Proving that $(abc)^2geqleft(frac{4Delta}{sqrt{3}}right)^3$, where $a$, $b$, $c$ are the sides, and $Delta$...












1












$begingroup$


Let $a$, $b$, $c$ be the sides of $triangle ABC$.




Prove $$(abc)^2geqfrac{4Delta}{sqrt{3}}$$
where $Delta$ is the area of the triangle.



(Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2geqleft(frac{4Delta}{sqrt{3}}right)^3$$ I incorporated the correction into the title. —@Blue)





  • Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$

  • I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.

  • Then $angle AFB = angle AFC = angle BFC=120^circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.

  • Observe that
    $$xy+yz+zx=frac{4Delta}{sqrt{3}}$$

  • So, we have a big inequality to prove:
    $$left(x^2+xy+y^2right)left(y^2+yz+z^2right)left(z^2+zx+x^2right)geq xy+yz+zx$$ but I can't prove it.


Thanks for help






geometry inequality







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$endgroup$

















    1












    $begingroup$


    Let $a$, $b$, $c$ be the sides of $triangle ABC$.




    Prove $$(abc)^2geqfrac{4Delta}{sqrt{3}}$$
    where $Delta$ is the area of the triangle.



    (Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2geqleft(frac{4Delta}{sqrt{3}}right)^3$$ I incorporated the correction into the title. —@Blue)





    • Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$

    • I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.

    • Then $angle AFB = angle AFC = angle BFC=120^circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.

    • Observe that
      $$xy+yz+zx=frac{4Delta}{sqrt{3}}$$

    • So, we have a big inequality to prove:
      $$left(x^2+xy+y^2right)left(y^2+yz+z^2right)left(z^2+zx+x^2right)geq xy+yz+zx$$ but I can't prove it.


    Thanks for help






    geometry inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      Let $a$, $b$, $c$ be the sides of $triangle ABC$.




      Prove $$(abc)^2geqfrac{4Delta}{sqrt{3}}$$
      where $Delta$ is the area of the triangle.



      (Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2geqleft(frac{4Delta}{sqrt{3}}right)^3$$ I incorporated the correction into the title. —@Blue)





      • Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$

      • I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.

      • Then $angle AFB = angle AFC = angle BFC=120^circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.

      • Observe that
        $$xy+yz+zx=frac{4Delta}{sqrt{3}}$$

      • So, we have a big inequality to prove:
        $$left(x^2+xy+y^2right)left(y^2+yz+z^2right)left(z^2+zx+x^2right)geq xy+yz+zx$$ but I can't prove it.


      Thanks for help






      geometry inequality







      share|cite|improve this question











      $endgroup$




      Let $a$, $b$, $c$ be the sides of $triangle ABC$.




      Prove $$(abc)^2geqfrac{4Delta}{sqrt{3}}$$
      where $Delta$ is the area of the triangle.



      (Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2geqleft(frac{4Delta}{sqrt{3}}right)^3$$ I incorporated the correction into the title. —@Blue)





      • Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$

      • I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.

      • Then $angle AFB = angle AFC = angle BFC=120^circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.

      • Observe that
        $$xy+yz+zx=frac{4Delta}{sqrt{3}}$$

      • So, we have a big inequality to prove:
        $$left(x^2+xy+y^2right)left(y^2+yz+z^2right)left(z^2+zx+x^2right)geq xy+yz+zx$$ but I can't prove it.


      Thanks for help






      geometry inequality




      geometry inequality






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      edited Nov 28 '18 at 13:30









      Blue

      47.7k870151




      47.7k870151










      asked Nov 24 '13 at 4:34









      krishan actonkrishan acton

      420214




      420214






















          3 Answers
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          $begingroup$

          The actual inequality is $(abc)^2ge left(dfrac{4Delta}{sqrt{3}}right)^3$.I will show this one.





          • Lemma:



            $displaystyle sin{alpha}+sin{beta}+sin{gamma} le frac{3sqrt{3}}{2}$
            when $alpha$,$beta$,$gamma$ are angles of a triangle.Now we see
            that $sin x$ is concave in $(0,pi)$ so applying Jensen's inequality
            we get $dfrac{sin{alpha}+sin{beta}+sin{gamma}}{3}le
            sinleft({dfrac{alpha+beta+gamma}{3}}right)=dfrac{sqrt{3}}{2}$
            So our lemma is proved.




