Question 1207835
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Show that for all real numbers a and b, we have
|a| - |b| <= |a - b|

Hint:

Beginning with the identity a = (a - b) + b, take the absolute value of each side 
and then use the triangle inequality.
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I will strictly follow the given instructions.


<pre>
             <U>Step by step</U>



(1)  Start with the identity a = (a - b) + b.


(2)  Take absolute values

         |a| = |(a-b) + b|.


(3)  Apply the triangle inequality

          |a| = |(a-b) + b| <= |a-b| + |b|.


     So, you have 

          |a| <= |a-b| + |b|.


(4)  In the last equality, subtract |b| from both sides

     (it is the same as transfer term |b| from right to left).  You will get

          |a| - |b| <= |a-b|.


     It is precisely what they want you prove.


At this point, the proof is complete.
</pre>

Solved.