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Triangle inequality proof 
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#1
Mar3108, 06:36 PM

P: 239

1. The problem statement, all variables and given/known data
Prove that x+y</x+y 2. Relevant equations 3. The attempt at a solution Assume that x and y are real numbers. 5+2</5+7 7</7 52</5+2 7</7 I know that it is true by testing different numbers, but I'm not sure how to prove it. Could someone please show me how or give me a hint? Thank you very much 


#2
Mar3108, 06:56 PM

P: 1,633

since x<x<x, and y<y<y we add these to get
xy<x+y<x+y=>(x+y)<x+y<x+y, now from the abs value properties it immediately yeilds to: x+y<x+y read ''<" as greater or equal to. 


#3
Mar3108, 07:01 PM

P: 1,633

Or we could prove it this way also: to prove it it is also sufficient to prove that
[tex]x+y^{2}\leq (x+y)^{2} ???[/tex] so we have [tex]x+y^{2}=(x+y)^{2}=x^{2}+2xy+y^{2}=x^{2}+2xy+y^{2}<x^{2}+2xy +y^{2}=(x+y)^{2}[/tex] hence we are done. 


#4
Mar3108, 07:44 PM

P: 239

Triangle inequality proof
Thank you very much
Regards 


#5
Mar1810, 07:30 AM

P: 1




#6
Mar1810, 08:03 AM

HW Helper
P: 3,348

I don't think its valid in general to add two inequalities like that. But if you accept up to there, then to see the final step is just seeing that if A < a < A, then we can say a < A.



#7
Mar2310, 06:33 AM

HW Helper
P: 3,348

Correction and sincere apologies to sutupidmath, you CAN add inequalities as long as they are pointing the same same direction which in this case in true. And that goes to make an extremely simple proof!



#8
Mar2410, 08:36 AM

P: 15

I think vrdfx wasn't asking how x<x<x + y<y<y = xy<x+y<x+y => (x+y)<x+y<x+y , but how (x+y)<x+y<x+y yields x+y<x+y.
Does (x+y) = x+y or something? I don't see how it could... 


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