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Max and Min Inequality

Inequalities and floors.Max/Min ProblemAbsolute value function inequalityDirect proof of inequality between arithmetic and harmonic meaninequality with min on both sidesSolving a quadratic inequality geometricallyMaximum difference inequalityEvaluating absolute inequalitiesTriangle Inequality IssueHow to show that $min{alpha,max{beta,gamma}}=max { min{alpha,beta},min{alpha,gamma}}.$

0

$begingroup$

I have to show that for $a,b,c,d in mathbb{R}$ $$|avee b - cvee d| leq |a-c|vee |b-d|$$
I know this can be showed using cases, but I need help with a proof that doesn't involve cases. I found out this inequality can be written as:
$$|avee b - cvee d| leq (avee c - a wedge c) vee (b vee d - b wedge d)$$
Since $avee c= frac{(a+c) + |a-c|}{2}$ and $awedge c= frac{(a+c) - |a-c|}{2}$

I guess the proof uses this form of the inequality, but I don't know how to write such proof.

real-analysis inequality absolute-value

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edited Mar 15 at 17:39

MMM

asked Mar 15 at 17:26

MMMMMM

84

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add a comment | 

0

$begingroup$

I have to show that for $a,b,c,d in mathbb{R}$ $$|avee b - cvee d| leq |a-c|vee |b-d|$$
I know this can be showed using cases, but I need help with a proof that doesn't involve cases. I found out this inequality can be written as:
$$|avee b - cvee d| leq (avee c - a wedge c) vee (b vee d - b wedge d)$$
Since $avee c= frac{(a+c) + |a-c|}{2}$ and $awedge c= frac{(a+c) - |a-c|}{2}$

I guess the proof uses this form of the inequality, but I don't know how to write such proof.

real-analysis inequality absolute-value

share|cite|improve this question

edited Mar 15 at 17:39

MMM

asked Mar 15 at 17:26

MMMMMM

84

$endgroup$

add a comment | 

$begingroup$

I have to show that for $a,b,c,d in mathbb{R}$ $$|avee b - cvee d| leq |a-c|vee |b-d|$$
I know this can be showed using cases, but I need help with a proof that doesn't involve cases. I found out this inequality can be written as:
$$|avee b - cvee d| leq (avee c - a wedge c) vee (b vee d - b wedge d)$$
Since $avee c= frac{(a+c) + |a-c|}{2}$ and $awedge c= frac{(a+c) - |a-c|}{2}$

I guess the proof uses this form of the inequality, but I don't know how to write such proof.

real-analysis inequality absolute-value

share|cite|improve this question

edited Mar 15 at 17:39

MMM

asked Mar 15 at 17:26

MMMMMM

84

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:39

MMM

asked Mar 15 at 17:26

MMMMMM

84

share|cite|improve this question

edited Mar 15 at 17:39

MMM

asked Mar 15 at 17:26

MMMMMM

84

asked Mar 15 at 17:26

MMMMMM

84

asked Mar 15 at 17:26

MMMMMM

84

1 Answer
1

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0

$begingroup$

This is basically just repeated applications of the triangle inequality; namely, we have
$$ lvert max(a,b) - max(c,d) rvert = bigglvert frac{(a-c) + (b-d) + lvert a - b rvert - lvert c - d rvert}{2}biggrvert leq frac{(lvert a - c rvert + lvert b - d rvert) + lvert lvert a-b rvert - lvert c - d rvertrvert}{2}.$$
But,
$$lvert lvert a-b rvert - lvert c - d rvert rvert leq lvert lvert a-c rvert + lvert b-c rvert - (lvert c-b rvert + lvert b-d rvert) rvert = lvert lvert a-c rvert - lvert b - d rvert rvert.$$
That should give the desired result.

