showing that $fraca+b2cdotfracc+d2 =big( fraca+b+c+d4big)^2$ [on hold]

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I could expand the right side of the equation below to show that the sides are equal but I was wondering if there was some equality that would have allowed me to see that the left side equals the right? and $a,b,c,d >0$

$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 = Big( fraca+b+c+d4 Big)^3 $$

Here's the entire proof. The problem is from Arthur Engel's problem solving strategies in the section on inequalities. The question was to prove the inequality below

$3$. We have

$$ fracabc + abd + acd + bcd4 = frac12 Big( ab fracc+d2 + cd fraca+b2 Big) $$
$$ le frac12 Big( Big( fraca+b2 Big)^2 fracc+d2 + Big( fracc+d2 Big)^2 fraca+b2 Big) $$
$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 le Big( fraca+b+c+d4 Big)^3. $$

Hence,

$$ sqrt[3] fracabc + abd + acd + bcd4 le fraca+b+c+d4 le sqrt fraca^2+b^2+c^2+d^24. $$

(Original image 1)
(Original image 2)

proof-writing

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edited Aug 3 at 4:09

Somos

10.9k1831

asked Aug 2 at 13:05

john fowles

1,066817

put on hold as unclear what you're asking by mfl, John Ma, Isaac Browne, Taroccoesbrocco, Mostafa Ayaz Aug 3 at 7:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It cannot be true, you will never have a term in $a^3$ for the LHS for example.
  – nicomezi
  Aug 2 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is clearly false. The left hand vanishes if $a=-b$ but the right hand does not.
  – lulu
  Aug 2 at 13:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @mfl oh, it's meant to be an identity of polynomials. The restriction can't save it.
  – lulu
  Aug 2 at 13:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  You really want to show that $fraca+b2cdotfracc+d2 le big( fraca+b+c+d4big)^2$ because the right minus left is a square. This is AM-GM inequality.
  – Somos
  Aug 2 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it could be a typo. See @Somos's comment above.
  – Batominovski
  Aug 2 at 13:18
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

1

I could expand the right side of the equation below to show that the sides are equal but I was wondering if there was some equality that would have allowed me to see that the left side equals the right? and $a,b,c,d >0$

$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 = Big( fraca+b+c+d4 Big)^3 $$

Here's the entire proof. The problem is from Arthur Engel's problem solving strategies in the section on inequalities. The question was to prove the inequality below

$3$. We have

$$ fracabc + abd + acd + bcd4 = frac12 Big( ab fracc+d2 + cd fraca+b2 Big) $$
$$ le frac12 Big( Big( fraca+b2 Big)^2 fracc+d2 + Big( fracc+d2 Big)^2 fraca+b2 Big) $$
$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 le Big( fraca+b+c+d4 Big)^3. $$

Hence,

$$ sqrt[3] fracabc + abd + acd + bcd4 le fraca+b+c+d4 le sqrt fraca^2+b^2+c^2+d^24. $$

(Original image 1)
(Original image 2)

proof-writing

share|cite|improve this question

edited Aug 3 at 4:09

Somos

10.9k1831

asked Aug 2 at 13:05

john fowles

1,066817

put on hold as unclear what you're asking by mfl, John Ma, Isaac Browne, Taroccoesbrocco, Mostafa Ayaz Aug 3 at 7:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It cannot be true, you will never have a term in $a^3$ for the LHS for example.
  – nicomezi
  Aug 2 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is clearly false. The left hand vanishes if $a=-b$ but the right hand does not.
  – lulu
  Aug 2 at 13:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @mfl oh, it's meant to be an identity of polynomials. The restriction can't save it.
  – lulu
  Aug 2 at 13:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  You really want to show that $fraca+b2cdotfracc+d2 le big( fraca+b+c+d4big)^2$ because the right minus left is a square. This is AM-GM inequality.
  – Somos
  Aug 2 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it could be a typo. See @Somos's comment above.
  – Batominovski
  Aug 2 at 13:18
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

