Fdtxjr

How to pass a string to a command that expects a file?

A question on the ultrafilter number

Accountant/ lawyer will not return my call

Single word request: Harming the benefactor

Why the color red for the Republican Party

PTIJ: Why can't I eat anything?

Make a transparent 448*448 image

Replacing Windows 7 security updates with anti-virus?

The bar has been raised

How to create a hard link to an inode (ext4)?

2×2×2 rubik's cube corner is twisted!

Why does the negative sign arise in this thermodynamic relation?

Virginia employer terminated employee and wants signing bonus returned

In the late 1940’s to early 1950’s what technology was available that could melt a LOT of ice?

Space in array system equations

How do I deal with a powergamer in a game full of beginners in a school club?

How are such low op-amp input currents possible?

Are the terms "stab" and "staccato" synonyms?

Why is there a voltage between the mains ground and my radiator?

What is the likely impact of grounding an entire aircraft series?

Do f-stop and exposure time perfectly cancel?

What are the pros and cons of practising figure-eight landings on perpendicular runways?

How much stiffer are 23c tires over 28c?

Why is this plane circling around the Lucknow airport every day?

A nice IMO 1983 inequality from a stronger inequality

Prove variant of triangle inequality containing p-th power for 0 < p < 1A generalization of IMO 1983 problem 6How prove this inequality generalized from 1969 IMO problem 6Stronger Inequality than Holder InequalityStronger than Nesbitt inequalityNice InequalityA beautiful triangle inequalityIMO 1966 Shortlist InequalityA couple of big-O questions about a problem from number theoryInequality from AMM problems section

7

2

$begingroup$

If you are interested in IMO $1983$ please see:
$$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]geqq b(a+b-c)(a-c)(c-b),$$
where $a,b,c$ are three side-lengths of a triangle.

If $c≠rm mida,b,c$, the inequality is true, obviously! If $c=rm mid a,b,c$, we have $$(a - c)(c - b)= 0 Leftrightarrow c= fracc^2+aba+b.$$
I tried to prove that
$$f(c)- fleft(fracc^2+aba+bright)=(a - c)(c - b)Xgeqq 0,$$
where $$f(c)= 3[a^2b(a - b) + b^2c(b - c) + c^2a(c - a)]- b(a + b - c)(a - c)(c - b),$$
but without success! I found this inequality by using discriminant and some coefficient skills.

Hope you try to solve the rest on your own! Good luck everybody! Thank you so much!

inequality contest-math substitution geometric-inequalities buffalo-way

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks $!$ @Eevee Trainer $,$ so much $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It isn't same stronger with Bernhard Leeb $'$ solution $!$ See $:$ artofproblemsolving.com/community/c6h1175790p5672603 $.$ But both are nice $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:14
  
  

  
  
  
  

add a comment | 

7

2

$begingroup$

If you are interested in IMO $1983$ please see:
$$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]geqq b(a+b-c)(a-c)(c-b),$$
where $a,b,c$ are three side-lengths of a triangle.

If $c≠rm mida,b,c$, the inequality is true, obviously! If $c=rm mid a,b,c$, we have $$(a - c)(c - b)= 0 Leftrightarrow c= fracc^2+aba+b.$$
I tried to prove that
$$f(c)- fleft(fracc^2+aba+bright)=(a - c)(c - b)Xgeqq 0,$$
where $$f(c)= 3[a^2b(a - b) + b^2c(b - c) + c^2a(c - a)]- b(a + b - c)(a - c)(c - b),$$
but without success! I found this inequality by using discriminant and some coefficient skills.

Hope you try to solve the rest on your own! Good luck everybody! Thank you so much!

inequality contest-math substitution geometric-inequalities buffalo-way

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks $!$ @Eevee Trainer $,$ so much $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It isn't same stronger with Bernhard Leeb $'$ solution $!$ See $:$ artofproblemsolving.com/community/c6h1175790p5672603 $.$ But both are nice $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:14
  
  

  
  
  
  

add a comment | 

$begingroup$

If you are interested in IMO $1983$ please see:
$$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]geqq b(a+b-c)(a-c)(c-b),$$
where $a,b,c$ are three side-lengths of a triangle.

If $c≠rm mida,b,c$, the inequality is true, obviously! If $c=rm mid a,b,c$, we have $$(a - c)(c - b)= 0 Leftrightarrow c= fracc^2+aba+b.$$
I tried to prove that
$$f(c)- fleft(fracc^2+aba+bright)=(a - c)(c - b)Xgeqq 0,$$
where $$f(c)= 3[a^2b(a - b) + b^2c(b - c) + c^2a(c - a)]- b(a + b - c)(a - c)(c - b),$$
but without success! I found this inequality by using discriminant and some coefficient skills.

Hope you try to solve the rest on your own! Good luck everybody! Thank you so much!

inequality contest-math substitution geometric-inequalities buffalo-way

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

$endgroup$

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

1 Answer
1

active

oldest

votes

3

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 11:16
  

  
  
  
  

add a comment | 

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138409%2fa-nice-imo-1983-inequality-from-a-stronger-inequality%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 11:16
  

  
  
  
  

add a comment | 

3

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 11:16
  

  
  
  
  

add a comment | 

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200