SOLUTION: Show that
|a + b + c| <= |a| + |b| + |c|
for all real numbers a, b, and c.
Hint:
The left-hand side can be written
|a + (b + c)| .
Now use the triangle inequali
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-> SOLUTION: Show that
|a + b + c| <= |a| + |b| + |c|
for all real numbers a, b, and c.
Hint:
The left-hand side can be written
|a + (b + c)| .
Now use the triangle inequali
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Question 1207836: Show that
|a + b + c| <= |a| + |b| + |c|
for all real numbers a, b, and c.
Hint:
The left-hand side can be written
|a + (b + c)| .
Now use the triangle inequality. Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Show that
|a + b + c| <= |a| + |b| + |c|
for all real numbers a, b, and c.
Hint:
The left-hand side can be written
|a + (b + c)| .
Now use the triangle inequality.
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Step by step
(1) Make grouping |a + b + c| = |a + (b + c)|.
(2) Apply the triangle inequality
|a + (b + c)| <= |a| + |b + c|.
(3) Apply the triangle inequality again to the second absolute value addend
|a| + |b + c| <= |a| + |b| + |c|.
(4) Combine the steps into the final conclusion
|a + v + c| <= |a| + |b| + |c|.
At this point, the proof is fully completed.
