a inequality to be proved

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hero0765

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hi:

if we have :
a ≤ c ≤ b , how to prove |c| ≤ max{ |a|, |b| }.
 

Hi hero0765,
This seems more like an axiom...
How ever,
if we separate the cases the idea becomes evident
1) a=c=b, max{|a|,|b|}=a=b. and c=b. hence |c|=|b|
2) a<c=b, max{|a|,|b|}=b. and c=b. hence |c|=|b|

3) a=c<b, max{|a|,|b|}=b. and c<b. hence |c|<|b|

4) a<c<b, max{|a|,|b|}=b. and c<b. hence |c|<|b|
Otherwise you can use induction
Hope this helps
 

hi harsha_jois
your method has some errors.
when a<c=b , but we can't prove max{|a|,|b|}=b.
for example , when a =-2, c=b=-1, but |a| =2,|b| =1, so the max{|a|,|b|} is |a|.

Added after 20 minutes:

I have a method :
we can seperate the c into two cases to consider.
when c ≥ 0, and c ≤ b , we can prove |c| ≤ |b|, but if we were to prove |b| ≤ max{|a|,|b|}, we could prove |c| ≤ max{|a|,|b|} ,but how can we do that?

when c <0 ,and as a≤c ,so we can have 0 ≤ -c ≤ -a , and |c| ≤ |a|, but if we were to prove |a| ≤ max{|a|,|b|}, we could prove |c| ≤ max{|a|,|b|} ,but how can we do that?
 

Hi again hero0765,
My apologies... for the error, rather incompleteness of my response.
and thank you for pointing that out.
That will need one more case as,
if |a|=max[|a|,|b|] then |c|<|a| for the first case.
I hope you can deduce the second similarly.
Hope this helps.
regards
- Harsha S
 

thanks harsha_jois.
I have a question below :

If I were to prove |c| ≤ |b| and |b| ≤ max{|a|,|b|},
I could prove |c|≤ max{|a|,|b|} . but I can't deduce |b| ≤ max{|a|,|b|} with mathematics method.

similarly, If I were to prove |c| ≤ |a| and |a| ≤ max{|a|,|b|},I could prove |c|≤ max{|a|,|b|} . but I can't deduce |a| ≤ max{|a|,|b|} with mathematics method.
 

Since a ≤ c ≤ b then only four cases are possible:

1. a,b,c ≥ 0
2. b,c ≥ 0, a ≤ 0 so if |a| > |b| then |c| ≤ |a| (that is max{|a|,|b|}) otherwise |c| ≤ |b|
3. b ≥ 0, a,c ≤ 0 as before
4. a,b,c ≤ 0

1. e 4. are immediate.
 

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