Triangle Inequality Proven: Prove la + bl ≤ lal + lbl

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SUMMARY

The discussion focuses on proving the triangle inequality, specifically the statement that |a + b| ≤ |a| + |b|. The proof utilizes the dot product and properties of norms, particularly the identity |a|^2 = a · a and the expansion of |a + b|^2 = (a + b) · (a + b). Participants emphasize the importance of understanding these mathematical identities to derive the triangle inequality effectively.

PREREQUISITES
  • Understanding of vector dot products
  • Familiarity with norms and absolute values
  • Knowledge of mathematical identities involving vectors
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study vector dot product properties in detail
  • Learn about norms and their applications in geometry
  • Explore mathematical identities related to vector addition
  • Practice proving inequalities in vector spaces
USEFUL FOR

Students in mathematics, particularly those studying linear algebra or vector calculus, as well as educators looking for methods to teach the triangle inequality effectively.

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Homework Statement


I don't know if I'm posting in the right area, but here is the question:
From the inequality: la dot bl= lal lbl lcos(theta)l is less than or equal to lal lbl

Deduce the triangle inequality:
la + bl is less than or equal to lal +lbl


Homework Equations





The Attempt at a Solution



I am not even sure where to begin :(
I'm wondering if I could use numbers to prove it, but that doesn't seem right.
 
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Sorry, misread. Think about (|a| + |b|)^2. You also have a property available to you that says |a|^2 = a dot a.
 
Last edited:
|a+b|^2 =(a+b)dot(a+b) is a very nice identity application.
 

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