)$$ which is obvious after homogenization and expansion







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered May 24 '14 at 8:27









                          math110math110

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