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Question 1207835: 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.
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
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.
Step by step
(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.
Solved.
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