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Can I not use QM-AM inequality to solve this?

How to prove this inequality in Euclidean space?Inequality QuestionTrigonometric Inequality. $sin1+sin2+ldots+sinn <2$ .Proof of inequality containing a function and its integralTrigonometric inequality - x domainInequality proof for real numbers $x geq 2x^2$Solve this inequality with nested radicals (possibly by induction)Solve inequalityHow can we not use Muirhead's Inequality for proving the following inequality?I made an inequality to solve myself, but someone pointed out my solution is wrong.

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4

$begingroup$

I was doing my inequality homework and I encountered the following problem:

Show that $forall a,b,cinmathbbR^+$, $$sqrta^2+b^2+sqrtb^2+c^2+sqrtc^2+a^2ge(a+b+c)sqrt2.$$

I came up a prove using QM-AM inequality, as follows (you can also try to prove with QM-AM inequality for exercise).

But my problem is, can we not use the QM-AM inequality?

inequality alternative-proof

share|cite|improve this question

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean that you want to solve it without AM-QM?
  $endgroup$
  – Arthur
  Aug 12 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur Yes, the QM-AM proof is just for reference.
  $endgroup$
  – Culver Kwan
  Aug 12 at 8:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The two answers currently given below use essentially the same inequalities, but do not call them AM-QM. Is that good enough? Because "Do not use AM-QM" is a bit vague of a requirement.
  $endgroup$
  – Arthur
  Aug 12 at 8:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur In my worldview, QM/AM does not exist, only AM/GM.
  $endgroup$
  – Parcly Taxel
  Aug 12 at 8:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Consider a square whose one side has length $a+b+c$ and the other $b+c+a$ :).
  $endgroup$
  – Michal Adamaszek
  Aug 12 at 8:50
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

I was doing my inequality homework and I encountered the following problem:

Show that $forall a,b,cinmathbbR^+$, $$sqrta^2+b^2+sqrtb^2+c^2+sqrtc^2+a^2ge(a+b+c)sqrt2.$$

I came up a prove using QM-AM inequality, as follows (you can also try to prove with QM-AM inequality for exercise).

But my problem is, can we not use the QM-AM inequality?

inequality alternative-proof

share|cite|improve this question

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

2,11055 silver badges2727 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean that you want to solve it without AM-QM?
  $endgroup$
  – Arthur
  Aug 12 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur Yes, the QM-AM proof is just for reference.
  $endgroup$
  – Culver Kwan
  Aug 12 at 8:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The two answers currently given below use essentially the same inequalities, but do not call them AM-QM. Is that good enough? Because "Do not use AM-QM" is a bit vague of a requirement.
  $endgroup$
  – Arthur
  Aug 12 at 8:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur In my worldview, QM/AM does not exist, only AM/GM.
  $endgroup$
  – Parcly Taxel
  Aug 12 at 8:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Consider a square whose one side has length $a+b+c$ and the other $b+c+a$ :).
  $endgroup$
  – Michal Adamaszek
  Aug 12 at 8:50
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I was doing my inequality homework and I encountered the following problem:

Show that $forall a,b,cinmathbbR^+$, $$sqrta^2+b^2+sqrtb^2+c^2+sqrtc^2+a^2ge(a+b+c)sqrt2.$$

I came up a prove using QM-AM inequality, as follows (you can also try to prove with QM-AM inequality for exercise).

But my problem is, can we not use the QM-AM inequality?

inequality alternative-proof

share|cite|improve this question

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

2,11055 silver badges2727 bronze badges

$endgroup$

share|cite|improve this question

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

2,11055 silver badges2727 bronze badges

share|cite|improve this question

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

2,11055 silver badges2727 bronze badges

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

edited Aug 12 at 18:04

Parcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

2,11055 silver badges2727 bronze badges

asked Aug 12 at 8:16

Culver KwanCulver Kwan

2,11055 silver badges2727 bronze badges

4 Answers
4

active

oldest

votes

9

$begingroup$

Following the hint given in the comments, consider the following diagram:

The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.

share|cite|improve this answer

edited Aug 12 at 9:53

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Here is a purely visual proof based on Michal Adamaszek's comment:

share|cite|improve this answer

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

By CS:

$left(sumlimits_cycsqrta^2+b^2right)^2=2sumlimits_cyca^2+2sumlimits_cycsqrta^2+b^2sqrtc^2+a^2geq 2sumlimits_cyca^2+2sumlimits_cyc(ac+ab)=2(a+b+c)^2$

share|cite|improve this answer

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

By using Hölder’s Inequality,
$$ bigl(a^2+b^2bigr)^tfrac12bigl(1+1bigr)^tfrac12ge a+b \ sqrta^2+b^2mathstrut ge dfraca+bsqrt2$$
Similarly for others, and end it yourself

share|cite|improve this answer

edited Aug 12 at 9:36

Bernard

137k77 gold badges4343 silver badges126126 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

