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Prove that $sum_cycfraca^2ca^2 + 2c^2 ge 1$ [duplicate]

0

3

$begingroup$

This question already has an answer here:

• 
  How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
  
  1 answer
  
  

$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$

Here's what I did. My stupidity has reached a spiritual level.

We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.

$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$

$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$

$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$

$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$

I am done with my life.

inequality substitution cauchy-schwarz-inequality

share|cite|improve this question

edited Mar 23 at 18:08

StubbornAtom

6,37831440

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

$endgroup$

marked as duplicate by Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFan Mar 24 at 13:26

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
  

  
  
  
  
  
  

  
  $begingroup$
  is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 17:34
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
  $endgroup$
  – Word Shallow
  Mar 24 at 2:46
  
  

  
  
  
  

add a comment | 

0

3

$begingroup$

This question already has an answer here:

• 
  How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
  
  1 answer
  
  

$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$

Here's what I did. My stupidity has reached a spiritual level.

We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.

$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$

$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$

$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$

$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$

I am done with my life.

inequality substitution cauchy-schwarz-inequality

share|cite|improve this question

edited Mar 23 at 18:08

StubbornAtom

6,37831440

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

$endgroup$

marked as duplicate by Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFan Mar 24 at 13:26

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
  

  
  
  
  
  
  

  
  $begingroup$
  is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 23 at 17:34
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
  $endgroup$
  – Word Shallow
  Mar 24 at 2:46
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
  
  1 answer
  
  

$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$

Here's what I did. My stupidity has reached a spiritual level.

We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.

$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$

$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$

$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$

$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$

I am done with my life.

inequality substitution cauchy-schwarz-inequality

share|cite|improve this question

edited Mar 23 at 18:08

StubbornAtom

6,37831440

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

$endgroup$

share|cite|improve this question

edited Mar 23 at 18:08

StubbornAtom

6,37831440

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

share|cite|improve this question

edited Mar 23 at 18:08

StubbornAtom

6,37831440

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

edited Mar 23 at 18:08

StubbornAtom

6,37831440

edited Mar 23 at 18:08

StubbornAtom

6,37831440

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

asked Mar 23 at 17:29

Lê Thành ĐạtLê Thành Đạt

47313

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.

Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?

share|cite|improve this answer

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Power means works, right?
  $endgroup$
  – user574848
  Mar 24 at 7:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 7:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh I misread xyz=1
  $endgroup$
  – user574848
  Mar 24 at 7:41
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.

Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?

share|cite|improve this answer

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Power means works, right?
  $endgroup$
  – user574848
  Mar 24 at 7:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 7:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh I misread xyz=1
  $endgroup$
  – user574848
  Mar 24 at 7:41
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.

Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?

share|cite|improve this answer

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Power means works, right?
  $endgroup$
  – user574848
  Mar 24 at 7:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 7:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh I misread xyz=1
  $endgroup$
  – user574848
  Mar 24 at 7:41
  

  
  
  
  

add a comment | 

$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.

Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?

share|cite|improve this answer

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

share|cite|improve this answer

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 23 at 17:53

Michael RozenbergMichael Rozenberg

110k1896201