          Now to the actual problem.$a+b+c=2R(sin{alpha}+sin{beta}+sin{gamma})le 3sqrt{3}R$ (from the lemma).
          Now $$dfrac{4Delta}{sqrt{3}}=dfrac{abc}{Rsqrt{3}}le dfrac{3abc}{a+b+c}le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            Your inequality $(abc)^2 ge frac{4}{sqrt 3}triangle$ cannot be true in general, as a matter of simple dimensional analysis:



            If you shrink the triangle by a linear factor of $2$, the left-hand side $(abc)^2$ becomes $64$ times smaller, whereas the right-hand side only becomes $4$ times smaller. Since both sides of the inequality are nonzero, even if it happens to be true to begin with, you can make it false by considering a small enough triangle similar to the original one.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              I think you mean to the following:
              $$(abc)^{frac{2}{3}}geqfrac{4Delta}{sqrt3},$$ which is
              $$16Delta^2leq3sqrt[3]{a^4b^4c^4}$$ or
              $$sum_{cyc}(2a^2b^2-a^4)leqsqrt[3]{a^4b^4c^4}$$ or
              $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)geq0,$$ which is true by Schur and AM-GM:
              $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
              $$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$






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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2












                $begingroup$

                The actual inequality is $(abc)^2ge left(dfrac{4Delta}{sqrt{3}}right)^3$.I will show this one.





                • Lemma:



                  $displaystyle sin{alpha}+sin{beta}+sin{gamma} le frac{3sqrt{3}}{2}$
                  when $alpha$,$beta$,$gamma$ are angles of a triangle.Now we see
                  that $sin x$ is concave in $(0,pi)$ so applying Jensen's inequality
                  we get $dfrac{sin{alpha}+sin{beta}+sin{gamma}}{3}le
                  sinleft({dfrac{alpha+beta+gamma}{3}}right)=dfrac{sqrt{3}}{2}$
                  So our lemma is proved.




                Now to the actual problem.$a+b+c=2R(sin{alpha}+sin{beta}+sin{gamma})le 3sqrt{3}R$ (from the lemma).
                Now $$dfrac{4Delta}{sqrt{3}}=dfrac{abc}{Rsqrt{3}}le dfrac{3abc}{a+b+c}le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  The actual inequality is $(abc)^2ge left(dfrac{4Delta}{sqrt{3}}right)^3$.I will show this one.





                  • Lemma:



                    $displaystyle sin{alpha}+sin{beta}+sin{gamma} le frac{3sqrt{3}}{2}$
                    when $alpha$,$beta$,$gamma$ are angles of a triangle.Now we see
                    that $sin x$ is concave in $(0,pi)$ so applying Jensen's inequality
                    we get $dfrac{sin{alpha}+sin{beta}+sin{gamma}}{3}le
                    sinleft({dfrac{alpha+beta+gamma}{3}}right)=dfrac{sqrt{3}}{2}$
                    So our lemma is proved.




                  Now to the actual problem.$a+b+c=2R(sin{alpha}+sin{beta}+sin{gamma})le 3sqrt{3}R$ (from the lemma).
                  Now $$dfrac{4Delta}{sqrt{3}}=dfrac{abc}{Rsqrt{3}}le dfrac{3abc}{a+b+c}le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The actual inequality is $(abc)^2ge left(dfrac{4Delta}{sqrt{3}}right)^3$.I will show this one.





                    • Lemma:



                      $displaystyle sin{alpha}+sin{beta}+sin{gamma} le frac{3sqrt{3}}{2}$
                      when $alpha$,$beta$,$gamma$ are angles of a triangle.Now we see
                      that $sin x$ is concave in $(0,pi)$ so applying Jensen's inequality
                      we get $dfrac{sin{alpha}+sin{beta}+sin{gamma}}{3}le
                      sinleft({dfrac{alpha+beta+gamma}{3}}right)=dfrac{sqrt{3}}{2}$
                      So our lemma is proved.




                    Now to the actual problem.$a+b+c=2R(sin{alpha}+sin{beta}+sin{gamma})le 3sqrt{3}R$ (from the lemma).
                    Now $$dfrac{4Delta}{sqrt{3}}=dfrac{abc}{Rsqrt{3}}le dfrac{3abc}{a+b+c}le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.






                    share|cite|improve this answer











                    $endgroup$



                    The actual inequality is $(abc)^2ge left(dfrac{4Delta}{sqrt{3}}right)^3$.I will show this one.





                    • Lemma:



                      $displaystyle sin{alpha}+sin{beta}+sin{gamma} le frac{3sqrt{3}}{2}$
                      when $alpha$,$beta$,$gamma$ are angles of a triangle.Now we see
                      that $sin x$ is concave in $(0,pi)$ so applying Jensen's inequality
                      we get $dfrac{sin{alpha}+sin{beta}+sin{gamma}}{3}le
                      sinleft({dfrac{alpha+beta+gamma}{3}}right)=dfrac{sqrt{3}}{2}$
                      So our lemma is proved.




