Prove $Sigma_{cyc}(frac{a}{b-c}-3)^4ge193$
$begingroup$
The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."
$$Sigma_{cyc}left(frac{a}{b-c}-3right)^4ge193$$
Full expanding gives 12-degree polynomial with about 90 terms. It starts with $Sigma_{cyc}(a^{12}-16a^{11}b+8a^{11}c)$ and it does not look good for Muirhead or Schur.
Also I tried substitution of $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$. Then by $uvw$, it suffices to show when $x=y$ (See answer to linked question for details). That is, $frac{a}{b-c}=frac{b}{c-a}$ or $c=frac{a^2+b^2}{a+b}$, therefore either $a<c<b$ or $b<c<a$.
Given the constraints, it is clear that $x>0$ and substituting $y=x$, $z=-frac{1+x^2}{2x}$ gives nonnegative polynomial for $0<x$ (which is not nonnegative polynomial for all $x$).
However, it looks like I cannot deduce $x>0$ from the fact it is enough to consider $x=y$.
How can I prove it? Thank you!
inequality uvw
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
$endgroup$
add a comment |
$begingroup$
The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."
$$Sigma_{cyc}left(frac{a}{b-c}-3right)^4ge193$$
Full expanding gives 12-degree polynomial with about 90 terms. It starts with $Sigma_{cyc}(a^{12}-16a^{11}b+8a^{11}c)$ and it does not look good for Muirhead or Schur.
Also I tried substitution of $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$. Then by $uvw$, it suffices to show when $x=y$ (See answer to linked question for details). That is, $frac{a}{b-c}=frac{b}{c-a}$ or $c=frac{a^2+b^2}{a+b}$, therefore either $a<c<b$ or $b<c<a$.
Given the constraints, it is clear that $x>0$ and substituting $y=x$, $z=-frac{1+x^2}{2x}$ gives nonnegative polynomial for $0<x$ (which is not nonnegative polynomial for all $x$).
However, it looks like I cannot deduce $x>0$ from the fact it is enough to consider $x=y$.
How can I prove it? Thank you!
inequality uvw
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
$endgroup$
add a comment |
$begingroup$
The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."
$$Sigma_{cyc}left(frac{a}{b-c}-3right)^4ge193$$
Full expanding gives 12-degree polynomial with about 90 terms. It starts with $Sigma_{cyc}(a^{12}-16a^{11}b+8a^{11}c)$ and it does not look good for Muirhead or Schur.
Also I tried substitution of $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$. Then by $uvw$, it suffices to show when $x=y$ (See answer to linked question for details). That is, $frac{a}{b-c}=frac{b}{c-a}$ or $c=frac{a^2+b^2}{a+b}$, therefore either $a<c<b$ or $b<c<a$.
Given the constraints, it is clear that $x>0$ and substituting $y=x$, $z=-frac{1+x^2}{2x}$ gives nonnegative polynomial for $0<x$ (which is not nonnegative polynomial for all $x$).
However, it looks like I cannot deduce $x>0$ from the fact it is enough to consider $x=y$.
How can I prove it? Thank you!
inequality uvw
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
$endgroup$
The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."
$$Sigma_{cyc}left(frac{a}{b-c}-3right)^4ge193$$
Full expanding gives 12-degree polynomial with about 90 terms. It starts with $Sigma_{cyc}(a^{12}-16a^{11}b+8a^{11}c)$ and it does not look good for Muirhead or Schur.
Also I tried substitution of $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$. Then by $uvw$, it suffices to show when $x=y$ (See answer to linked question for details). That is, $frac{a}{b-c}=frac{b}{c-a}$ or $c=frac{a^2+b^2}{a+b}$, therefore either $a<c<b$ or $b<c<a$.
Given the constraints, it is clear that $x>0$ and substituting $y=x$, $z=-frac{1+x^2}{2x}$ gives nonnegative polynomial for $0<x$ (which is not nonnegative polynomial for all $x$).
However, it looks like I cannot deduce $x>0$ from the fact it is enough to consider $x=y$.
How can I prove it? Thank you!
inequality uvw
inequality uvw
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
asked Jan 14 at 15:34
didgognsdidgogns
3,218523
3,218523
add a comment |
add a comment |
1 Answer
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$begingroup$
It remains to make two steps only.
