This question already has an answer here:

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?


marked as duplicate by T. Bongers, Martin R, user147263, user228113, Leucippus Jan 9 '16 at 0:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is \langle? $\endgroup$ – Wojowu Jan 8 '16 at 21:37
  • $\begingroup$ If $x$ and $y$ have the same sign then you have equality. What happens if they have opposite signs? $\endgroup$ – Ian Jan 8 '16 at 21:44
  • $\begingroup$ We don't know the signs of x and y It can be both positive and negative $\endgroup$ – Manuela NIEVES Jan 8 '16 at 21:45
  • 1
    $\begingroup$ @ManalBouabdallaoui : I think Ian is suggesting that you examine separate cases depending on the signs of $x$ and $y$. $\endgroup$ – Joel Cohen Jan 8 '16 at 21:47
  • 1
    $\begingroup$ 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). $\endgroup$ – Brian Tung Jan 8 '16 at 21:47

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 $\lvert x\rvert^2=x^2$).

  • $\begingroup$ This answer contains more logic for my question I have to try it again to see if I well understood. $\endgroup$ – Manuela NIEVES Jan 8 '16 at 21:58

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

  • $\begingroup$ 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.) $\endgroup$ – Ian Jan 9 '16 at 1:08

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 .

  • $\begingroup$ Thank you I will try all the possible answers $\endgroup$ – Manuela NIEVES Jan 8 '16 at 21:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.