Scalar product to prove triangle inequality?

AI Thread Summary
To prove the triangle inequality |a+b| <= |a| + |b| using the scalar product, begin by expressing |a+b|^2 as (a+b)·(a+b). Expanding this gives |a|^2 + |b|^2 + 2(a·b). Utilize the inequality |a·b| <= |a||b| to bound the term 2(a·b). This leads to |a+b|^2 <= |a|^2 + |b|^2 + 2|a||b|, which simplifies to |a+b| <= |a| + |b| upon taking the square root. The triangle inequality is thus established.
8emnero8
Messages
1
Reaction score
0

Homework Statement


From the inequality

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

prove the triangle inequality:

|a+b| <= |a| + |b|

Homework Equations



a.b = |a|b| cos theta

The Attempt at a Solution



Making a triangle where side c = a+b. Don't know how to approach the question.

Thanks.
 
Physics news on Phys.org
I may not be understanding your question, but it seems like you should start with the Pythagorean Identities.
 
Also think about what values your trig function is between
 
You should start by knowing |a+b|^2=(a+b).(a+b). Now expand the right side.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top