Any sciencefags knowing how to prove that shit ?

Any sciencefags knowing how to prove that shit ?
I've been sitting on this for the past hour

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math.stackexchange.com/questions/1928229/absolute-value-inequalities-involving-multiple-variables
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triangle inequality

let ℝ be equal to ａ， ｂ， ｔ
if ℝ < ｔ | A then suck my dick faggot

>>>sci

been trying to use it earlier but couldnt figure the right way

Wait

k

bump ?

I solved it on paper here.
Start from this:

| a+( ) | + | a+( ) | > | [ a+ ( )] + [a + ( )] |

after that you'll get to:

(Stuff on the right) > 2|a + b|
.. > 2( |a|+|b| )

now because 0 < t < 1 you'll prove that t / ( 2 + t ) < 1
And use it there to get the result

I'm not sure I get what you are saying.
Could you take a picture of your work or be lil more specific ?

Also when you have 2|a+b| > (shit)*(|a|+|b|)
That doesnt have to be true as when you have a= -b you have that 0 > something positive

What is this a formula for OP?

nah just some useless uni homework.
but I agree, it is shit

Missed by one!!
For every a,b from ℝ you have :
|a + b| < |a| + |b|
That's basically the key to the problem

>triangle inequality

This. Note:
|a+(1+t)b| + |a+(1-t)b| >= |2a+(1+t)b+(1-t)b| = 2|a+b|

and from the original:
( 2+t )( |a+(1+t)b| + |a+(1-t)b| ) >= 2t(|a|+|b|)
so
LHS >= (2+t)*2|a+b| = 4|a+b| + 2t(|a+b|)

Ok, now I'm stuck too.

ya 1 sec


okay im stuck

No idiot
stop at the first line, then:
2 * |a + b| < 2 * ( |a| + |b| )
and t / ( 2 + t) < (2+t) / ( 2+t)=1

Yeah. Basicly everytime I tried to use triangle inequality I got sth like shit*|a+b|+shit>shit*(|a|+|b|)
and that doesnt hold when a+b = 0 /

Looks like the thread is about to die. Thanks for trying to help anyways

That proves nothing: LHS >= 2|a+b|

I got the same. Im not sure if the question is complete or im doing something wrong. Pic related

In other words, you have 4 cases to simplify separately: a,b>=0; a,b=0 & b

like he said You can convince yourself it's true, by separating the cases of a and b both positive, both negative, etc...
But there must be a smart way of proving it, I just can't figure it out.

it looks like you have 4 cases on rhs for a & b around 0 and all the cases on the left where a > / = / < (1+t)b / (1-t)b
So there like 24 cases (some of them may cover or be impossible tho)

It does feel like there's more elegant solution tho... :/

genuinely interesting question tbh

I got the same. Im not sure if the question is complete or im doing something wrong. Pic relatedYeah if both a and b have the same sign then the thing is easily proved, but if a and b have different signs then the question falls apart doesnt it?

You're truly retarded

it doesn't necessarily simplify to that though at all. That's a sufficient condition, but it's not necessary.

Download wolfram alpha. It helps you solve shit while showing you the work to get there. I used to use it to help when i was confused on a step in calc.

The full question was:
Prove that following inequality is true for every real a & b, where t is any real number between 0 and1

the answer is 25

I believe its a question from some olympiad so there shouldnt be a mistake.
Yeah I've used online version quite a lot It even solved for me some other olympic shit question

Yeah it doesnt necessarily simplify to that, but you can make that a condition and the question falls apart doesnt it? Isnt if a+b=0 a counterexample?

Yeah we could show that left side is smaller than something else and that something else is still bigger than the right side. Its not a simplification so it doesnt have to be true everytime you do this and for simple combination of left side it doesnt work.

It would be a good idea to just simplify it whitout changing its value but I dont see a way in which it can be done

>if a and b have different signs then the question falls apart doesnt it?
You have to consider if a+(1+t)b>=0 & a+(1-t)b

no, because that's a sufficient condition but not a necessary condition

Remember that your choice for a and b applies both to the LHS and RHS, and also there is sign invariance for the pair (a,b). That means 2 cases to start, with the possibility of another 2 that for whether the sign of a+(1+/-t)b goes negative for b

So this is what STEMlords do for fun...

I think I'll stick to saving the world please.

>you're welcome

>I think I'll stick to saving the world please.
That's what they actually believe...

Sorry if l8

line 2 is wrong

When you're splitting of |a| at second line the whole things gets bigger, not smaller. Therefore its wrond :/
Thanks for trying to help tho

Which part?
|x + y| >= |x| + |y|

no sry

Are you asking or telling ?
The triangular inequality states that |x+y|

what's the source of the problem

Oh right, my bad

I thinkwe are not going in the right direction. When combing values with triangular we are basicly ignoring the t constant so we would get general solution and Im pretty sure they specified t's domainso it is solvable

The expressions are invariant under
a->-a, b->-b. So you only need to check at most 4 sign cases (dependent on relative sizes)... but there must be a better way

uni teacher gave us some olympic exercises I belive. Got the photo of it so everything in picture is right, but I dont have a source for where its exactly from

Is this for a real analysis course?

yeah there's something special about t

this seems like some sort of Jewish problem tbh. it'll be easy when we see the trick but the trick currently evades us.

One interesting thing is that we can bring factors like 1±t in and out of the absolute values, since we know that they're positive

possibly. We had like a big revision so there were all the topics so Im not so sure

Yeah but you will still have a/(1+-t) inside and pulling it out doesnt seem to help

agreed- not saying necessarily that that forms part of the answer but atm i'm not seeing any other way of making use of our t domain

Post this shit on the maths stack exchange then post us the link here. Someone insane will come along and solve it

on my way

Well, we can do alot without fucking this up.
If any of a,b or t is 0 then this becomes easy as fuck so lets asume that a,b,t != 0.
If sgn(a) == sgn(b) this is also trivial so lets skip to the opposite case.
If a and b have opposite signs and are nonzero we can assume that a is positive and b is equal to -a + e.
Now we only have to cases to solve for e > 0 and e < 0.
Someone tell me if that's wrong.

that's pretty much correct but you'll probably need to do multiple cases.. since if a + (1-t)b >0, we still don't know if a + (1+t)b >0

the problem is just that while we've identified above that the case based strategies exist, we're looking for something a bit better

Alright I posted it on mathexchange as propposed.
Here's the link math.stackexchange.com/questions/1928229/absolute-value-inequalities-involving-multiple-variables

Those absolute values are imposible to break you have to go through all the different cases systematically

This is wrong

Nice trips

that is actually a really good point. Also check em

pity the analysis is wrong

i think there's some theorem we're supposed to apply to do this; without knowing the result it'd be research-level mathematics

That's what was said earlier, except when writing the cases, you put |a+(1+t)b|>=0 instead of a+(1+t)b>=0

Well I've been sitting on this problem for so long that I no longer see other way than checking all the cases

i think the cases are a little more fiddly than described

also you've got |b|

hi skrzelik is my brother, post cp

Well it looks like thread is about to die. We didnt prove it but still thanks everyone for help