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Supremum of $a^ab+b^bc+c^cd+d^da$ with $a+b+c+d=4$ [closed]

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2

1

$begingroup$

Let $a,b,c,d>0$ I want to find the supremum of :

$$a^ab+b^bc+c^cd+d^da$$
With $a+b+c+d=4$

I claim that the supremum has the following form :
$$a^ab+3$$
With $a+b=4$

In fact it remains to prove the following theorem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $a>b>c>d$ then we have :
$$a^ab+b^bc+c^cd+d^da< a^ab+3$$

All of this is just an intuition and I'm really stuck to prove this...

If you have hints it will be nice .

Thanks in advance for your time .

real-analysis inequality

share|cite|improve this question

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

$endgroup$

closed as off-topic by Martin R, mrtaurho, Saad, Xander Henderson, RRL Mar 14 at 16:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, mrtaurho, Saad, Xander Henderson, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you heard of Lagrange multipliers?
  $endgroup$
  – enedil
  Mar 14 at 8:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $(a, b, c, d) = (1.99, 1.1, 0.9, 0.01)$ is a counter-example for your “theorem.”
  $endgroup$
  – Martin R
  Mar 14 at 9:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Some numerics: $a^ab$ has a max value of about 32.7037, for $a+b=4$ (wolframalpha.com/input/?i=max+x%5E(4x-x%5E2),+x+from+0+to+4). So the desired supremum is at least 32.7037+3.
  $endgroup$
  – bonsoon
  Mar 14 at 12:44
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Let $a,b,c,d>0$ I want to find the supremum of :

$$a^ab+b^bc+c^cd+d^da$$
With $a+b+c+d=4$

I claim that the supremum has the following form :
$$a^ab+3$$
With $a+b=4$

In fact it remains to prove the following theorem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $a>b>c>d$ then we have :
$$a^ab+b^bc+c^cd+d^da< a^ab+3$$

All of this is just an intuition and I'm really stuck to prove this...

If you have hints it will be nice .

Thanks in advance for your time .

real-analysis inequality

share|cite|improve this question

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

$endgroup$

closed as off-topic by Martin R, mrtaurho, Saad, Xander Henderson, RRL Mar 14 at 16:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, mrtaurho, Saad, Xander Henderson, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you heard of Lagrange multipliers?
  $endgroup$
  – enedil
  Mar 14 at 8:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $(a, b, c, d) = (1.99, 1.1, 0.9, 0.01)$ is a counter-example for your “theorem.”
  $endgroup$
  – Martin R
  Mar 14 at 9:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Some numerics: $a^ab$ has a max value of about 32.7037, for $a+b=4$ (wolframalpha.com/input/?i=max+x%5E(4x-x%5E2),+x+from+0+to+4). So the desired supremum is at least 32.7037+3.
  $endgroup$
  – bonsoon
  Mar 14 at 12:44
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $a,b,c,d>0$ I want to find the supremum of :

$$a^ab+b^bc+c^cd+d^da$$
With $a+b+c+d=4$

I claim that the supremum has the following form :
$$a^ab+3$$
With $a+b=4$

In fact it remains to prove the following theorem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $a>b>c>d$ then we have :
$$a^ab+b^bc+c^cd+d^da< a^ab+3$$

All of this is just an intuition and I'm really stuck to prove this...

If you have hints it will be nice .

Thanks in advance for your time .

real-analysis inequality

share|cite|improve this question

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

$endgroup$

share|cite|improve this question

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

share|cite|improve this question

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

asked Mar 14 at 8:16

FatsWallersFatsWallers

1307

1 Answer
1

active

oldest

votes

1

$begingroup$

Your conjecture is not true. I just generated 4 random numbers satisfying the condition $a+b+c+d=4$ and immediately found counter-examples.

share|cite|improve this answer

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A concrete counter-example would be more helpful ...
  $endgroup$
  – Martin R
  Mar 14 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PierreCarre Your counterexample does not satisfy the requirement $a>b>c>d$.
  $endgroup$
  – Arnaud D.
  Mar 14 at 11:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct @ArnaudD. I'll remove or edit the counter-example.
  $endgroup$
  – PierreCarre
  Mar 14 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $a=2.52799, b= 1.25971, c = 0.208747, d = 0.0035519$ is one counter-example obtained by simple simulation.
  $endgroup$
  – PierreCarre
  Mar 14 at 11:36
  

  
  
  
  

add a comment | 

1

$begingroup$

Your conjecture is not true. I just generated 4 random numbers satisfying the condition $a+b+c+d=4$ and immediately found counter-examples.

share|cite|improve this answer

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A concrete counter-example would be more helpful ...
  $endgroup$
  – Martin R
  Mar 14 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PierreCarre Your counterexample does not satisfy the requirement $a>b>c>d$.
  $endgroup$
  – Arnaud D.
  Mar 14 at 11:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct @ArnaudD. I'll remove or edit the counter-example.
  $endgroup$
  – PierreCarre
  Mar 14 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $a=2.52799, b= 1.25971, c = 0.208747, d = 0.0035519$ is one counter-example obtained by simple simulation.
  $endgroup$
  – PierreCarre
  Mar 14 at 11:36
  

  
  
  
  

add a comment | 

1

$begingroup$

Your conjecture is not true. I just generated 4 random numbers satisfying the condition $a+b+c+d=4$ and immediately found counter-examples.

share|cite|improve this answer

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A concrete counter-example would be more helpful ...
  $endgroup$
  – Martin R
  Mar 14 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PierreCarre Your counterexample does not satisfy the requirement $a>b>c>d$.
  $endgroup$
  – Arnaud D.
  Mar 14 at 11:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct @ArnaudD. I'll remove or edit the counter-example.
  $endgroup$
  – PierreCarre
  Mar 14 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $a=2.52799, b= 1.25971, c = 0.208747, d = 0.0035519$ is one counter-example obtained by simple simulation.
  $endgroup$
  – PierreCarre
  Mar 14 at 11:36
  

  
  
  
  

add a comment | 

$begingroup$

Your conjecture is not true. I just generated 4 random numbers satisfying the condition $a+b+c+d=4$ and immediately found counter-examples.

share|cite|improve this answer

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211

$endgroup$

share|cite|improve this answer

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211

answered Mar 14 at 8:53

PierreCarrePierreCarre

1,467211