For $x$, $y$, $z$ the sides of a triangle, show...












0












$begingroup$


in $triangle ABC$, let $AB=z,BC=x,AC=y$,show that
$$sum_{cyc}frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}gefrac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$$



by well kown Iran 96 inequality
$$(xy+yz+xz)left(frac{1}{(x+y)^2}+frac{1}{(y+z)^2}+dfrac{1}{(x+z)^2}right)gedfrac{9}{4}$$






inequality symmetric-polynomials sos







share|cite|improve this question











$endgroup$












  • $begingroup$
    I have a proof by C-S and uvw,
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:03










  • $begingroup$
    maybe have without uvw methods?
    $endgroup$
    – communnites
    Dec 19 '18 at 8:04










  • $begingroup$
    We can use SOS here, but it's very ugly.
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:05






  • 1




    $begingroup$
    I want to see a solution which use SOS
    $endgroup$
    – Word Shallow
    Dec 19 '18 at 8:11










  • $begingroup$
    From where does this inequality come?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 9:00
















0












$begingroup$


in $triangle ABC$, let $AB=z,BC=x,AC=y$,show that
$$sum_{cyc}frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}gefrac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$$



by well kown Iran 96 inequality
$$(xy+yz+xz)left(frac{1}{(x+y)^2}+frac{1}{(y+z)^2}+dfrac{1}{(x+z)^2}right)gedfrac{9}{4}$$






inequality symmetric-polynomials sos







share|cite|improve this question











$endgroup$












  • $begingroup$
    I have a proof by C-S and uvw,
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:03










  • $begingroup$
    maybe have without uvw methods?
    $endgroup$
    – communnites
    Dec 19 '18 at 8:04










  • $begingroup$
    We can use SOS here, but it's very ugly.
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:05






  • 1




    $begingroup$
    I want to see a solution which use SOS
    $endgroup$
    – Word Shallow
    Dec 19 '18 at 8:11










  • $begingroup$
    From where does this inequality come?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 9:00














0












0








0


2



$begingroup$


in $triangle ABC$, let $AB=z,BC=x,AC=y$,show that
$$sum_{cyc}frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}gefrac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$$



by well kown Iran 96 inequality
$$(xy+yz+xz)left(frac{1}{(x+y)^2}+frac{1}{(y+z)^2}+dfrac{1}{(x+z)^2}right)gedfrac{9}{4}$$






inequality symmetric-polynomials sos







share|cite|improve this question











$endgroup$




in $triangle ABC$, let $AB=z,BC=x,AC=y$,show that
$$sum_{cyc}frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}gefrac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$$



by well kown Iran 96 inequality
$$(xy+yz+xz)left(frac{1}{(x+y)^2}+frac{1}{(y+z)^2}+dfrac{1}{(x+z)^2}right)gedfrac{9}{4}$$






inequality symmetric-polynomials sos




inequality symmetric-polynomials sos






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 16:45









Michael Rozenberg

105k1892197




105k1892197










asked Dec 19 '18 at 6:29









communnitescommunnites

1,1761535




1,1761535












  • $begingroup$
    I have a proof by C-S and uvw,
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:03










  • $begingroup$
    maybe have without uvw methods?
    $endgroup$
    – communnites
    Dec 19 '18 at 8:04










  • $begingroup$
    We can use SOS here, but it's very ugly.
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:05






  • 1




    $begingroup$
    I want to see a solution which use SOS
    $endgroup$
    – Word Shallow
    Dec 19 '18 at 8:11










  • $begingroup$
    From where does this inequality come?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 9:00


















  • $begingroup$
    I have a proof by C-S and uvw,
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:03










  • $begingroup$
    maybe have without uvw methods?
    $endgroup$
    – communnites
    Dec 19 '18 at 8:04










  • $begingroup$
    We can use SOS here, but it's very ugly.
    $endgroup$
    – Michael Rozenberg
    Dec 19 '18 at 8:05






  • 1




    $begingroup$
    I want to see a solution which use SOS
    $endgroup$
    – Word Shallow
    Dec 19 '18 at 8:11










  • $begingroup$
    From where does this inequality come?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 19 '18 at 9:00
















$begingroup$
I have a proof by C-S and uvw,
$endgroup$
– Michael Rozenberg
Dec 19 '18 at 8:03




$begingroup$
I have a proof by C-S and uvw,
$endgroup$
– Michael Rozenberg
Dec 19 '18 at 8:03












