Absolute value : Freshman exercise

Question

Prove that each of the group R :

$$|x+y| \leq\ |x| + |y|$$

Of course I'm not a lazy person I tried solving before since the teacher gave this exercise to us in the exam. I wanted to know the right answer because we still haven't corrected the test.

Thanks and Sorry for any faults. This is my first question in this website. How would I know if it's duplicate?

If $x$ and $y$ have the same sign then you have equality. What happens if they have opposite signs? — Ian, Jan 08 '16 at 21:44
We don't know the signs of x and y It can be both positive and negative — Manuela NIEVES, Jan 08 '16 at 21:45
@ManalBouabdallaoui : I think Ian is suggesting that you examine separate cases depending on the signs of $x$ and $y$. — Joel Cohen, Jan 08 '16 at 21:47
I think Ian is suggesting that you consider that case explicitly and see what happens if $x$ is positive and $y$ is negative (or vice versa). — Brian Tung, Jan 08 '16 at 21:47

score 4 · Answer 1 · answered Jan 08 '16 at 21:51

4

If $x+y\geq0$ then $$|x+y|=x+y\leq|x|+|y|.$$ If $x+y<0$ then $$|x+y|=-x-y\leq|x|+|y|.$$

answered Jan 08 '16 at 21:51

Spenser

20,135

A remark on the second case: you have $-x-y=(-x)+(-y) \leq |-x|+|-y|=|x|+|y|$. (It took me a moment to catch this.) – Ian Jan 09 '16 at 01:08

Bernard · Accepted Answer · 2016-01-13T20:23:02.740

1

As both sides are non-negative numbers, it is the same to prove \begin{align*} \lvert x+y\rvert^2\le(\lvert x\rvert+\lvert y\rvert)^2&\iff (x+y)^2\le x^2+y^2+2\lvert x\rvert\,\lvert y\rvert\\&\iff 2xy\le 2\lvert x\rvert\,\lvert y\rvert\iff xy\le\lvert xy\rvert \end{align*}

(We used $\lvert x\rvert^2=x^2$).

edited Jan 13 '16 at 20:23

answered Jan 08 '16 at 21:54

Bernard

179,256

This answer contains more logic for my question I have to try it again to see if I well understood. – Manuela NIEVES Jan 08 '16 at 21:58

score 0 · Answer 3 · 2016-02-17T19:32:40.670

0

Another method is to square it :

$$|x|+|y| \geq |x+y|$$ if and only if :

$$(|x|+|y|)^2 \geq (x+y)^2$$ $$x^2+2|xy|+y^2 \geq x^2+2xy+y^2$$

$$2|xy| \geq 2xy$$ but this is obvious .

edited Feb 17 '16 at 19:32

answered Jan 08 '16 at 21:55

Thank you I will try all the possible answers – Manuela NIEVES Jan 08 '16 at 21:59

Absolute value : Freshman exercise

3 Answers3