how do I prove this? (linear) ||u+v|| < ||u|| + ||v||

>>8935899
Im guessing counterexample, c'mon bro that's textbook

>>8935901
can't. has to be generalized.

>>8935902
what fucks me up is that
<u+v,u+v> = <u,u> + 2<u,v> + <v,v> which is > than <u,u> + <v,v>
so how can their magnitudes be smaller
logically it makes sense to me but the math is fucking me up

>>8935899
Of course I can, this is indeed a simple application of the definition of a metric. You just need to start by pointing the pertinent properties which are...

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HNNGGGG

I could prove that but I instead think I will jerk off to that sweet sweet booty. Thanks OP.

Don't you mean <= ?

You can square both sides and then introduce a middle man to prove the inequality

>>8935905
Oh my god <u,v><=||u||*||v||. Just substitute and factor like you learned in grade school. You should square the original inner product as well. This is basic stuff boss.

>>8935899
use integrals?

>>8935899
Use the usual inner product definition as ||u||*||v||*cos(A) and work from there

>>8935899
got more of dat booty?

>>8935899
oh my god that is one sweet theorem

>>8935899
square everything, both sides are greater than 0

>>8935901
This is the generalized triangle inequality, it's definitely true.

That said, OP, like others in the thread said, just thinking about different representations of ||u +v||, and after some playing around it should be obvious. And I mean that, the proof of this shouldn't be too long so if you find you're writing a lot you're probably thinking about it incorrectly.

>>8935899
Sauce for the ass?

>>8935899
brap

Are you retarded?

what the fuck is a cauchy schwarz?

All you have to do is write False.
Both sets contain the same number of elements

>>8935899
10/10 ass
7/10 feet
Breddy gud.