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 tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top