$begingroup$
maybe have without uvw methods?
$endgroup$
– communnites
Dec 19 '18 at 8:04




$begingroup$
maybe have without uvw methods?
$endgroup$
– communnites
Dec 19 '18 at 8:04












$begingroup$
We can use SOS here, but it's very ugly.
$endgroup$
– Michael Rozenberg
Dec 19 '18 at 8:05




$begingroup$
We can use SOS here, but it's very ugly.
$endgroup$
– Michael Rozenberg
Dec 19 '18 at 8:05




1




1




$begingroup$
I want to see a solution which use SOS
$endgroup$
– Word Shallow
Dec 19 '18 at 8:11




$begingroup$
I want to see a solution which use SOS
$endgroup$
– Word Shallow
Dec 19 '18 at 8:11












$begingroup$
From where does this inequality come?
$endgroup$
– Dr. Sonnhard Graubner
Dec 19 '18 at 9:00




$begingroup$
From where does this inequality come?
$endgroup$
– Dr. Sonnhard Graubner
Dec 19 '18 at 9:00










1 Answer
1






active

oldest

votes


















4












$begingroup$

We need to prove that:
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geqfrac{9}{4xyz(x+y+z)}.$$
Now, by C-S
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqsum_{cyc}frac{y^2z^2}{yz(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqfrac{(xy+xz+yz)^2}{sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)}.$$
Thus, it's enough to prove that
$$4xyz(x+y+z)(xy+xz+yz)^2geq9sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)$$ or
$$sum_{sym}(9x^7y-18x^6y^2+27x^5y^3-18x^4y^4-9x^5y^2z+4x^4y^3z+4x^4y^2z^2+x^3y^3z^2)geq0$$ or
$$9sum_{cyc}xy(x^6-2x^5y+3x^4y^2-4x^3y^3+3x^2y^4-2xy^5+y^6)-$$
$$-9xyzsum_{cyc}xy(x^3-x^2y-xy^2+y^3)-xyzsum_{cyc}(5x^3y^2+5x^3z^2-8x^3yz-2x^2y^2z)geq0$$ or
$$sum_{cyc}(x-y)^2xy(9(x^2+y^2)^2-9xyz(x+y)-5z^4-xyz^2)geq0$$ or
$$7sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)+$$
$$+sum_{cyc}(x-y)^2xy(2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4))geq0.$$



We'll prove that $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq0.$$
Indeed, let $xgeq ygeq z$.



Thus, $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq$$
$$geq(x-z)^2xz(x^4+z^4-y^4)+(y-z)^2yz(y^4+z^4-x^4)geq$$
$$geq(y-z)^2xz(x^4-y^4)+(y-z)^2yz(y^4-x^4)=$$
$$=z(y-z)^2(x^4-y^4)(x-y)geq0.$$
Thus, it's enough to prove that:
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq0.$$
We'll prove that:
$$sqrt[4]{frac{x^4+9x^2y^2+y^4}{11}}geqsqrt[3]{frac{xy(x+y)}{2}}.$$
Let $x^2+y^2=2uxy$.



Thus, $ugeq1$ and we need to prove that:
$$left(frac{x^4+9x^2y^2+y^4}{11}right)^3geqleft(frac{xy(x+y)}{2}right)^4$$ or
$$left(frac{4u^2-2+9}{11}right)^3geqfrac{(2u+2)^2}{16}$$ or $f(u)geq0,$ where
$$f(u)=3ln(4u^2+7)-2ln(u+1)+2ln2-3ln11.$$
But $$f'(u)=frac{24u}{4u^2+7}-frac{2}{u+1}>0,$$ which says $f(u)geq f(1)=0.$



Also, by AM-GM $$sqrt[3]{frac{xy(x+y)}{2}}geqsqrt{xy}.$$



Now, let $z=tsqrt[3]{frac{xy(x+y)}{2}}.$



Thus,
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq$$
$$geq2z^4-left(sqrt[3]{frac{xy(x+y)}{2}}right)^2z^2-18left(sqrt[3]{frac{xy(x+y)}{2}}right)^3z+22left(sqrt[3]{frac{xy(x+y)}{2}}right)^4=$$
$$=left(sqrt[3]{frac{xy(x+y)}{2}}right)^4(2t^4-t^2-18t+22)geq0.$$
Can you end it now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
    $endgroup$
    – function sug
    Dec 20 '18 at 6:33










