Prove Triangle Inequality: ||a|| - ||b|| ≤ ||a - b||

cscott
Messages
778
Reaction score
1
"Use the triangle inequality to prove that:
||\vec{a}|| - ||\vec{b}|| \le ||\vec{a} - \vec{b}||"

I can start from that expression and prove it true using the Cauchy-Shwarz inequality but I don't think that's what's asked. Any hints?
 
Physics news on Phys.org
Do they mean the following triangle inequality:

||\vec{a}|| + ||\vec{b}|| \le ||\vec{a} + \vec{b}||

because it can be proved from that by picking a and b correctly.
 
StatusX said:
Do they mean the following triangle inequality:

||\vec{a}|| + ||\vec{b}|| \le ||\vec{a} + \vec{b}||

because it can be proved from that by picking a and b correctly.


Yeah but it should be ||a|| + ||b|| >= ||a + b||
 
Start with
||\vec{u} + \vec{v}|| - ||\vec{v}|| \le ||\vec{u}||

and choose

\vec{u} + \vec{v} = \vec{a}

The rest should be pretty self evident after that.
 
cscott said:
Yeah but it should be ||a|| + ||b|| >= ||a + b||

Right, sorry. So you got the answer?
 
river_rat said:
Start with
||\vec{u} + \vec{v}|| - ||\vec{v}|| \le ||\vec{u}||

and choose

\vec{u} + \vec{v} = \vec{a}

The rest should be pretty self evident after that.

I got it but how do I justify saying that?
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top