Algebraic Inequality involving AM-GM-HM

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

If $$a,b,c ;epsilon ;R^+$$
Then show, $$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.

I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$fracbcx-a + fracabx-c + fracacx-b leq frac12 Bigl(xBigl)$$
This didn't help so instead I wrote it as $$2Biggl(frac 1a + frac1b + frac1cBiggl ) leq frac12 Bigl(a+b+cBigl)$$
Not sure what to to do next.
Any help would be appreciated.

inequality a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

asked Jul 24 at 7:51

Prakhar Nagpal

428315

add a comment | 

up vote
2
down vote

favorite

If $$a,b,c ;epsilon ;R^+$$
Then show, $$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.

I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$fracbcx-a + fracabx-c + fracacx-b leq frac12 Bigl(xBigl)$$
This didn't help so instead I wrote it as $$2Biggl(frac 1a + frac1b + frac1cBiggl ) leq frac12 Bigl(a+b+cBigl)$$
Not sure what to to do next.
Any help would be appreciated.

inequality a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

asked Jul 24 at 7:51

Prakhar Nagpal

428315

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

If $$a,b,c ;epsilon ;R^+$$
Then show, $$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.

I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$fracbcx-a + fracabx-c + fracacx-b leq frac12 Bigl(xBigl)$$
This didn't help so instead I wrote it as $$2Biggl(frac 1a + frac1b + frac1cBiggl ) leq frac12 Bigl(a+b+cBigl)$$
Not sure what to to do next.
Any help would be appreciated.

inequality a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

asked Jul 24 at 7:51

Prakhar Nagpal

428315

If $$a,b,c ;epsilon ;R^+$$
Then show, $$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.

I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$fracbcx-a + fracabx-c + fracacx-b leq frac12 Bigl(xBigl)$$
This didn't help so instead I wrote it as $$2Biggl(frac 1a + frac1b + frac1cBiggl ) leq frac12 Bigl(a+b+cBigl)$$
Not sure what to to do next.
Any help would be appreciated.

inequality a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

asked Jul 24 at 7:51

Prakhar Nagpal

428315

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

edited Jul 24 at 9:43

Michael Rozenberg

87.9k1578180

87.9k1578180

asked Jul 24 at 7:51

Prakhar Nagpal

428315

asked Jul 24 at 7:51

Prakhar Nagpal

428315

asked Jul 24 at 7:51

Prakhar Nagpal

428315

428315

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
3
down vote

Hint: Remember that $HM(a, b)$ can also be written as $dfrac 2aba+b$, which looks a lot like the LHS...

share|cite|improve this answer

answered Jul 24 at 7:55

Ovi

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, alright thank you so much.
  – Prakhar Nagpal
  Jul 24 at 7:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PrakharNagpal You're welcome!
  – Ovi
  Jul 24 at 7:57
  

  
  
  
  

add a comment | 

up vote
1
down vote

Proof

Just by $H_n leq A_n$, we have

beginalign*
fracbcb+c + fracaba+b + fracaca+c&=frac12left(frac2frac1b+frac1c+frac2frac1a+frac1b+frac2frac1a+frac1cright)\&leq frac12left(fracb+c2+fraca+b2+fraca+c2right)\&=fraca+b+c2.
endalign*

share|cite|improve this answer

edited Jul 24 at 10:58

answered Jul 24 at 10:53

mengdie1982

2,877216

add a comment | 

up vote
1
down vote

Multiply both sides by $4$ and rearrange:
$$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl) Rightarrow \
left(a+b-frac4aba+bright)+left(b+c-frac4bcb+cright)+left(c+a-frac4cac+aright)ge 0 Rightarrow \
frac(a-b)^2a+b+frac(b-c)^2b+c+frac(c-a)^2c+age 0.$$

share|cite|improve this answer

answered Jul 24 at 12:33

farruhota

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 Nice, refreshing way to do it
  – Ovi
  Jul 24 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ovi, thank you
  – farruhota
  Jul 24 at 18:03
  

  
  
  
  

add a comment | 

up vote
0
down vote

By C-S
$$sum_cycfracbcb+cleqsum_cycfracbc(1+1)^2left(frac1^2b+frac1^2cright)=fraca+b+c2.$$

share|cite|improve this answer

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861098%2falgebraic-inequality-involving-am-gm-hm%23new-answer', 'question_page');

);

