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If $a+b+c=1$ and a,b,c ＞0 prove $dfracb^2a+b^2+dfracc^2b+c^2+dfraca^2c+a^2 geqslant dfrac34$ [duplicate]

inequality with three variables and conditionHow prove this inequality $dfracRrgedfracbc+dfraccb$Symmetric inequality for a rational function of three variablessymetric inequality for a rational function of three variablesInequality for a rational function of three variablesIf $xyz=1$, prove $frac1y(x+y)+frac1z(y+z)+frac1x(z+x) geqslant frac32$Prove: $abcgeqslant 162$How to prove $(frac 1n)^n+(frac 2n)^n+cdots+(frac nn)^ngeqslantfrac3n+12n+2$inequality with three variables and conditionInequality with condition $a^2+b^2+c^2=3$Proving that $dfracabc^3+dfracbca^3+dfraccab^3> dfrac1a+dfrac1b+dfrac1c$

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  inequality with three variables and condition
  
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If $a+b+c=1$ and a,b,c＞0 prove $dfracb^2a+b^2+dfracc^2b+c^2+dfraca^2c+a^2 geqslant dfrac34$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can someone helpp?

inequality cauchy-schwarz-inequality rearrangement-inequality

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edited Mar 21 at 23:34

chaos

asked Mar 21 at 21:58

chaoschaos

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marked as duplicate by Martin R, Macavity inequality
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Mar 22 at 8:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  You asked the identical question three years ago (and got an answer) ...
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  – Martin R
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add a comment | 

-1

0

$begingroup$

This question already has an answer here:

• 
  inequality with three variables and condition
  
  1 answer
  
  

If $a+b+c=1$ and a,b,c＞0 prove $dfracb^2a+b^2+dfracc^2b+c^2+dfraca^2c+a^2 geqslant dfrac34$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can someone helpp?

inequality cauchy-schwarz-inequality rearrangement-inequality

share|cite|improve this question

edited Mar 21 at 23:34

chaos

asked Mar 21 at 21:58

chaoschaos

23918

$endgroup$

marked as duplicate by Martin R, Macavity inequality
Users with the  inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.

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Mar 22 at 8:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You asked the identical question three years ago (and got an answer) ...
  $endgroup$
  – Martin R
  Mar 22 at 8:39
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  inequality with three variables and condition
  
  1 answer
  
  

If $a+b+c=1$ and a,b,c＞0 prove $dfracb^2a+b^2+dfracc^2b+c^2+dfraca^2c+a^2 geqslant dfrac34$. I tried with CS Engel form,homogenization but ina anyway i can't prove inequality. Can someone helpp?

inequality cauchy-schwarz-inequality rearrangement-inequality

share|cite|improve this question

edited Mar 21 at 23:34

chaos

asked Mar 21 at 21:58

chaoschaos

23918

$endgroup$

share|cite|improve this question

edited Mar 21 at 23:34

chaos

asked Mar 21 at 21:58

chaoschaos

23918

share|cite|improve this question

edited Mar 21 at 23:34

chaos

asked Mar 21 at 21:58

chaoschaos

23918

asked Mar 21 at 21:58

chaoschaos

23918

asked Mar 21 at 21:58

chaoschaos

23918

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2

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It's wrong for real numbers.

For positive variables by C-S
$$sum_cycfraca^2a^2+c=sum_cycfraca^2(a+b)^2(c(a+b+c)+a^2)(a+b)^2geqfracleft(sumlimits_cyc(a^2+ab)right)^2sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2.$$
Thus, it's enough to prove that
$$4left(sumlimits_cyc(a^2+ab)right)^2geq3sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2$$ or
$$sum_cyc(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)geq0,$$ which is true because
$$sum_cyc(a^4-a^3b)geq0$$ by Rearrangement;
$$sum_cyca^3cgeqsum_cyca^2bc$$ it's
$$sum_cycfraca^2bgeqsum_cyca,$$ which is true by Rearrangement again and
$$sum_cyc(a^2b^2-a^2bc)=frac12sum_cycc^2(a-b)^2geq0.$$

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answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for the good efforts, even though duplicate.
  $endgroup$
  – Macavity
  Mar 22 at 8:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Macavity: Michael already answered the original question: math.stackexchange.com/a/1602340/42969 :)
  $endgroup$
  – Martin R
  Mar 22 at 8:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Macavity Thank you! I usually don't remember what I proved.
  $endgroup$
  – Michael Rozenberg
  Mar 22 at 9:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg: It is quite easy with Approach0 – it takes less than a minute to find that this question has been asked and answered before.
  $endgroup$
  – Martin R
  Mar 22 at 10:06
  
  

  
  
  
  

add a comment | 

2

$begingroup$

It's wrong for real numbers.