  • $begingroup$
    because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
    $endgroup$
    – function sug
    Dec 20 '18 at 6:36












  • $begingroup$
    use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
    $endgroup$
    – function sug
    Dec 20 '18 at 6:39






  • 1




    $begingroup$
    @function sug I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 20 '18 at 9:09










  • $begingroup$
    Thanks you,I have understand
    $endgroup$
    – function sug
    Dec 20 '18 at 12:05











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1 Answer
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1 Answer
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active

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active

oldest

votes









4












$begingroup$

We need to prove that:
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geqfrac{9}{4xyz(x+y+z)}.$$
Now, by C-S
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqsum_{cyc}frac{y^2z^2}{yz(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqfrac{(xy+xz+yz)^2}{sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)}.$$
Thus, it's enough to prove that
$$4xyz(x+y+z)(xy+xz+yz)^2geq9sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)$$ or
$$sum_{sym}(9x^7y-18x^6y^2+27x^5y^3-18x^4y^4-9x^5y^2z+4x^4y^3z+4x^4y^2z^2+x^3y^3z^2)geq0$$ or
$$9sum_{cyc}xy(x^6-2x^5y+3x^4y^2-4x^3y^3+3x^2y^4-2xy^5+y^6)-$$
$$-9xyzsum_{cyc}xy(x^3-x^2y-xy^2+y^3)-xyzsum_{cyc}(5x^3y^2+5x^3z^2-8x^3yz-2x^2y^2z)geq0$$ or
$$sum_{cyc}(x-y)^2xy(9(x^2+y^2)^2-9xyz(x+y)-5z^4-xyz^2)geq0$$ or
$$7sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)+$$
$$+sum_{cyc}(x-y)^2xy(2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4))geq0.$$



We'll prove that $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq0.$$
Indeed, let $xgeq ygeq z$.



Thus, $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq$$
$$geq(x-z)^2xz(x^4+z^4-y^4)+(y-z)^2yz(y^4+z^4-x^4)geq$$
$$geq(y-z)^2xz(x^4-y^4)+(y-z)^2yz(y^4-x^4)=$$
$$=z(y-z)^2(x^4-y^4)(x-y)geq0.$$
Thus, it's enough to prove that:
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq0.$$
We'll prove that:
$$sqrt[4]{frac{x^4+9x^2y^2+y^4}{11}}geqsqrt[3]{frac{xy(x+y)}{2}}.$$
Let $x^2+y^2=2uxy$.



Thus, $ugeq1$ and we need to prove that:
$$left(frac{x^4+9x^2y^2+y^4}{11}right)^3geqleft(frac{xy(x+y)}{2}right)^4$$ or
$$left(frac{4u^2-2+9}{11}right)^3geqfrac{(2u+2)^2}{16}$$ or $f(u)geq0,$ where
$$f(u)=3ln(4u^2+7)-2ln(u+1)+2ln2-3ln11.$$
But $$f'(u)=frac{24u}{4u^2+7}-frac{2}{u+1}>0,$$ which says $f(u)geq f(1)=0.$



Also, by AM-GM $$sqrt[3]{frac{xy(x+y)}{2}}geqsqrt{xy}.$$



Now, let $z=tsqrt[3]{frac{xy(x+y)}{2}}.$



Thus,
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq$$
$$geq2z^4-left(sqrt[3]{frac{xy(x+y)}{2}}right)^2z^2-18left(sqrt[3]{frac{xy(x+y)}{2}}right)^3z+22left(sqrt[3]{frac{xy(x+y)}{2}}right)^4=$$
$$=left(sqrt[3]{frac{xy(x+y)}{2}}right)^4(2t^4-t^2-18t+22)geq0.$$
Can you end it now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
    $endgroup$
    – function sug
    Dec 20 '18 at 6:33










  • $begingroup$
    because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
    $endgroup$
    – function sug
    Dec 20 '18 at 6:36












  • $begingroup$
    use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
    $endgroup$
    – function sug
    Dec 20 '18 at 6:39






  • 1




    $begingroup$
    @function sug I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 20 '18 at 9:09










  • $begingroup$
    Thanks you,I have understand
    $endgroup$
    – function sug
    Dec 20 '18 at 12:05
















