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
$$ab + bc + ca geq 4sqrt3S?$$
geometry
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Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$
geometry
3
Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
â Jack D'Aurizioâ¦
Aug 14 at 11:13
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up vote
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down vote
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Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$
geometry
Is the Weitzenbock triangle inequality
$$a^2 + b^2 + c^2 geq 4sqrt3S$$
equivalent to the Roland inequality
$$ab + bc + ca geq 4sqrt3S?$$
geometry
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.
â Jack D'Aurizioâ¦
Aug 14 at 11:13
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3
Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
â Jack D'Aurizioâ¦
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.
â Jack D'Aurizioâ¦
Aug 14 at 11:13
Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
â Jack D'Aurizioâ¦
Aug 14 at 11:13
add a comment |Â
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 $(0,pi)$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 $(0,pi)$.
add a comment |Â
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 $(0,pi)$.
add a comment |Â
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 $(0,pi)$.
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 $(0,pi)$.
answered Aug 14 at 13:55
Jack D'Aurizioâ¦
271k31266632
271k31266632
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3
Roland inequality is sharper, since $a^2+b^2+c^2geq ab+bc+ac$ holds by the rearrangement inequality.
â Jack D'Aurizioâ¦
Aug 14 at 11:13