Elementary Analysis, Triangle Inequality Help

In summary, the problem requires proving the inequality ||a|-|b||\leq |a-b| for all real numbers a and b. This can be done by considering four possible cases and proving the inequality for each one. The triangle inequality is not necessary to solve this problem.
  • #1
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Homework Statement


Prove that ||a|-|b||[tex]\leq[/tex] |a-b| for all a,b in the reals


Homework Equations


I know we have to use the triangle inequality, which states:
|a+b|[tex]\leq[/tex] |a|+|b|.

Also, we proved in another problem that |b|[tex]\leq[/tex] a iff -a[tex]\leq[/tex]b[tex]\leq[/tex]a


The Attempt at a Solution


Using the other problem we solved, we can say -|a-b| [tex]\leq[/tex] ||a|-|b||[tex]\leq[/tex] |a-b|

then I said...
|b|=|(b-a)+a|
then I use the Triangle Inequality"
|(b-a)+a|[tex]\leq[/tex] |b-a| + |a|

|b-a| + |a| [tex]\leftrightarrow[/tex] |a-b| + |a|

and someone said this proved the first side of the inequality, but I have no idea why.

All help is appreciated ! Thanks !
 
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  • #2
I think it will be hard to use the triangle inequality, because whenever you use a minus sign, the expression will always have to be in parentheses. In other words, I don't think the triangle inequality will help you go from, say, |a-b| to something where there's a minus sign outside parentheses. I was able to solve the problem by proving the inequality for each possible case, namely
Case 1: a and b are positive
Case 2: a and b are negative, a>b
Case 3: a and b are negative, a<b
Case 4: a is positive and b is negative. (You don't need to have case 5 where b is positive and a is negative, since both cases turn out to be the same.)
The triangle inequality was not necessary when I did the problem.
 

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