- For $$frac{a}{b-c}=frac{b}{c-a}$$ or
$$c=frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2left(frac{a}{b-frac{a^2+b^2}{a+b}}-3right)^2+left(frac{a^2+b^2}{a^2-b^2}-3right)^4geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2left(frac{t}{b-frac{t^2+1}{t+1}}-3right)^2+left(frac{t^2+1}{t^2-1}-3right)^4geq193$$ or
$$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95geq0,$$ which is obviously true for $t>0$.
- For $w^3rightarrow0$ let $frac{c}{a-b}rightarrow0$.
Thus, we need to prove that
$$left(frac{a}{b}-3right)^4+left(frac{b}{-a}-3right)^4+81geq193$$ or
$$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)geq0,$$ which is obvious again.
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
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add a comment |
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$begingroup$
It remains to make two steps only.
- For $$frac{a}{b-c}=frac{b}{c-a}$$ or
$$c=frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2left(frac{a}{b-frac{a^2+b^2}{a+b}}-3right)^2+left(frac{a^2+b^2}{a^2-b^2}-3right)^4geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2left(frac{t}{b-frac{t^2+1}{t+1}}-3right)^2+left(frac{t^2+1}{t^2-1}-3right)^4geq193$$ or
$$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95geq0,$$ which is obviously true for $t>0$.
- For $w^3rightarrow0$ let $frac{c}{a-b}rightarrow0$.
Thus, we need to prove that
$$left(frac{a}{b}-3right)^4+left(frac{b}{-a}-3right)^4+81geq193$$ or
$$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)geq0,$$ which is obvious again.
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
add a comment |
$begingroup$
It remains to make two steps only.
- For $$frac{a}{b-c}=frac{b}{c-a}$$ or
$$c=frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2left(frac{a}{b-frac{a^2+b^2}{a+b}}-3right)^2+left(frac{a^2+b^2}{a^2-b^2}-3right)^4geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2left(frac{t}{b-frac{t^2+1}{t+1}}-3right)^2+left(frac{t^2+1}{t^2-1}-3right)^4geq193$$ or
$$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95geq0,$$ which is obviously true for $t>0$.
- For $w^3rightarrow0$ let $frac{c}{a-b}rightarrow0$.
Thus, we need to prove that
$$left(frac{a}{b}-3right)^4+left(frac{b}{-a}-3right)^4+81geq193$$ or
$$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)geq0,$$ which is obvious again.
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
add a comment |
$begingroup$
It remains to make two steps only.
- For $$frac{a}{b-c}=frac{b}{c-a}$$ or
$$c=frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2left(frac{a}{b-frac{a^2+b^2}{a+b}}-3right)^2+left(frac{a^2+b^2}{a^2-b^2}-3right)^4geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2left(frac{t}{b-frac{t^2+1}{t+1}}-3right)^2+left(frac{t^2+1}{t^2-1}-3right)^4geq193$$ or
$$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95geq0,$$ which is obviously true for $t>0$.
- For $w^3rightarrow0$ let $frac{c}{a-b}rightarrow0$.
Thus, we need to prove that
$$left(frac{a}{b}-3right)^4+left(frac{b}{-a}-3right)^4+81geq193$$ or
$$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)geq0,$$ which is obvious again.
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
It remains to make two steps only.
- For $$frac{a}{b-c}=frac{b}{c-a}$$ or
$$c=frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2left(frac{a}{b-frac{a^2+b^2}{a+b}}-3right)^2+left(frac{a^2+b^2}{a^2-b^2}-3right)^4geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2left(frac{t}{b-frac{t^2+1}{t+1}}-3right)^2+left(frac{t^2+1}{t^2-1}-3right)^4geq193$$ or
$$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95geq0,$$ which is obviously true for $t>0$.
- For $w^3rightarrow0$ let $frac{c}{a-b}rightarrow0$.
Thus, we need to prove that
$$left(frac{a}{b}-3right)^4+left(frac{b}{-a}-3right)^4+81geq193$$ or
$$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)geq0,$$ which is obvious again.
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
answered Jan 14 at 17:57
Michael RozenbergMichael Rozenberg
101k1591193
101k1591193
add a comment |
add a comment |
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