4












$begingroup$

We need to prove that:
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geqfrac{9}{4xyz(x+y+z)}.$$
Now, by C-S
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqsum_{cyc}frac{y^2z^2}{yz(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqfrac{(xy+xz+yz)^2}{sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)}.$$
Thus, it's enough to prove that
$$4xyz(x+y+z)(xy+xz+yz)^2geq9sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)$$ or
$$sum_{sym}(9x^7y-18x^6y^2+27x^5y^3-18x^4y^4-9x^5y^2z+4x^4y^3z+4x^4y^2z^2+x^3y^3z^2)geq0$$ or
$$9sum_{cyc}xy(x^6-2x^5y+3x^4y^2-4x^3y^3+3x^2y^4-2xy^5+y^6)-$$
$$-9xyzsum_{cyc}xy(x^3-x^2y-xy^2+y^3)-xyzsum_{cyc}(5x^3y^2+5x^3z^2-8x^3yz-2x^2y^2z)geq0$$ or
$$sum_{cyc}(x-y)^2xy(9(x^2+y^2)^2-9xyz(x+y)-5z^4-xyz^2)geq0$$ or
$$7sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)+$$
$$+sum_{cyc}(x-y)^2xy(2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4))geq0.$$



We'll prove that $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq0.$$
Indeed, let $xgeq ygeq z$.



Thus, $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq$$
$$geq(x-z)^2xz(x^4+z^4-y^4)+(y-z)^2yz(y^4+z^4-x^4)geq$$
$$geq(y-z)^2xz(x^4-y^4)+(y-z)^2yz(y^4-x^4)=$$
$$=z(y-z)^2(x^4-y^4)(x-y)geq0.$$
Thus, it's enough to prove that:
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq0.$$
We'll prove that:
$$sqrt[4]{frac{x^4+9x^2y^2+y^4}{11}}geqsqrt[3]{frac{xy(x+y)}{2}}.$$
Let $x^2+y^2=2uxy$.



Thus, $ugeq1$ and we need to prove that:
$$left(frac{x^4+9x^2y^2+y^4}{11}right)^3geqleft(frac{xy(x+y)}{2}right)^4$$ or
$$left(frac{4u^2-2+9}{11}right)^3geqfrac{(2u+2)^2}{16}$$ or $f(u)geq0,$ where
$$f(u)=3ln(4u^2+7)-2ln(u+1)+2ln2-3ln11.$$
But $$f'(u)=frac{24u}{4u^2+7}-frac{2}{u+1}>0,$$ which says $f(u)geq f(1)=0.$



Also, by AM-GM $$sqrt[3]{frac{xy(x+y)}{2}}geqsqrt{xy}.$$



Now, let $z=tsqrt[3]{frac{xy(x+y)}{2}}.$



Thus,
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq$$
$$geq2z^4-left(sqrt[3]{frac{xy(x+y)}{2}}right)^2z^2-18left(sqrt[3]{frac{xy(x+y)}{2}}right)^3z+22left(sqrt[3]{frac{xy(x+y)}{2}}right)^4=$$
$$=left(sqrt[3]{frac{xy(x+y)}{2}}right)^4(2t^4-t^2-18t+22)geq0.$$
Can you end it now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
    $endgroup$
    – function sug
    Dec 20 '18 at 6:33










  • $begingroup$
    because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
    $endgroup$
    – function sug
    Dec 20 '18 at 6:36












  • $begingroup$
    use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
    $endgroup$
    – function sug
    Dec 20 '18 at 6:39






  • 1




    $begingroup$
    @function sug I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 20 '18 at 9:09










  • $begingroup$
    Thanks you,I have understand
    $endgroup$
    – function sug
    Dec 20 '18 at 12:05














4












4








4





$begingroup$

We need to prove that:
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geqfrac{9}{4xyz(x+y+z)}.$$
Now, by C-S
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqsum_{cyc}frac{y^2z^2}{yz(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqfrac{(xy+xz+yz)^2}{sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)}.$$
Thus, it's enough to prove that
$$4xyz(x+y+z)(xy+xz+yz)^2geq9sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)$$ or
$$sum_{sym}(9x^7y-18x^6y^2+27x^5y^3-18x^4y^4-9x^5y^2z+4x^4y^3z+4x^4y^2z^2+x^3y^3z^2)geq0$$ or
$$9sum_{cyc}xy(x^6-2x^5y+3x^4y^2-4x^3y^3+3x^2y^4-2xy^5+y^6)-$$
$$-9xyzsum_{cyc}xy(x^3-x^2y-xy^2+y^3)-xyzsum_{cyc}(5x^3y^2+5x^3z^2-8x^3yz-2x^2y^2z)geq0$$ or
$$sum_{cyc}(x-y)^2xy(9(x^2+y^2)^2-9xyz(x+y)-5z^4-xyz^2)geq0$$ or
$$7sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)+$$
$$+sum_{cyc}(x-y)^2xy(2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4))geq0.$$



We'll prove that $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq0.$$
Indeed, let $xgeq ygeq z$.