                    Now to the actual problem.$a+b+c=2R(sin{alpha}+sin{beta}+sin{gamma})le 3sqrt{3}R$ (from the lemma).
                    Now $$dfrac{4Delta}{sqrt{3}}=dfrac{abc}{Rsqrt{3}}le dfrac{3abc}{a+b+c}le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jul 3 '14 at 11:29

























                    answered Nov 24 '13 at 5:44









                    shadow10shadow10

                    2,855931




                    2,855931























                        3












                        $begingroup$

                        Your inequality $(abc)^2 ge frac{4}{sqrt 3}triangle$ cannot be true in general, as a matter of simple dimensional analysis:



                        If you shrink the triangle by a linear factor of $2$, the left-hand side $(abc)^2$ becomes $64$ times smaller, whereas the right-hand side only becomes $4$ times smaller. Since both sides of the inequality are nonzero, even if it happens to be true to begin with, you can make it false by considering a small enough triangle similar to the original one.






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          Your inequality $(abc)^2 ge frac{4}{sqrt 3}triangle$ cannot be true in general, as a matter of simple dimensional analysis:



                          If you shrink the triangle by a linear factor of $2$, the left-hand side $(abc)^2$ becomes $64$ times smaller, whereas the right-hand side only becomes $4$ times smaller. Since both sides of the inequality are nonzero, even if it happens to be true to begin with, you can make it false by considering a small enough triangle similar to the original one.






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            Your inequality $(abc)^2 ge frac{4}{sqrt 3}triangle$ cannot be true in general, as a matter of simple dimensional analysis:



                            If you shrink the triangle by a linear factor of $2$, the left-hand side $(abc)^2$ becomes $64$ times smaller, whereas the right-hand side only becomes $4$ times smaller. Since both sides of the inequality are nonzero, even if it happens to be true to begin with, you can make it false by considering a small enough triangle similar to the original one.






                            share|cite|improve this answer









                            $endgroup$



                            Your inequality $(abc)^2 ge frac{4}{sqrt 3}triangle$ cannot be true in general, as a matter of simple dimensional analysis:



                            If you shrink the triangle by a linear factor of $2$, the left-hand side $(abc)^2$ becomes $64$ times smaller, whereas the right-hand side only becomes $4$ times smaller. Since both sides of the inequality are nonzero, even if it happens to be true to begin with, you can make it false by considering a small enough triangle similar to the original one.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 '13 at 4:40









                            Henning MakholmHenning Makholm

                            239k16303540




                            239k16303540























                                0












                                $begingroup$

                                I think you mean to the following:
                                $$(abc)^{frac{2}{3}}geqfrac{4Delta}{sqrt3},$$ which is
                                $$16Delta^2leq3sqrt[3]{a^4b^4c^4}$$ or
                                $$sum_{cyc}(2a^2b^2-a^4)leqsqrt[3]{a^4b^4c^4}$$ or
                                $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)geq0,$$ which is true by Schur and AM-GM:
                                $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
                                $$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  I think you mean to the following:
                                  $$(abc)^{frac{2}{3}}geqfrac{4Delta}{sqrt3},$$ which is
                                  $$16Delta^2leq3sqrt[3]{a^4b^4c^4}$$ or
                                  $$sum_{cyc}(2a^2b^2-a^4)leqsqrt[3]{a^4b^4c^4}$$ or
                                  $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)geq0,$$ which is true by Schur and AM-GM:
                                  $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
                                  $$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    I think you mean to the following:
                                    $$(abc)^{frac{2}{3}}geqfrac{4Delta}{sqrt3},$$ which is
                                    $$16Delta^2leq3sqrt[3]{a^4b^4c^4}$$ or
                                    $$sum_{cyc}(2a^2b^2-a^4)leqsqrt[3]{a^4b^4c^4}$$ or
                                    $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)geq0,$$ which is true by Schur and AM-GM:
                                    $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
                                    $$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    I think you mean to the following:
                                    $$(abc)^{frac{2}{3}}geqfrac{4Delta}{sqrt3},$$ which is
                                    $$16Delta^2leq3sqrt[3]{a^4b^4c^4}$$ or
                                    $$sum_{cyc}(2a^2b^2-a^4)leqsqrt[3]{a^4b^4c^4}$$ or
                                    $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)geq0,$$ which is true by Schur and AM-GM:
                                    $$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
                                    $$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 28 '18 at 13:02









                                    Michael RozenbergMichael Rozenberg

                                    97.6k1589188




                                    97.6k1589188






























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