I could expand the right side of the equation below to show that the sides are equal but I was wondering if there was some equality that would have allowed me to see that the left side equals the right? and $a,b,c,d >0$

$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 = Big( fraca+b+c+d4 Big)^3 $$

Here's the entire proof. The problem is from Arthur Engel's problem solving strategies in the section on inequalities. The question was to prove the inequality below

$3$. We have

$$ fracabc + abd + acd + bcd4 = frac12 Big( ab fracc+d2 + cd fraca+b2 Big) $$
$$ le frac12 Big( Big( fraca+b2 Big)^2 fracc+d2 + Big( fracc+d2 Big)^2 fraca+b2 Big) $$
$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 le Big( fraca+b+c+d4 Big)^3. $$

Hence,

$$ sqrt[3] fracabc + abd + acd + bcd4 le fraca+b+c+d4 le sqrt fraca^2+b^2+c^2+d^24. $$

(Original image 1)
(Original image 2)

proof-writing

share|cite|improve this question

edited Aug 3 at 4:09

Somos

10.9k1831

asked Aug 2 at 13:05

john fowles

1,066817

I could expand the right side of the equation below to show that the sides are equal but I was wondering if there was some equality that would have allowed me to see that the left side equals the right? and $a,b,c,d >0$

$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 = Big( fraca+b+c+d4 Big)^3 $$

Here's the entire proof. The problem is from Arthur Engel's problem solving strategies in the section on inequalities. The question was to prove the inequality below

$3$. We have

$$ fracabc + abd + acd + bcd4 = frac12 Big( ab fracc+d2 + cd fraca+b2 Big) $$
$$ le frac12 Big( Big( fraca+b2 Big)^2 fracc+d2 + Big( fracc+d2 Big)^2 fraca+b2 Big) $$
$$ = fraca+b2 cdot fracc+d2 cdot fraca+b+c+d4 le Big( fraca+b+c+d4 Big)^3. $$

Hence,

$$ sqrt[3] fracabc + abd + acd + bcd4 le fraca+b+c+d4 le sqrt fraca^2+b^2+c^2+d^24. $$

(Original image 1)
(Original image 2)

proof-writing

share|cite|improve this question

edited Aug 3 at 4:09

Somos

10.9k1831

asked Aug 2 at 13:05

john fowles

1,066817

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 4:09

Somos

10.9k1831

edited Aug 3 at 4:09

Somos

10.9k1831

edited Aug 3 at 4:09

Somos

10.9k1831

10.9k1831

asked Aug 2 at 13:05

john fowles

1,066817

asked Aug 2 at 13:05

john fowles

1,066817

asked Aug 2 at 13:05

john fowles

1,066817

1,066817

put on hold as unclear what you're asking by mfl, John Ma, Isaac Browne, Taroccoesbrocco, Mostafa Ayaz Aug 3 at 7:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as unclear what you're asking by mfl, John Ma, Isaac Browne, Taroccoesbrocco, Mostafa Ayaz Aug 3 at 7:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It cannot be true, you will never have a term in $a^3$ for the LHS for example.
  – nicomezi
  Aug 2 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is clearly false. The left hand vanishes if $a=-b$ but the right hand does not.
  – lulu
  Aug 2 at 13:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @mfl oh, it's meant to be an identity of polynomials. The restriction can't save it.
  – lulu
  Aug 2 at 13:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  You really want to show that $fraca+b2cdotfracc+d2 le big( fraca+b+c+d4big)^2$ because the right minus left is a square. This is AM-GM inequality.
  – Somos
  Aug 2 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it could be a typo. See @Somos's comment above.
  – Batominovski
  Aug 2 at 13:18
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It cannot be true, you will never have a term in $a^3$ for the LHS for example.
  – nicomezi
  Aug 2 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is clearly false. The left hand vanishes if $a=-b$ but the right hand does not.
  – lulu
  Aug 2 at 13:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @mfl oh, it's meant to be an identity of polynomials. The restriction can't save it.
  – lulu
  Aug 2 at 13:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  You really want to show that $fraca+b2cdotfracc+d2 le big( fraca+b+c+d4big)^2$ because the right minus left is a square. This is AM-GM inequality.
  – Somos
  Aug 2 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it could be a typo. See @Somos's comment above.
  – Batominovski
  Aug 2 at 13:18
  

