For $x$, $y$, $z$ the sides of a triangle, show...
$begingroup$
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
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
$endgroup$
add a comment |
$begingroup$
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
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
$endgroup$
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
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
edited Dec 19 '18 at 16:45
Michael Rozenberg
105k1892197
105k1892197
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
asked Dec 19 '18 at 6:29
communnitescommunnites
1,1761535
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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?
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
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
|
show 2 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046073%2ffor-x-y-z-the-sides-of-a-triangle-show-sum-cyc-fracyzyz2-x2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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?
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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?
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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?
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
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?
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
edited Dec 22 '18 at 6:21
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
answered Dec 19 '18 at 15:31
Michael RozenbergMichael Rozenberg
105k1892197
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046073%2ffor-x-y-z-the-sides-of-a-triangle-show-sum-cyc-fracyzyz2-x2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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