Is the Weitzenbock inequality equivalent to the Roland inequality?

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Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$







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    Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
    – Jack D'Aurizio♦
    Aug 14 at 11:13














up vote
0
down vote

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Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$







share|cite|improve this question
















  • 3




    Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
    – Jack D'Aurizio♦
    Aug 14 at 11:13












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$







share|cite|improve this question












Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 14 at 11:11









Davidmath7

714




714







  • 3




    Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
    – Jack D'Aurizio♦
    Aug 14 at 11:13












  • 3




    Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
    – Jack D'Aurizio♦
    Aug 14 at 11:13







3




3




Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
– Jack D'Aurizio♦
Aug 14 at 11:13




Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
– Jack D'Aurizio♦
Aug 14 at 11:13










1 Answer
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As stated in the comments, Roland inequality is sharper since $a^2+b^2+c^2geq ab+ac+bc$ holds by the rearrangement inequality or simply from $(a-b)^2+(a-c)^2+(b-c)^2geq 0$. About proofs, I have outlined some proofs of Weitzenbock inequality here (among them, a nice proof without words). The inequality



$$ ab+ac+bc geq 4Deltasqrt3 $$
can be proved by invoking Heron's formula, Ravi substitution and Muirhead's inequality, or simply by exploiting $ab=frac2Deltasin C$ and the convexity of $frac1sintheta$ over $(0,pi)$.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    As stated in the comments, Roland inequality is sharper since $a^2+b^2+c^2geq ab+ac+bc$ holds by the rearrangement inequality or simply from $(a-b)^2+(a-c)^2+(b-c)^2geq 0$. About proofs, I have outlined some proofs of Weitzenbock inequality here (among them, a nice proof without words). The inequality



    $$ ab+ac+bc geq 4Deltasqrt3 $$
    can be proved by invoking Heron's formula, Ravi substitution and Muirhead's inequality, or simply by exploiting $ab=frac2Deltasin C$ and the convexity of $frac1sintheta$ over $(0,pi)$.






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      As stated in the comments, Roland inequality is sharper since $a^2+b^2+c^2geq ab+ac+bc$ holds by the rearrangement inequality or simply from $(a-b)^2+(a-c)^2+(b-c)^2geq 0$. About proofs, I have outlined some proofs of Weitzenbock inequality here (among them, a nice proof without words). The inequality



      $$ ab+ac+bc geq 4Deltasqrt3 $$
      can be proved by invoking Heron's formula, Ravi substitution and Muirhead's inequality, or simply by exploiting $ab=frac2Deltasin C$ and the convexity of $frac1sintheta$ over $(0,pi)$.






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        As stated in the comments, Roland inequality is sharper since $a^2+b^2+c^2geq ab+ac+bc$ holds by the rearrangement inequality or simply from $(a-b)^2+(a-c)^2+(b-c)^2geq 0$. About proofs, I have outlined some proofs of Weitzenbock inequality here (among them, a nice proof without words). The inequality



        $$ ab+ac+bc geq 4Deltasqrt3 $$
        can be proved by invoking Heron's formula, Ravi substitution and Muirhead's inequality, or simply by exploiting $ab=frac2Deltasin C$ and the convexity of $frac1sintheta$ over $(0,pi)$.






        share|cite|improve this answer












        As stated in the comments, Roland inequality is sharper since $a^2+b^2+c^2geq ab+ac+bc$ holds by the rearrangement inequality or simply from $(a-b)^2+(a-c)^2+(b-c)^2geq 0$. About proofs, I have outlined some proofs of Weitzenbock inequality here (among them, a nice proof without words). The inequality



        $$ ab+ac+bc geq 4Deltasqrt3 $$
        can be proved by invoking Heron's formula, Ravi substitution and Muirhead's inequality, or simply by exploiting $ab=frac2Deltasin C$ and the convexity of $frac1sintheta$ over $(0,pi)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 14 at 13:55









        Jack D'Aurizio♦

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