Proving that $(abc)^2geqleft(frac{4Delta}{sqrt{3}}right)^3$, where $a$, $b$, $c$ are the sides, and $Delta$...
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited Nov 28 '18 at 13:30
Blue
47.7k870151
47.7k870151
asked Nov 24 '13 at 4:34
krishan actonkrishan acton
420214
420214
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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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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
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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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Jul 3 '14 at 11:29
answered Nov 24 '13 at 5:44
shadow10shadow10
2,855931
2,855931
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Nov 24 '13 at 4:40
Henning MakholmHenning Makholm
239k16303540
239k16303540
add a comment |
add a comment |
$begingroup$
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:
$$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
$$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$
$endgroup$
add a comment |
$begingroup$
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:
$$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
$$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$
$endgroup$
add a comment |
$begingroup$
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:
$$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
$$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$
$endgroup$
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:
$$sum_{cyc}left(a^4-2a^2b^2+sqrt[3]{a^4b^4c^4}right)=sum_{cyc}left(sqrt[3]{a^{12}}+sqrt[3]{a^4b^4c^4}-2a^2b^2right)geq$$
$$geqsum_{cyc}left(sqrt[3]{a^9b^3}+sqrt[3]{a^9c^3}-2a^2b^2right)=sum_{cyc}(a^3b+ab^3-2a^2b^2)geq0.$$
answered Nov 28 '18 at 13:02
Michael RozenbergMichael Rozenberg
97.6k1589188
97.6k1589188
add a comment |
add a comment |
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