Post as a guest

Name Email

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Hint: Remember that $HM(a, b)$ can also be written as $dfrac 2aba+b$, which looks a lot like the LHS...

share|cite|improve this answer

answered Jul 24 at 7:55

Ovi

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, alright thank you so much.
  – Prakhar Nagpal
  Jul 24 at 7:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PrakharNagpal You're welcome!
  – Ovi
  Jul 24 at 7:57
  

  
  
  
  

add a comment | 

up vote
3
down vote

Hint: Remember that $HM(a, b)$ can also be written as $dfrac 2aba+b$, which looks a lot like the LHS...

share|cite|improve this answer

answered Jul 24 at 7:55

Ovi

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, alright thank you so much.
  – Prakhar Nagpal
  Jul 24 at 7:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PrakharNagpal You're welcome!
  – Ovi
  Jul 24 at 7:57
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Hint: Remember that $HM(a, b)$ can also be written as $dfrac 2aba+b$, which looks a lot like the LHS...

share|cite|improve this answer

answered Jul 24 at 7:55

Ovi

11.3k935105

Hint: Remember that $HM(a, b)$ can also be written as $dfrac 2aba+b$, which looks a lot like the LHS...

share|cite|improve this answer

answered Jul 24 at 7:55

Ovi

11.3k935105

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 7:55

Ovi

11.3k935105

answered Jul 24 at 7:55

Ovi

11.3k935105

answered Jul 24 at 7:55

Ovi

11.3k935105

11.3k935105

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, alright thank you so much.
  – Prakhar Nagpal
  Jul 24 at 7:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PrakharNagpal You're welcome!
  – Ovi
  Jul 24 at 7:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, alright thank you so much.
  – Prakhar Nagpal
  Jul 24 at 7:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PrakharNagpal You're welcome!
  – Ovi
  Jul 24 at 7:57
  

  
  
  
  

Oh, alright thank you so much.
– Prakhar Nagpal
Jul 24 at 7:57

Oh, alright thank you so much.
– Prakhar Nagpal
Jul 24 at 7:57

@PrakharNagpal You're welcome!
– Ovi
Jul 24 at 7:57

@PrakharNagpal You're welcome!
– Ovi
Jul 24 at 7:57

add a comment | 

up vote
1
down vote

Proof

Just by $H_n leq A_n$, we have

beginalign*
fracbcb+c + fracaba+b + fracaca+c&=frac12left(frac2frac1b+frac1c+frac2frac1a+frac1b+frac2frac1a+frac1cright)\&leq frac12left(fracb+c2+fraca+b2+fraca+c2right)\&=fraca+b+c2.
endalign*

share|cite|improve this answer

edited Jul 24 at 10:58

answered Jul 24 at 10:53

mengdie1982

2,877216

add a comment | 

up vote
1
down vote

Proof

Just by $H_n leq A_n$, we have

beginalign*
fracbcb+c + fracaba+b + fracaca+c&=frac12left(frac2frac1b+frac1c+frac2frac1a+frac1b+frac2frac1a+frac1cright)\&leq frac12left(fracb+c2+fraca+b2+fraca+c2right)\&=fraca+b+c2.
endalign*

share|cite|improve this answer

edited Jul 24 at 10:58

answered Jul 24 at 10:53

mengdie1982

2,877216

add a comment | 

up vote
1
down vote

up vote
1
down vote

Proof

Just by $H_n leq A_n$, we have

beginalign*
fracbcb+c + fracaba+b + fracaca+c&=frac12left(frac2frac1b+frac1c+frac2frac1a+frac1b+frac2frac1a+frac1cright)\&leq frac12left(fracb+c2+fraca+b2+fraca+c2right)\&=fraca+b+c2.
endalign*

share|cite|improve this answer

edited Jul 24 at 10:58

answered Jul 24 at 10:53

mengdie1982

2,877216

Proof

Just by $H_n leq A_n$, we have

beginalign*
fracbcb+c + fracaba+b + fracaca+c&=frac12left(frac2frac1b+frac1c+frac2frac1a+frac1b+frac2frac1a+frac1cright)\&leq frac12left(fracb+c2+fraca+b2+fraca+c2right)\&=fraca+b+c2.
endalign*