Thus, $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq$$
$$geq(x-z)^2xz(x^4+z^4-y^4)+(y-z)^2yz(y^4+z^4-x^4)geq$$
$$geq(y-z)^2xz(x^4-y^4)+(y-z)^2yz(y^4-x^4)=$$
$$=z(y-z)^2(x^4-y^4)(x-y)geq0.$$
Thus, it's enough to prove that:
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq0.$$
We'll prove that:
$$sqrt[4]{frac{x^4+9x^2y^2+y^4}{11}}geqsqrt[3]{frac{xy(x+y)}{2}}.$$
Let $x^2+y^2=2uxy$.



Thus, $ugeq1$ and we need to prove that:
$$left(frac{x^4+9x^2y^2+y^4}{11}right)^3geqleft(frac{xy(x+y)}{2}right)^4$$ or
$$left(frac{4u^2-2+9}{11}right)^3geqfrac{(2u+2)^2}{16}$$ or $f(u)geq0,$ where
$$f(u)=3ln(4u^2+7)-2ln(u+1)+2ln2-3ln11.$$
But $$f'(u)=frac{24u}{4u^2+7}-frac{2}{u+1}>0,$$ which says $f(u)geq f(1)=0.$



Also, by AM-GM $$sqrt[3]{frac{xy(x+y)}{2}}geqsqrt{xy}.$$



Now, let $z=tsqrt[3]{frac{xy(x+y)}{2}}.$



Thus,
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq$$
$$geq2z^4-left(sqrt[3]{frac{xy(x+y)}{2}}right)^2z^2-18left(sqrt[3]{frac{xy(x+y)}{2}}right)^3z+22left(sqrt[3]{frac{xy(x+y)}{2}}right)^4=$$
$$=left(sqrt[3]{frac{xy(x+y)}{2}}right)^4(2t^4-t^2-18t+22)geq0.$$
Can you end it now?






share|cite|improve this answer











$endgroup$



We need to prove that:
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geqfrac{9}{4xyz(x+y+z)}.$$
Now, by C-S
$$sum_{cyc}frac{yz}{(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqsum_{cyc}frac{y^2z^2}{yz(y^2+z^2)^2(x+y-z)(x+z-y)}geq$$
$$geqfrac{(xy+xz+yz)^2}{sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)}.$$
Thus, it's enough to prove that
$$4xyz(x+y+z)(xy+xz+yz)^2geq9sumlimits_{cyc}yz(y^2+z^2)^2(x+y-z)(x+z-y)$$ or
$$sum_{sym}(9x^7y-18x^6y^2+27x^5y^3-18x^4y^4-9x^5y^2z+4x^4y^3z+4x^4y^2z^2+x^3y^3z^2)geq0$$ or
$$9sum_{cyc}xy(x^6-2x^5y+3x^4y^2-4x^3y^3+3x^2y^4-2xy^5+y^6)-$$
$$-9xyzsum_{cyc}xy(x^3-x^2y-xy^2+y^3)-xyzsum_{cyc}(5x^3y^2+5x^3z^2-8x^3yz-2x^2y^2z)geq0$$ or
$$sum_{cyc}(x-y)^2xy(9(x^2+y^2)^2-9xyz(x+y)-5z^4-xyz^2)geq0$$ or
$$7sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)+$$
$$+sum_{cyc}(x-y)^2xy(2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4))geq0.$$



We'll prove that $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq0.$$
Indeed, let $xgeq ygeq z$.