  
  
  
  

It cannot be true, you will never have a term in $a^3$ for the LHS for example.
– nicomezi
Aug 2 at 13:07

It cannot be true, you will never have a term in $a^3$ for the LHS for example.
– nicomezi
Aug 2 at 13:07

This is clearly false. The left hand vanishes if $a=-b$ but the right hand does not.
– lulu
Aug 2 at 13:09

This is clearly false. The left hand vanishes if $a=-b$ but the right hand does not.
– lulu
Aug 2 at 13:09

1

1

@mfl oh, it's meant to be an identity of polynomials. The restriction can't save it.
– lulu
Aug 2 at 13:11

@mfl oh, it's meant to be an identity of polynomials. The restriction can't save it.
– lulu
Aug 2 at 13:11

5

5

You really want to show that $fraca+b2cdotfracc+d2 le big( fraca+b+c+d4big)^2$ because the right minus left is a square. This is AM-GM inequality.
– Somos
Aug 2 at 13:14

You really want to show that $fraca+b2cdotfracc+d2 le big( fraca+b+c+d4big)^2$ because the right minus left is a square. This is AM-GM inequality.
– Somos
Aug 2 at 13:14

1

1

I think it could be a typo. See @Somos's comment above.
– Batominovski
Aug 2 at 13:18

I think it could be a typo. See @Somos's comment above.
– Batominovski
Aug 2 at 13:18

 | 
show 1 more comment

2 Answers
2

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votes

up vote
2
down vote



accepted

Let

• $x=fraca+b2$


• $y= fracc+d2$

then

$$fraca+b2cdotfracc+d2 =big( fraca+b+c+d4big)^2iff xy=left(fracx+y2right)^2$$

which is false in general indeed by AM-GM

$$fracx+y2ge sqrtxy iff xyle left(fracx+y2right)^2$$

and equality holds if and only if $x=y$.

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answered Aug 2 at 13:23

gimusi

63.8k73480

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up vote
-1
down vote

The given statement is false.

$$beginalign
fraca+b2fracc+d2~&=~left(fraca+b+c+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2+fracc+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)2\
0~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)4\
endalign$$

Hence you gave the restriction $a,b,c,d>0$ the L.H.S. cannot equal zero because we got two squares, which are never zero for any number $ne0$, plus a product of positive numbers which does not equal zero aswell.

Since the original question is about inequalites you might edit the question title to make clear what you are looking for.

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answered Aug 2 at 13:16

mrtaurho

619117

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Let

• $x=fraca+b2$


• $y= fracc+d2$

then

$$fraca+b2cdotfracc+d2 =big( fraca+b+c+d4big)^2iff xy=left(fracx+y2right)^2$$

which is false in general indeed by AM-GM

$$fracx+y2ge sqrtxy iff xyle left(fracx+y2right)^2$$

and equality holds if and only if $x=y$.

share|cite|improve this answer

answered Aug 2 at 13:23

gimusi

63.8k73480

add a comment | 

up vote
2
down vote



accepted

Let

• $x=fraca+b2$


• $y= fracc+d2$

then

$$fraca+b2cdotfracc+d2 =big( fraca+b+c+d4big)^2iff xy=left(fracx+y2right)^2$$

which is false in general indeed by AM-GM

$$fracx+y2ge sqrtxy iff xyle left(fracx+y2right)^2$$

and equality holds if and only if $x=y$.