For positive variables by C-S
$$sum_cycfraca^2a^2+c=sum_cycfraca^2(a+b)^2(c(a+b+c)+a^2)(a+b)^2geqfracleft(sumlimits_cyc(a^2+ab)right)^2sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2.$$
Thus, it's enough to prove that
$$4left(sumlimits_cyc(a^2+ab)right)^2geq3sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2$$ or
$$sum_cyc(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)geq0,$$ which is true because
$$sum_cyc(a^4-a^3b)geq0$$ by Rearrangement;
$$sum_cyca^3cgeqsum_cyca^2bc$$ it's
$$sum_cycfraca^2bgeqsum_cyca,$$ which is true by Rearrangement again and
$$sum_cyc(a^2b^2-a^2bc)=frac12sum_cycc^2(a-b)^2geq0.$$

share|cite|improve this answer

answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for the good efforts, even though duplicate.
  $endgroup$
  – Macavity
  Mar 22 at 8:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Macavity: Michael already answered the original question: math.stackexchange.com/a/1602340/42969 :)
  $endgroup$
  – Martin R
  Mar 22 at 8:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Macavity Thank you! I usually don't remember what I proved.
  $endgroup$
  – Michael Rozenberg
  Mar 22 at 9:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg: It is quite easy with Approach0 – it takes less than a minute to find that this question has been asked and answered before.
  $endgroup$
  – Martin R
  Mar 22 at 10:06
  
  

  
  
  
  

add a comment | 

2

$begingroup$

It's wrong for real numbers.

For positive variables by C-S
$$sum_cycfraca^2a^2+c=sum_cycfraca^2(a+b)^2(c(a+b+c)+a^2)(a+b)^2geqfracleft(sumlimits_cyc(a^2+ab)right)^2sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2.$$
Thus, it's enough to prove that
$$4left(sumlimits_cyc(a^2+ab)right)^2geq3sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2$$ or
$$sum_cyc(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)geq0,$$ which is true because
$$sum_cyc(a^4-a^3b)geq0$$ by Rearrangement;
$$sum_cyca^3cgeqsum_cyca^2bc$$ it's
$$sum_cycfraca^2bgeqsum_cyca,$$ which is true by Rearrangement again and
$$sum_cyc(a^2b^2-a^2bc)=frac12sum_cycc^2(a-b)^2geq0.$$

share|cite|improve this answer

answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for the good efforts, even though duplicate.
  $endgroup$
  – Macavity
  Mar 22 at 8:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Macavity: Michael already answered the original question: math.stackexchange.com/a/1602340/42969 :)
  $endgroup$
  – Martin R
  Mar 22 at 8:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Macavity Thank you! I usually don't remember what I proved.
  $endgroup$
  – Michael Rozenberg
  Mar 22 at 9:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg: It is quite easy with Approach0 – it takes less than a minute to find that this question has been asked and answered before.
  $endgroup$
  – Martin R
  Mar 22 at 10:06
  
  

  
  
  
  

add a comment | 

$begingroup$

It's wrong for real numbers.

For positive variables by C-S
$$sum_cycfraca^2a^2+c=sum_cycfraca^2(a+b)^2(c(a+b+c)+a^2)(a+b)^2geqfracleft(sumlimits_cyc(a^2+ab)right)^2sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2.$$
Thus, it's enough to prove that
$$4left(sumlimits_cyc(a^2+ab)right)^2geq3sumlimits_cyc(a^2+c^2+ac+bc)(a+b)^2$$ or
$$sum_cyc(a^4-a^3b+5a^3c+3a^2b^2-8a^2bc)geq0,$$ which is true because
$$sum_cyc(a^4-a^3b)geq0$$ by Rearrangement;
$$sum_cyca^3cgeqsum_cyca^2bc$$ it's
$$sum_cycfraca^2bgeqsum_cyca,$$ which is true by Rearrangement again and
$$sum_cyc(a^2b^2-a^2bc)=frac12sum_cycc^2(a-b)^2geq0.$$

share|cite|improve this answer

answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201

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share|cite|improve this answer

answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 21 at 22:37

Michael RozenbergMichael Rozenberg

110k1896201