Thus, $$sum_{cyc}(x-y)^2xy(x^4+y^4-z^4)geq$$
$$geq(x-z)^2xz(x^4+z^4-y^4)+(y-z)^2yz(y^4+z^4-x^4)geq$$
$$geq(y-z)^2xz(x^4-y^4)+(y-z)^2yz(y^4-x^4)=$$
$$=z(y-z)^2(x^4-y^4)(x-y)geq0.$$
Thus, it's enough to prove that:
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq0.$$
We'll prove that:
$$sqrt[4]{frac{x^4+9x^2y^2+y^4}{11}}geqsqrt[3]{frac{xy(x+y)}{2}}.$$
Let $x^2+y^2=2uxy$.



Thus, $ugeq1$ and we need to prove that:
$$left(frac{x^4+9x^2y^2+y^4}{11}right)^3geqleft(frac{xy(x+y)}{2}right)^4$$ or
$$left(frac{4u^2-2+9}{11}right)^3geqfrac{(2u+2)^2}{16}$$ or $f(u)geq0,$ where
$$f(u)=3ln(4u^2+7)-2ln(u+1)+2ln2-3ln11.$$
But $$f'(u)=frac{24u}{4u^2+7}-frac{2}{u+1}>0,$$ which says $f(u)geq f(1)=0.$



Also, by AM-GM $$sqrt[3]{frac{xy(x+y)}{2}}geqsqrt{xy}.$$



Now, let $z=tsqrt[3]{frac{xy(x+y)}{2}}.$



Thus,
$$2z^4-xyz^2-9xy(x+y)z+2(x^4+9x^2y^2+y^4)geq$$
$$geq2z^4-left(sqrt[3]{frac{xy(x+y)}{2}}right)^2z^2-18left(sqrt[3]{frac{xy(x+y)}{2}}right)^3z+22left(sqrt[3]{frac{xy(x+y)}{2}}right)^4=$$
$$=left(sqrt[3]{frac{xy(x+y)}{2}}right)^4(2t^4-t^2-18t+22)geq0.$$
Can you end it now?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 6:21

























answered Dec 19 '18 at 15:31









Michael RozenbergMichael Rozenberg

105k1892197




105k1892197












  • $begingroup$
    at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
    $endgroup$
    – function sug
    Dec 20 '18 at 6:33










  • $begingroup$
    because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
    $endgroup$
    – function sug
    Dec 20 '18 at 6:36












  • $begingroup$
    use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
    $endgroup$
    – function sug
    Dec 20 '18 at 6:39






  • 1




    $begingroup$
    @function sug I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 20 '18 at 9:09










  • $begingroup$
    Thanks you,I have understand
    $endgroup$
    – function sug
    Dec 20 '18 at 12:05


















  • $begingroup$
    at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
    $endgroup$
    – function sug
    Dec 20 '18 at 6:33










  • $begingroup$
    because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
    $endgroup$
    – function sug
    Dec 20 '18 at 6:36












  • $begingroup$
    use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
    $endgroup$
    – function sug
    Dec 20 '18 at 6:39






  • 1




    $begingroup$
    @function sug I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 20 '18 at 9:09










  • $begingroup$
    Thanks you,I have understand
    $endgroup$
    – function sug
    Dec 20 '18 at 12:05
















$begingroup$
at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
$endgroup$
– function sug
Dec 20 '18 at 6:33




$begingroup$
at last inequality how to get $2t^4-t^2-18t+22$,becase I found you maybe something is not right,
$endgroup$
– function sug
Dec 20 '18 at 6:33












$begingroup$
because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
$endgroup$
– function sug
Dec 20 '18 at 6:36






$begingroup$
because it must to show this $2z^4-xyz^2-9xyz(x+y)z+2(x^4+9x^2y^2+y^4)ge 0$,and I think the following it equaliment something wrong
$endgroup$
– function sug
Dec 20 '18 at 6:36














$begingroup$
use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
$endgroup$
– function sug
Dec 20 '18 at 6:39




$begingroup$
use you it must to show $2z^4-xyz^2-9xy(x+y)z+22(xy)^2ge 0$,then How to following step?
$endgroup$
– function sug
Dec 20 '18 at 6:39




1




1




$begingroup$
@function sug I added something. See now.
$endgroup$
– Michael Rozenberg
Dec 20 '18 at 9:09




$begingroup$
@function sug I added something. See now.
$endgroup$
– Michael Rozenberg
Dec 20 '18 at 9:09












$begingroup$
Thanks you,I have understand
$endgroup$
– function sug
Dec 20 '18 at 12:05




$begingroup$
Thanks you,I have understand
$endgroup$
– function sug
Dec 20 '18 at 12:05


















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