a inequality to be proved

[hero0765]

hi:

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

 

[harsha_jois]

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

 

[hero0765]

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?

 

[harsha_jois]

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

 

[hero0765]

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.

 

[albbg]

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.