share|cite|improve this answer

edited Jul 24 at 10:58

answered Jul 24 at 10:53

mengdie1982

2,877216

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 10:58

edited Jul 24 at 10:58

edited Jul 24 at 10:58

answered Jul 24 at 10:53

mengdie1982

2,877216

answered Jul 24 at 10:53

mengdie1982

2,877216

answered Jul 24 at 10:53

mengdie1982

2,877216

2,877216

add a comment | 

add a comment | 

up vote
1
down vote

Multiply both sides by $4$ and rearrange:
$$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl) Rightarrow \
left(a+b-frac4aba+bright)+left(b+c-frac4bcb+cright)+left(c+a-frac4cac+aright)ge 0 Rightarrow \
frac(a-b)^2a+b+frac(b-c)^2b+c+frac(c-a)^2c+age 0.$$

share|cite|improve this answer

answered Jul 24 at 12:33

farruhota

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 Nice, refreshing way to do it
  – Ovi
  Jul 24 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ovi, thank you
  – farruhota
  Jul 24 at 18:03
  

  
  
  
  

add a comment | 

up vote
1
down vote

Multiply both sides by $4$ and rearrange:
$$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl) Rightarrow \
left(a+b-frac4aba+bright)+left(b+c-frac4bcb+cright)+left(c+a-frac4cac+aright)ge 0 Rightarrow \
frac(a-b)^2a+b+frac(b-c)^2b+c+frac(c-a)^2c+age 0.$$

share|cite|improve this answer

answered Jul 24 at 12:33

farruhota

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 Nice, refreshing way to do it
  – Ovi
  Jul 24 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ovi, thank you
  – farruhota
  Jul 24 at 18:03
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Multiply both sides by $4$ and rearrange:
$$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl) Rightarrow \
left(a+b-frac4aba+bright)+left(b+c-frac4bcb+cright)+left(c+a-frac4cac+aright)ge 0 Rightarrow \
frac(a-b)^2a+b+frac(b-c)^2b+c+frac(c-a)^2c+age 0.$$

share|cite|improve this answer

answered Jul 24 at 12:33

farruhota

13.6k2632

Multiply both sides by $4$ and rearrange:
$$fracbcb+c + fracaba+b + fracaca+c leq frac12 Bigl(a+b+cBigl) Rightarrow \
left(a+b-frac4aba+bright)+left(b+c-frac4bcb+cright)+left(c+a-frac4cac+aright)ge 0 Rightarrow \
frac(a-b)^2a+b+frac(b-c)^2b+c+frac(c-a)^2c+age 0.$$

share|cite|improve this answer

answered Jul 24 at 12:33

farruhota

13.6k2632

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 12:33

farruhota

13.6k2632

answered Jul 24 at 12:33

farruhota

13.6k2632

answered Jul 24 at 12:33

farruhota

13.6k2632

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 Nice, refreshing way to do it
  – Ovi
  Jul 24 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ovi, thank you
  – farruhota
  Jul 24 at 18:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 Nice, refreshing way to do it
  – Ovi
  Jul 24 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ovi, thank you
  – farruhota
  Jul 24 at 18:03
  

  
  
  
  

+1 Nice, refreshing way to do it
– Ovi
Jul 24 at 17:55

+1 Nice, refreshing way to do it
– Ovi
Jul 24 at 17:55

@Ovi, thank you
– farruhota
Jul 24 at 18:03

@Ovi, thank you
– farruhota
Jul 24 at 18:03

add a comment | 

up vote
0
down vote

By C-S
$$sum_cycfracbcb+cleqsum_cycfracbc(1+1)^2left(frac1^2b+frac1^2cright)=fraca+b+c2.$$

share|cite|improve this answer

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

add a comment | 

up vote
0
down vote

By C-S
$$sum_cycfracbcb+cleqsum_cycfracbc(1+1)^2left(frac1^2b+frac1^2cright)=fraca+b+c2.$$

share|cite|improve this answer

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

add a comment | 

up vote
0
down vote

up vote
0
down vote

By C-S
$$sum_cycfracbcb+cleqsum_cycfracbc(1+1)^2left(frac1^2b+frac1^2cright)=fraca+b+c2.$$

share|cite|improve this answer

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

By C-S
$$sum_cycfracbcb+cleqsum_cycfracbc(1+1)^2left(frac1^2b+frac1^2cright)=fraca+b+c2.$$

share|cite|improve this answer

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

answered Jul 24 at 9:40

Michael Rozenberg

87.9k1578180

87.9k1578180

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861098%2falgebraic-inequality-involving-am-gm-hm%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email