Using Triangle Inequality

[nycmathguy]

Set 1.2
Questions 62 & 63

See attachment.

Work out 62 and 63. A few years back, I tried showing the prove for both questions and simply gave up after several endeavors. I think MarkFL explained how to show this prove but I was banned from the site containing Mark's notes.

 

[MathLover1]

62.

|a| - |b|≤|a - b|

a=(a-b) + b

then we have |a|=|(a-b)+b|

using the triangular ineaquality, this implies that

|a|≤|a-b|+|b|

thus |a|-|b|≤|a-b|

63.
|a+b+c| ≤ |a|+|b|+|c|

start with
a+b+c=(a+b)+c
then
|a+b+c|=|(a+b)|+|c|...........since |a| + |b|≤|a+ b|
|a+b+c|≤|(a+b)|+|c|...........since |a+ b| =|a| + |b|
|a+b+c|≤|a| + |b|+|c|

 

[nycmathguy]

62.

|a| - |b|≤|a - b|

a=(a-b) + b

then we have |a|=|(a-b)+b|

using the triangular ineaquality, this implies that

|a|≤|a-b|+|b|

thus |a|-|b|≤|a-b|

63.
|a+b+c| ≤ |a|+|b|+|c|

start with
a+b+c=(a+b)+c
then
|a+b+c|=|(a+b)|+|c|...........since |a| + |b|≤|a+ b|
|a+b+c|≤|(a+b)|+|c|...........since |a+ b| =|a| + |b|
|a+b+c|≤|a| + |b|+|c|

Thanks. I couldn't do this alone. I tried. I failed.