2,60611 silver badge2626 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is essentially the same proof.
  $endgroup$
  – Arnaud D.
  Aug 12 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, there is a better answer. So I accepted it.
  $endgroup$
  – Culver Kwan
  Aug 13 at 4:06
  

  
  
  
  

add a comment
 | 

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9

$begingroup$

Following the hint given in the comments, consider the following diagram:

The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.

share|cite|improve this answer

edited Aug 12 at 9:53

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

$endgroup$

add a comment
 | 

9

$begingroup$

Following the hint given in the comments, consider the following diagram:

The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.

share|cite|improve this answer

edited Aug 12 at 9:53

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Following the hint given in the comments, consider the following diagram:

The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.

share|cite|improve this answer

edited Aug 12 at 9:53

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

$endgroup$

share|cite|improve this answer

edited Aug 12 at 9:53

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

answered Aug 12 at 8:35

Parcly TaxelParcly Taxel

53.4k1313 gold badges8181 silver badges120120 bronze badges

5

$begingroup$

Here is a purely visual proof based on Michal Adamaszek's comment:

share|cite|improve this answer

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Here is a purely visual proof based on Michal Adamaszek's comment:

share|cite|improve this answer

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Here is a purely visual proof based on Michal Adamaszek's comment:

share|cite|improve this answer

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

answered Aug 12 at 9:52

Toby MakToby Mak

6,30622 gold badges1515 silver badges3030 bronze badges

4

$begingroup$

By CS:

$left(sumlimits_cycsqrta^2+b^2right)^2=2sumlimits_cyca^2+2sumlimits_cycsqrta^2+b^2sqrtc^2+a^2geq 2sumlimits_cyca^2+2sumlimits_cyc(ac+ab)=2(a+b+c)^2$

share|cite|improve this answer

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

By CS:

$left(sumlimits_cycsqrta^2+b^2right)^2=2sumlimits_cyca^2+2sumlimits_cycsqrta^2+b^2sqrtc^2+a^2geq 2sumlimits_cyca^2+2sumlimits_cyc(ac+ab)=2(a+b+c)^2$

share|cite|improve this answer

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

By CS:

$left(sumlimits_cycsqrta^2+b^2right)^2=2sumlimits_cyca^2+2sumlimits_cycsqrta^2+b^2sqrtc^2+a^2geq 2sumlimits_cyca^2+2sumlimits_cyc(ac+ab)=2(a+b+c)^2$

share|cite|improve this answer

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

answered Aug 12 at 8:32

AO1992AO1992

1,3831010 bronze badges

3

$begingroup$

By using Hölder’s Inequality,
$$ bigl(a^2+b^2bigr)^tfrac12bigl(1+1bigr)^tfrac12ge a+b \ sqrta^2+b^2mathstrut ge dfraca+bsqrt2$$
Similarly for others, and end it yourself

share|cite|improve this answer

edited Aug 12 at 9:36

Bernard

137k77 gold badges4343 silver badges126126 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

2,60611 silver badge2626 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is essentially the same proof.
  $endgroup$
  – Arnaud D.
  Aug 12 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, there is a better answer. So I accepted it.
  $endgroup$
  – Culver Kwan
  Aug 13 at 4:06
  

  
  
  
  

add a comment
 | 

3

$begingroup$

By using Hölder’s Inequality,
$$ bigl(a^2+b^2bigr)^tfrac12bigl(1+1bigr)^tfrac12ge a+b \ sqrta^2+b^2mathstrut ge dfraca+bsqrt2$$
Similarly for others, and end it yourself

share|cite|improve this answer

edited Aug 12 at 9:36

Bernard

137k77 gold badges4343 silver badges126126 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

2,60611 silver badge2626 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is essentially the same proof.
  $endgroup$
  – Arnaud D.
  Aug 12 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, there is a better answer. So I accepted it.
  $endgroup$
  – Culver Kwan
  Aug 13 at 4:06
  

  
  
  
  

add a comment
 | 

$begingroup$

By using Hölder’s Inequality,
$$ bigl(a^2+b^2bigr)^tfrac12bigl(1+1bigr)^tfrac12ge a+b \ sqrta^2+b^2mathstrut ge dfraca+bsqrt2$$
Similarly for others, and end it yourself

share|cite|improve this answer

edited Aug 12 at 9:36

Bernard

137k77 gold badges4343 silver badges126126 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

2,60611 silver badge2626 bronze badges

$endgroup$

share|cite|improve this answer

edited Aug 12 at 9:36

Bernard

137k77 gold badges4343 silver badges126126 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

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edited Aug 12 at 9:36

Bernard

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edited Aug 12 at 9:36

Bernard

137k77 gold badges4343 silver badges126126 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

2,60611 silver badge2626 bronze badges

answered Aug 12 at 8:36

Isaac YIU Math StudioIsaac YIU Math Studio

2,60611 silver badge2626 bronze badges