share|cite|improve this answer

answered Aug 2 at 13:23

gimusi

63.8k73480

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Let

• $x=fraca+b2$


• $y= fracc+d2$

then

$$fraca+b2cdotfracc+d2 =big( fraca+b+c+d4big)^2iff xy=left(fracx+y2right)^2$$

which is false in general indeed by AM-GM

$$fracx+y2ge sqrtxy iff xyle left(fracx+y2right)^2$$

and equality holds if and only if $x=y$.

share|cite|improve this answer

answered Aug 2 at 13:23

gimusi

63.8k73480

Let

• $x=fraca+b2$


• $y= fracc+d2$

then

$$fraca+b2cdotfracc+d2 =big( fraca+b+c+d4big)^2iff xy=left(fracx+y2right)^2$$

which is false in general indeed by AM-GM

$$fracx+y2ge sqrtxy iff xyle left(fracx+y2right)^2$$

and equality holds if and only if $x=y$.

share|cite|improve this answer

answered Aug 2 at 13:23

gimusi

63.8k73480

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 13:23

gimusi

63.8k73480

answered Aug 2 at 13:23

gimusi

63.8k73480

answered Aug 2 at 13:23

gimusi

63.8k73480

63.8k73480

add a comment | 

add a comment | 

up vote
-1
down vote

The given statement is false.

$$beginalign
fraca+b2fracc+d2~&=~left(fraca+b+c+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2+fracc+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)2\
0~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)4\
endalign$$

Hence you gave the restriction $a,b,c,d>0$ the L.H.S. cannot equal zero because we got two squares, which are never zero for any number $ne0$, plus a product of positive numbers which does not equal zero aswell.

Since the original question is about inequalites you might edit the question title to make clear what you are looking for.

share|cite|improve this answer

answered Aug 2 at 13:16

mrtaurho

619117

add a comment | 

up vote
-1
down vote

The given statement is false.

$$beginalign
fraca+b2fracc+d2~&=~left(fraca+b+c+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2+fracc+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)2\
0~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)4\
endalign$$

Hence you gave the restriction $a,b,c,d>0$ the L.H.S. cannot equal zero because we got two squares, which are never zero for any number $ne0$, plus a product of positive numbers which does not equal zero aswell.

Since the original question is about inequalites you might edit the question title to make clear what you are looking for.

share|cite|improve this answer

answered Aug 2 at 13:16

mrtaurho

619117

add a comment | 

up vote
-1
down vote

up vote
-1
down vote

The given statement is false.

$$beginalign
fraca+b2fracc+d2~&=~left(fraca+b+c+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2+fracc+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)2\
0~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)4\
endalign$$

Hence you gave the restriction $a,b,c,d>0$ the L.H.S. cannot equal zero because we got two squares, which are never zero for any number $ne0$, plus a product of positive numbers which does not equal zero aswell.

Since the original question is about inequalites you might edit the question title to make clear what you are looking for.

share|cite|improve this answer

answered Aug 2 at 13:16

mrtaurho

619117

The given statement is false.

$$beginalign
fraca+b2fracc+d2~&=~left(fraca+b+c+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2+fracc+d2right)^2\
fraca+b2fracc+d2~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)2\
0~&=~left(fraca+b2right)^2+left(fracc+d2right)^2+frac(a+b)(c+d)4\
endalign$$

Hence you gave the restriction $a,b,c,d>0$ the L.H.S. cannot equal zero because we got two squares, which are never zero for any number $ne0$, plus a product of positive numbers which does not equal zero aswell.

Since the original question is about inequalites you might edit the question title to make clear what you are looking for.

share|cite|improve this answer

answered Aug 2 at 13:16

mrtaurho

619117

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 13:16

mrtaurho

619117

answered Aug 2 at 13:16

mrtaurho

619117

answered Aug 2 at 13:16

mrtaurho

619117

619117

add a comment | 

add a comment |