Question

How can I prove that: abs(x+y) less than or equal to abs(x) + abs (y) please?

Answer 1

think of some negative values ...

Answer 2

|x+y| <= |x|+|y|

let x=3, and let y = -5

|3-5| <= |3|+|-5|

|-2| <= 3+5

Answer 3

I have tried saying that if x>y>0, then abs(x+y) = abs(x)+abs(y)
If x>0>y, then abs(x+y)<abs(x)+abs(y)
and then also for 0>x>y, abs(x+y) = abs(x)+abs(y)

Does not feel like a proof though

Answer 4

it wold prolly be fine with cases.

let x,y be >= 0; then |x+y| = |x|+|y| = x+y

let x,y be < 0; then |-x-y| = |-x|+|-y| = x+y

the distance between same signed values is the same either way ... its when we get into different signs that we run into troubles

let x < 0 and let y > 0; then |-x+y| < |-x|+|y| = x+y

but trying to iterate it in another manner eludes me at the time

Answer 5

Hey I gotta dash... what do you think though. Is that enough? Thanks for your help though Amistre. I'll buy you a beer tonight, but I will have to drink it for you too!

Answer 6

Or just two cases... either same sign or different signs :D

Answer 7

oh... you already said that... nvm :D

Answer 8

|-5+3| = |-2| = 2, which is less than 8
|5-3| = |2| = 2, which is less than 8

Answer 9

Really gotta dash but many thanks for the ideas.

Answer 10

like signs increase distances; an additive effect

different signs decreas distances; an "subtractive" effect

Answer 11

maybe some vector notations :)

Answer 12

note that \(xy\le |xy|\)\[|x+y|^2=(x+y)^2=x^2+2xy+y^2=|x|^2+2\color\red{xy}+|y|^2 \]\[|x|^2+2\color\red{xy}+|y|^2 \le |x|^2+2\color\red{|x||y|}+|y|^2=(|x|+|y|)^2\]it gives\[|x+y|^2 \le (|x|+|y|)^2\]take square root\[|x+y|\le |x|+|y|\]

Answer 13

are we allowed to just note that xy <= |xy| ? or would we have to prove that as well?

Answer 14

Thanks Mukushla, that's a nice way to look at it... Have you got a source for that proof?

Answer 15

I agree with Amistre too - proof needed for previous result.
Thanks guys!

Answer 16

No source. Your welcome.