Edit: For the last step, I used
$$lvert a - b rvert = lvert a - c + c - b rvert leq lvert a-c rvert + lvert b - c rvert.$$
You use that same trick on both terms in the absolute value.

share|cite|improve this answer

edited Mar 15 at 18:26

answered Mar 15 at 17:58

Gary MoonGary Moon

84616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What form of the triangle inequality you used at that last step?
  $endgroup$
  – MMM
  Mar 15 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added an edit to (hopefully) address your question.
  $endgroup$
  – Gary Moon
  Mar 15 at 18:26
  

  
  
  
  

add a comment | 

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0

$begingroup$

This is basically just repeated applications of the triangle inequality; namely, we have
$$ lvert max(a,b) - max(c,d) rvert = bigglvert frac{(a-c) + (b-d) + lvert a - b rvert - lvert c - d rvert}{2}biggrvert leq frac{(lvert a - c rvert + lvert b - d rvert) + lvert lvert a-b rvert - lvert c - d rvertrvert}{2}.$$
But,
$$lvert lvert a-b rvert - lvert c - d rvert rvert leq lvert lvert a-c rvert + lvert b-c rvert - (lvert c-b rvert + lvert b-d rvert) rvert = lvert lvert a-c rvert - lvert b - d rvert rvert.$$
That should give the desired result.

Edit: For the last step, I used
$$lvert a - b rvert = lvert a - c + c - b rvert leq lvert a-c rvert + lvert b - c rvert.$$
You use that same trick on both terms in the absolute value.

share|cite|improve this answer

edited Mar 15 at 18:26

answered Mar 15 at 17:58

Gary MoonGary Moon

84616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What form of the triangle inequality you used at that last step?
  $endgroup$
  – MMM
  Mar 15 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added an edit to (hopefully) address your question.
  $endgroup$
  – Gary Moon
  Mar 15 at 18:26
  

  
  
  
  

add a comment | 

0

$begingroup$

This is basically just repeated applications of the triangle inequality; namely, we have
$$ lvert max(a,b) - max(c,d) rvert = bigglvert frac{(a-c) + (b-d) + lvert a - b rvert - lvert c - d rvert}{2}biggrvert leq frac{(lvert a - c rvert + lvert b - d rvert) + lvert lvert a-b rvert - lvert c - d rvertrvert}{2}.$$
But,
$$lvert lvert a-b rvert - lvert c - d rvert rvert leq lvert lvert a-c rvert + lvert b-c rvert - (lvert c-b rvert + lvert b-d rvert) rvert = lvert lvert a-c rvert - lvert b - d rvert rvert.$$
That should give the desired result.

Edit: For the last step, I used
$$lvert a - b rvert = lvert a - c + c - b rvert leq lvert a-c rvert + lvert b - c rvert.$$
You use that same trick on both terms in the absolute value.

share|cite|improve this answer

edited Mar 15 at 18:26

answered Mar 15 at 17:58

Gary MoonGary Moon

84616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What form of the triangle inequality you used at that last step?
  $endgroup$
  – MMM
  Mar 15 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added an edit to (hopefully) address your question.
  $endgroup$
  – Gary Moon
  Mar 15 at 18:26
  

  
  
  
  

add a comment | 

$begingroup$

This is basically just repeated applications of the triangle inequality; namely, we have
$$ lvert max(a,b) - max(c,d) rvert = bigglvert frac{(a-c) + (b-d) + lvert a - b rvert - lvert c - d rvert}{2}biggrvert leq frac{(lvert a - c rvert + lvert b - d rvert) + lvert lvert a-b rvert - lvert c - d rvertrvert}{2}.$$
But,
$$lvert lvert a-b rvert - lvert c - d rvert rvert leq lvert lvert a-c rvert + lvert b-c rvert - (lvert c-b rvert + lvert b-d rvert) rvert = lvert lvert a-c rvert - lvert b - d rvert rvert.$$
That should give the desired result.

Edit: For the last step, I used
$$lvert a - b rvert = lvert a - c + c - b rvert leq lvert a-c rvert + lvert b - c rvert.$$
You use that same trick on both terms in the absolute value.

share|cite|improve this answer

edited Mar 15 at 18:26

answered Mar 15 at 17:58

Gary MoonGary Moon

84616

$endgroup$

share|cite|improve this answer

edited Mar 15 at 18:26

answered Mar 15 at 17:58

Gary MoonGary Moon

84616

answered Mar 15 at 17:58

Gary MoonGary Moon

84616

answered Mar 15 at 17:58

Gary MoonGary Moon

84616